Squares & Cubes
It All Begins Here
Calculation League has two complex categories that relate exclusively to exponents:
Cubes/Squares category: exponents are limited to ^2 and ^3.
Exponents category: exponents range from ^2 through ^9.
In Calculation League Season 3, the playoff-stage “exponent” category was the most difficult category in the competition. Further modifications will be made to decrease the difficulty while still preserving the purpose of the category.
A later post will be made regarding the exponent category after further determinations regarding the revision of the category.
Example 1: 35^3 (easy level)
Our first example is simply cubing a two-digit number.
One often helpful approach for squaring a number is to recognize that:
(35 − 5) × (35 + 5) + 5^2 = 35^2
This is an instance of the identity:
(x − y)(x + y) + y^2 = x^2
In the case of a number that ends in 5, the application of this becomes extremely easy:
The square of all numbers ending in 5 ends in 25.
The preceding digits are a × (a + 1).
So:
35^2 = 3 × 4 = 12, append 25 → 1225
Now we multiply by 35 to cube:
35^3 = 35 × 35^2 = 1225 × 35
1225 × 35 = 1225 × (30 + 5) = 36750 + 6125 = 42875
Answer: 42875
Example 2: (8.7)^2 (easy level)
Now we have the square of a two-digit number, which is simply a 2×2 multiplication (after scaling).
Compute 87^2 first:
87^2 = (90 × 84) + 3^2 = 7560 + 9 = 7569
Since we are actually squaring 8.7, not 87, we divide by 10^2:
(8.7)^2 = 7569 / 100 = 75.69
So the answer is 75.69 (or 76 if rounded to the nearest integer).
Example 3: 354348^2 (advanced level)
Now we have the square of a six-digit number. Using algebraic shortcuts on the full number is going to be difficult. Most likely, the most effective approach is to treat this as a 6×6 multiplication, using your preferred multiplication strategy.
That said, the number has a nice structure: 354 and 348. We can take advantage of this by using three-digit blocks.
Step 1: Compute the key 3-digit squares/products
We start with a nearby reference:
350^2 = 122500 (as shown earlier)
We can build from 351^2:
351^2 = 350^2 + (2×350×1) + 1^2
351^2 = 122500 + 700 + 1 = 123201
Now adjust to 354 and 348.
For 354^2: compare 354 to 351 (difference +3)
(351 + 3)^2 = 351^2 + 2×351×3 + 3^2
2×351×3 = 2106, plus 9 → 2115
354^2 = 123201 + 2115 = 125316
For 348^2: compare 348 to 351 (difference −3)
(351 − 3)^2 = 351^2 − 2×351×3 + 3^2
351^2 − 2106 + 9 = 123201 − 2097 = 121104
For 348×354: note the symmetry around 351:
(351 − 3)(351 + 3) = 351^2 − 3^2 = 123201 − 9 = 123192
Step 2: Cross-multiply using three-digit blocks
Treat 354348 as 354|348.
Then:
354348^2 = (354×1000 + 348)^2
= (354^2)×10^6 + 2×(354×348)×10^3 + (348^2)
Plug in the computed pieces:
354^2 × 10^6 = 125316000000
2×(354×348)×10^3 = 2×123192×1000 = 246384000
348^2 = 121104
Now add:
125316000000
+000246384000
+000000121104
= 125562505104
Answer: 125562505104
Example 4: (93.5)^3 (advanced level)
Now we have the cube of a 3-digit number (after scaling).
A helpful first step is squaring 935 using the “ends in 5” shortcut:
935^2 → 93 × 94 with 25 appended
93 × 94 = 8742
so 935^2 = 874225
But we are working with 93.5, not 935, so:
(93.5)^2 = 874225 / 10^2 = 8742.25
Now we multiply to cube:
(93.5)^3 = 8742.25 × 93.5
A workable way to think about ×93.5 is:
×93 + ×0.5
Step 1: 8742 × 93
First do 87 × 93:
87 × 93 is 9 less than 90^2 (because 87×93 = (90−3)(90+3) = 8100 − 9)
so 87×93 = 8091
Then scale:
8700 × 93 = 809100
42 × 93 = 3906
809100 + 3906 = 813006
Step 2: Add the “half” and decimal adjustments
We need to add (8742/2), (93/4), and (1/8). This corresponds to the extra contributions coming from the .5 in 93.5 and the .25 in 8742.25.
Compute:
8742 / 2 = 4371
93 / 4 = 23.25
1 / 8 = 0.125
Sum: 4371 + 23.25 + 0.125 = 4394.375
Now total:
813006 + 4394.375 = 817400.375
In Calculation League, if we only need the nearest integer:
817400 is sufficient.
Example 5: (4559936)^2 (expert level)
This is the square of a seven-digit number. The default approach is to treat it as a 7×7 multiplication using cross multiplication or standard multiplication (if you are extremely accomplished).
But we can also use the Binomial Theorem for (x + y)^2 by picking a convenient nearby base.
Let:
x = 4,560,000
y = −64
Then:
(x + y)^2 = x^2 + 2xy + y^2
Step 1: x^2
x = 456 × 10^4, so:
x^2 = (456^2) × 10^8
Compute 456^2:
456^2 = (500×412) + 44^2
500×412 = 206000
44^2 = 1936
456^2 = 207936
So:
x^2 = 207936 × 10^8 = 20793600000000
Step 2: 2xy
2xy = 2×(4,560,000)×(−64) = −(2×64×4,560,000)
Note: 64 = 2^6, so 2×64 = 2^7 = 128.
So we want 456 × 128, then scale by 10^4:
Doubling 456 seven times:
456 → 912 → 1824 → 3648 → 7296 → 14592 → 29184 → 58368
So:
456×128 = 58368
therefore 4,560,000×128 = 58368×10^4 = 583680000
Because y is negative, this middle term is −583,680,000.
Step 3: y^2
y^2 = (−64)^2 = 64^2 = 4096
Combine terms
So:
(4,559,936)^2 = 20793600000000 − 583680000 + 4096
= 20793600000000 − 583675904
= 20793016324096
Answer: 20793016324096
To execute this quickly, you need to be comfortable chunking the subtraction—e.g., writing the leading digits from x^2, then subtracting the 9-digit middle adjustment cleanly.
Example 6: 17.7^2 (expert level)
On this last example, the question generation was unusually friendly. There are many ways to square a three-digit number (or do a 3×3 multiplication). Here are several approaches that all reach the same square:
Method A: Symmetric product + add-back
(177 − 77)(177 + 77) + 77^2
= 100×254 + 5929
= 25400 + 5929 = 31329
Method B: Another symmetric choice
(177 + 23)(177 − 23) + 23^2
= 200×154 + 529
= 30800 + 529 = 31329
Method C: Binomial Theorem
(170 + 7)^2 = 170^2 + 2×170×7 + 7^2
= 28900 + 2380 + 49 = 31329
Method D: Alternate binomial direction
(180 − 3)^2 = 180^2 − 2×180×3 + 3^2
= 32400 − 1080 + 9 = 31329
Method E: Factorization shortcut
177^2 = (59^2)(3^2) = 3481×9 = 31329
Now shift the decimal (since 17.7 = 177/10):
17.7^2 = 31329 / 100 = 313.29
Answer: 313.29
Deep Roots
It All Begins Here
4th, 5th, 6th, 7th, and 8th Roots
Calculation League has five complex categories covering deep roots. All of these questions follow a format of:
x^(1/y) × 10^z,
where y is between 4 and 8, and z is between 0 and 3. The effect of multiplying by 10^z is to increase the number of significant digits you need to compute in the final answer.
Generally:
The prime deep roots (5th and 7th) are most often done through direct estimation and refinement techniques.
The composite deep roots can often be done through repeated simpler roots:
4th roots: square root twice
6th roots: square root + cube root (in either order)
8th roots: square root three times
Chunking Rule (Very Important)
In either situation, it helps to divide x into chunks of y digits, starting from the right, and then work from left to right.
(If x is small enough, this may be obvious and can be skipped.)
The number of chunks corresponds to the number of digits in the integer part of the answer.
This is a practical “place-value” way to predict how large the root will be before doing any detailed calculation.
Finally: logarithms can also be used to calculate deep roots. That technique is covered in the decimal/irregular exponents and roots section.
4th Roots
4th Roots — Example #1 (easy level)
550859^(1/4)
Because 10^4 < 550859 < 100^4, the answer will be two digits.
If we chunk the number into 4-digit blocks from the right, we get:
55 | 0859
Now we estimate the first chunk:
2^4 = 16 and 3^4 = 81, so 2^4 < 55 < 3^4
⇒ the answer is between 20 and 30.
We can narrow it more:
(5/2)^4 = 2.5^4 = 39.0625, and 39 < 55 < 81
⇒ the answer is between 25 and 30.
For Calculation League purposes (with up to five attempts), that estimate alone is often sufficient to land a correct answer quickly.
Direct Method (Square Root Twice)
Use the identity x^(y·z) = (x^y)^z. Here, 1/4 = (1/2)(1/2), so we take the square root twice:
550859^(1/2) ≈ 742
742^(1/2) ≈ 27.24
Even without computing decimal digits in the first square root, we get a correct answer to roughly four significant digits. In fact, we could have rounded the first square root to two significant digits (e.g., 740) and still arrived at the correct result.
4th Roots — Example #2 (advanced level)
25628856193368^(1/4)
This is a 14-digit number, so we chunk into 4-digit blocks:
25 | 6288 | 5619 | 3368
From the first chunk (25), we can already tell:
We will need four significant digits in the final answer.
The first digit of the answer is 2, since 2^4 = 16 < 25 < 3^4 = 81.
How many leading digits do we need?
If we only use 25:
25^(1/2) = 5, and 5^(1/2) = 2.236
This is not accurate enough, because we ignored too much information.
If we use 25.6:
25.6^(1/2) ≈ 5.06 (more digits help here)
5.06^(1/2) ≈ 2.249
This is close to the correct answer (about 2.250), but still slightly short depending on rounding.
A better refinement is to use 25.62:
25.62^(1/2) ≈ 5.062
5.062^(1/2) ≈ 2.250
Practical takeaway (what you actually need)
To get a correct answer for this example, it is enough to:
Use the first 4 digits of the number (25.62… as the leading chunk),
Compute the first square root to about 4 digits,
Compute the second square root to about 4 digits.
4th Roots — Example #3 (expert level)
44449882998507^(1/4) × 100
Here we again have a 14-digit number, but multiplying by 100 increases the number of required significant digits. We now need six significant digits. The correct answer is:
258207
Chunking into 4-digit blocks:
44 | 4498 | 8299 | 8507
If we only use the earlier “minimum precision” approach, we get:
44.44^(1/2) ≈ 6.666
6.666^(1/2) ≈ 2.58186
That is close, but not enough—because the next digits matter (especially the “98” later in the number).
A better choice is to use 44.450:
44.450^(1/2) ≈ 6.6671
6.6671^(1/2) ≈ 2.58207
Now we match the correct digits.
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 5 digits of the number,
Compute the first square root to about 5 digits,
Compute the second square root to about 6 digits.
8th Roots
For 8th roots, we use the fact that:
1/8 = (1/2)(1/2)(1/2)
So we take the square root three times.
8th Roots — Example #1 (easy level)
2493261732392^(1/8)
Chunk into 8-digit blocks:
24932 | 61732392
For any number between 9 digits and 16 digits, the integer part of the 8th root will be two digits.
Now apply three square roots using the leading chunk:
24932^(1/2) ≈ 158
158^(1/2) ≈ 13
13^(1/2) ≈ 3.6
This suggests about 3.6, and since the true answer is a two-digit number, this corresponds to about 36.
If we compute the middle step more accurately:
158^(1/2) ≈ 12.6
12.6^(1/2) ≈ 3.55
So the answer is about 35, and the correct answer is 35.
You could also round earlier:
24932 ≈ 25000
25000^(1/2) ≈ 158.1 ≈ 160
160^(1/2) ≈ 12.65 ≈ 12.6
12.6^(1/2) ≈ 3.55 → 35
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 2 digits of the number,
Compute the first square root to 2 digits,
Compute the second square root to 3 digits,
Compute the third square root to 2 digits.
Optional “power bound” check
We can also note:
3^8 < 24932 < 4^8
3^8 = ((3^2)^2)^2 = 81^2 = 6561
4^8 = ((4^2)^2)^2 = 256^2 = 65536
So the answer must be between 30 and 40, and 35 is very plausible.
8th Roots — Example #2 (advanced level)
7961721564417^(1/8) × 10
Chunk into 8-digit blocks:
79617 | 21564417
We now need three significant digits, since we multiply by 10.
A quick run using slightly rounded values:
80000^(1/2) ≈ 280
280^(1/2) ≈ 16.7
16.7^(1/2) ≈ 4.09
The correct answer is 410 (so 4.10 × 10^2 in the scaled interpretation).
All we need is one more digit in the first square root:
80000^(1/2) ≈ 283
283^(1/2) ≈ 16.8
16.8^(1/2) ≈ 4.10
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 2 digits of the number,
Compute the first square root to 3 digits,
Compute the second square root to 3 digits,
Compute the third square root to 3 digits.
8th Roots — Example #3 (expert level)
38294827023^(1/8) × 100
Chunk:
382 | 94827023
We now need four significant digits.
Using the same “minimum process”:
380^(1/2) ≈ 19.5
19.5^(1/2) ≈ 4.42
4.42^(1/2) ≈ 2.102
The correct answer is 2.103 (and then ×100 gives 210.3 in the scaled interpretation), so this is already extremely close.
To stabilize the fourth digit, expand the first step slightly:
382^(1/2) ≈ 19.54
19.545^(1/2) ≈ 4.421
4.421^(1/2) ≈ 2.103
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 3–4 digits of the number,
Compute the first square root to 4–5 digits,
Compute the second square root to 4 digits,
Compute the third square root to 4 digits.
6th Roots
For 6th roots, we use:
1/6 = (1/2)(1/3)
So we combine a square root and a cube root, in either order.
6th Roots — Example #1 (easy level)
782069951488^(1/6)
Chunk into 6-digit blocks:
782069 | 951488
We only need two significant digits here.
A quick anchor fact:
9^6 = 531441
(You can estimate this quickly by noting 9^3 = 729 and squaring ~729^2 ≈ 530,000.)
Since 782,069 is well above 531,441, the answer is clearly above 90, and likely in the mid-90s.
Square root then cube root (rough is allowed on easy level)
800000^(1/2) ≈ 890
890^(1/3) ≈ 9.6
Cube root then square root (also works)
800000^(1/3) ≈ 93
93^(1/2) ≈ 9.6
A key idea here: when only two significant digits are required, many very large ranges of integers will share the same rounded sixth root. For example, all twelve-digit numbers between 758,612,910,511 and 807,539,696,082 will have a sixth root that rounds to 96 (nearly 49 billion integers).
That is very different from smaller-root situations: numbers whose sixth root rounds to 22 lie between 129,746,338 and 168,425,239, which is under 39 million integers.
This matters because it shows why the leading chunk and the required accuracy heavily affect how much computation is necessary.
6th Roots — Example #2 (advanced level)
295765318037310^(1/6) × 10
Now we need the sixth root of a 15-digit number to the nearest 0.1.
Chunk into 6-digit blocks:
295 | 765318 | 037310
Trying the very rough approach from easy level:
300^(1/2) ≈ 17
17^(1/3) ≈ 2.571
Not accurate enough.
Other order:
300^(1/3) ≈ 6.7
6.7^(1/2) ≈ 2.588
Closer, but still not enough.
Use the actual first chunk more accurately:
295^(1/2) ≈ 17.2
17.2^(1/3) ≈ 2.581 (correct)
Or:
295^(1/3) ≈ 6.66
6.66^(1/2) ≈ 2.581 (also correct)
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 3 digits of the number,
Compute the square root or cube root to 3 digits,
Compute the remaining cube root or square root to 4 digits.
A common pattern in these examples is that intermediate results often need fewer significant digits than the final answer, but precision requirements increase as the calculation proceeds. That is one reason it can be advantageous to do the cube root first when possible: it allows more rounding earlier, and the more demanding step uses “seen” numbers rather than unseen ones.
6th Roots — Example #3 (expert level)
26652307644715^(1/6) × 100 (Correct answer: 172.83)
Now we need the sixth root of a 14-digit number to the nearest 0.01.
Chunk into 6-digit blocks:
26 | 652307 | 644715
Try the “square root then cube root” route with a bit more accuracy:
26.6^(1/2) ≈ 5.16
5.16^(1/3) ≈ 1.7280 (close)
Other order:
26.6^(1/3) ≈ 2.99
2.99^(1/2) ≈ 1.7292 (not quite close enough)
Now refine the first chunk slightly:
26.65^(1/2) ≈ 5.162
5.162^(1/3) ≈ 1.7282
Or:
26.65^(1/3) ≈ 2.987
2.987^(1/2) ≈ 1.7283 (this one matches)
Practical takeaway
To get a correct answer for this example, it is enough to:
Use the first 4 digits of the number,
Compute the square root or cube root to 4 digits,
Compute the remaining cube root or square root to 5 digits.
At this accuracy level, storing a cube root to four significant digits and then computing a square root to five significant digits is difficult—this is exactly why it is expert level. Even for sixth roots, once accuracy becomes demanding enough, direct calculation techniques begin to matter more than “rough” estimation.
5th Roots
5th roots are prime deep roots, and usually require either:
logarithms, or
refinement methods (linearized correction / Newton / Halley)
5th Roots — Example #1 (easy level)
88704604192282^(1/5)
This example requires three significant digits.
Chunk into 5-digit blocks:
8870 | 46041 | 92282
Because 5th roots are hard to do directly, some competitors prefer logarithms (see the decimal/irregular exponents and roots section). For many people, that will be the most effective method unless the number is very “friendly.”
Here, though, we outline refinement.
First, we need small fifth-power milestones:
6^5 = 7776
7^5 = 16807
Since 8870 is between 7776 and 16807, the first digit is between 6 and 7.
Linear Interpolation (very rough)
Difference:
7^5 − 6^5 = 16807 − 7776 = 9031
Position:
(8870 − 7776) / 9031 = 1094 / 9031 ≈ 0.12
So a rough estimate is:
6.12… → interpreted as 612 for the scaled number (since chunking implies the answer is three digits)
Because the slope of x^5 increases, this interpolation tends to be an underestimate. For Calculation League purposes (five attempts), it would typically be enough to get you into the correct neighborhood.
Linearized Power Correction
This improves accuracy without full Newton computation.
Compute the relative error:
(8870 − 7776) / 7776 = 1094 / 7776 ≈ 0.14
Call this value y.
Adjust the estimate using the 5th-root correction:
Multiply by (1 + y/5)
So:
y/5 ≈ 0.028
1 + y/5 ≈ 1.028
600 × 1.028 = 616.8
This is better, but still may not reliably produce the full three significant digits in every case.
Newton Refinement (one iteration, still using estimation)
We now refine the scale factor.
Compute (1.028)^2 using binomial reasoning:
(1 + 0.028)^2 ≈ 1 + 2(0.028) + (0.028)^2
≈ 1 + 0.056 + 0.001 ≈ 1.057
Square again (to approximate the 4th power effect):
(1.057)^2 ≈ 1 + 2(0.057) + (0.057)^2
≈ 1 + 0.114 + 0.003 ≈ 1.117
Form the adjustment:
1.14 / 1.117 ≈ 1.02
Move one-fifth of the way from 1.028 to 1.020:
1.0264
Multiply by 600:
600 × 1.0264 = 615.864
Now we have three significant digits.
5th Roots — Example #2 (advanced level)
78010745^(1/5) × 100
Advanced level increases the demand from three to four significant digits.
Chunk:
780 | 10745
We know:
3^5 = 243 < 780 < 4^5 = 1024
Linear interpolation suggests the base is around 3.7.
Estimating 3.7^5 (rough but workable)
We can do:
3.7^2 ≈ 13.69 → use 13.7
13.7^2 ≈ 187.7 → use 188
188 × 3.7 ≈ 695 (note: rounding choices affect whether this is slightly high or low)
Halley Increment
Halley’s increment for a 5th root can be written as:
A × (N − A^5) / (3A^5 + 2N)
Here:
A = 3.7
N = 780
A^5 ≈ 695
Compute:
numerator: 780 − 695 = 85
denominator: 3(695) + 2(780) = 2085 + 1560 = 3645
increment: 3.7 × 85 / 3645 ≈ 0.086
So the refined estimate is:
3.7 + 0.086 = 3.786
If we improve A^5 slightly (e.g., 693 or 694), we get:
694 → increment ≈ 0.087 → 3.787
693 → increment ≈ 0.088 → 3.788
And 3.788 is the correct four-digit estimate.
5th Roots — Example #3 (expert level)
120682^(1/5) × 10^3
Now we need five significant digits, but the numbers are often more cooperative.
Chunk:
1 | 20682
We know:
1^5 < 1.20682 < 2^5
So the first digit is 1.
We can also see the second digit is 0, since:
1.1^5 already exceeds 1.20682 (and 1.1^2 = 1.21 gives a quick warning)
So let’s start with 1.03:
1.03^2 = 1.0609 → use 1.061
1.061^2 ≈ 1.125721 → use 1.1257
1.1257 × 1.03 ≈ 1.1595 (approximate 1.03^5)
Now apply Halley increment (scaled form used in the draft):
10.3 × (120682 − 115950) / (347850 + 241364)
Compute the pieces:
120682 − 115950 = 4732
347850 + 241364 = 589214
10.3 × 4732 ≈ 48740
48740 / 589214 ≈ 0.0827
So the estimate is:
10.3 + 0.0827 = 10.3827
That is correct to five significant digits.
A note on the division step: in a quotient like 48740 / 589214, you only need as many significant digits as you are comfortable with. Depending on context, you can either refine the numerator/denominator slightly or choose whether to guess upward or downward based on whether your earlier approximations were underestimates or overestimates.
7th Roots
7th roots are another prime deep root, usually requiring estimation + refinement.
7th Roots — Example #1 (easy level)
446634503003^(1/7)
Easy level requires only two significant digits.
Chunk into 7-digit blocks:
44663 | 4503003
We know:
4^7 = 16384 < 44663 < 5^7 = 78125
For two significant digits with five attempts, a fast approach is linear interpolation:
(44663 − 16384) / (78125 − 16384)
This suggests a value around 44.6, so a reasonable first submission is 45, with follow-up guesses if needed.
First-step Newton correction (illustrative)
If we start with a guess A, the Newton increment is:
(N − A^7) / (7A^6)
Using A = 50:
numerator: 44663 − 78125 = −33462
denominator: 7 × 50^6 = 7 × 15625 = 109375
increment ≈ −0.305
revised guess ≈ 49.695 → which is not ideal here, because the true value is near the midpoint and the error behavior is not friendly.
Halley refinement (illustrative)
Halley’s refinement for 7th roots (as written here) is:
((N − A^7) / (3A^7 + 4N)) × A
Using A = 5 (as the draft sets it up):
((44663 − 78125) / (234375 + 178652)) × 5
(−33462 / 413027) × 5
≈ −0.4 × 5? (in rough magnitude)
This supports an answer around 46.
7th Roots — Example #2 (advanced level)
63122223304^(1/7)
Even at advanced level, we still only need two significant digits.
Chunk:
6312 | 2223304
We know:
3^7 = 2187 < 6312 < 4^7 = 16384
When a computationally difficult root only requires two significant digits, estimation is typically preferred.
Halley refinement with A = 3:
((6312 − 2187) / (6561 + 25248)) × 3
(4125 / 31809) × 3
≈ 0.389 × 3 ≈ 1.17
This points toward 3 + 1.17 ≈ 4.17 → interpreted as 41.7, but the correct answer is 34.90, showing how sensitive the method can be when A is small.
Halley refinement with A = 4:
((6312 − 16384) / (49152 + 25248)) × 4
(−10072 / 74400) × 4
≈ −0.542 × 4 ≈ −2.17
This points toward 4 − 2.17 ≈ 1.83 → interpreted as 18.3, also not stable.
This illustrates a practical point: refinement methods (Newton/Halley/log-style estimates) tend to behave better when the computations involve larger, more stable quantities. When the true answer lies near the midpoint and the guessed power is far off, it can be more effective to choose an overestimate and correct downward.
7th Roots — Example #3 (expert level)
3397123818961^(1/7) × 10
Now we need three significant digits.
Chunk:
339712 | 3818961
We know:
6^7 = 279936 < 339712 < 7^7 = 823543
Apply Halley refinement with A = 6 (as written in the draft):
((339712 − 279936) / (839808 + 1358848)) × 6
(59776 / 2198656) × 6
358656 / 2198656 ≈ 0.163
So the estimate becomes:6.163 → interpreted as 616.3, while the correct answer is 616.8
This is already quite close, but to reliably hit three significant digits mentally, you generally need at least one of the following:
Time and perseverance
Memorized milestones
The ability to compute a two-digit number to the 7th power (or approximate it closely)
Strong intuition about refinement and error direction
Decimal & Irregular Exponents & Roots
It All Begins Here
Consider the following question:
(x)^(y/z)
Some variation of this question — where the variables are not “easy numbers” — is common in finance, engineering, and STEM fields.
While precise calculation of the answer to many significant digits is typically unreasonable, estimation to at least 2 (and often 3 or 4) significant digits is plausible using basic logarithmic principles — plus a little bit of memorization.
Basic Principles
To estimate (or calculate) logarithms effectively, it helps to know a modest set of log values. These are particularly useful:
log(2) = .301
log(3) = .477
log(7) = .845
(For finance and STEM fields, an estimate of log(e) = .4343 is also helpful.)
The more log figures you know — and the more digits you know for each figure — the more accurate your results will be. However, the estimates above can still be used with impressive effectiveness because of the following core rules:
log(A×B) = log(A) + log(B)
log(A/B) = log(A) − log(B)
This means it is most practical to memorize logarithms for prime numbers, because logs of composite numbers can be computed by factoring and adding logs of the prime factors.
Note: log(5) is skipped above because it can be computed easily:
log(5) = log(10) − log(2) = 1 − .301 = .699
General Rule for Estimating (x)^(y/z)
To estimate (or calculate) x^(y/z) using logarithms, apply:
Estimate log(x).
Multiply your result by (y/z).
Compute 10^(result) (usually by estimation).
In the last step, linear interpolation can be used. More accurate methods exist, but are not as simple (and typically rely on calculus ideas — e.g., quadratic interpolation / second-degree Lagrange interpolation).
Example 1: 216536^.36
Step 1: Estimate log(216536).
216536 is between 10^5 and 10^6, so the integer part is 5. More precisely:
log(216536) = 5 + log(2.16536)
Now estimate log(2.16536).
2.16536 is about 1/6 of the way from 2 to 3, so we can estimate:
log(2.16536) ≈ .33 (or .333 for slightly more accuracy)
(The true value will be slightly higher because the slope of log(x) decreases as x increases.)
So:
log(216536) ≈ 5.333
Step 2: Multiply by .36.
5.333 × .36 ≈ 1.92
Step 3: Estimate 10^1.92.
10^1.92 = 10^1 × 10^.92 = 10 × 10^.92
Now estimate 10^.92.
We can bracket .92 using logs we can compute quickly:
log(8) ≈ 3×log(2) = 3×.301 = .903
log(9) ≈ 2×log(3) = 2×.477 = .954
.92 is about 1/3 of the way from .903 to .954.
So the value should be about 1/3 of the way from 8 to 9, i.e. roughly:
10^.92 ≈ 8.33
Therefore:
10^1.92 ≈ 10 × 8.33 = 83.33
This is correct to about four significant digits.
Example 2: 42^(100/57)
Step 1: Estimate log(42).
42 is between 10^1 and 10^2, so the integer part is 1:
log(42) = 1 + log(4.2)
Now estimate log(4.2).
4.2 is about 1/5 of the way from 4 to 5.
log(4) = 2×log(2) = 2×.301 = .602
log(5) = 1 − log(2) = 1 − .301 = .699
So a quick linear estimate is:
log(4.2) ≈ .622
(Again, the true value will be slightly higher because of decreasing slope.)
So:
log(42) ≈ 1.622
Step 2: Multiply by (100/57).
1.622 × (100/57) = 162.2/57 ≈ 2.845614
Step 3: Estimate 10^2.845614.
10^2.845614 = 10^2 × 10^.845614 = 100 × 10^.845614
Now compare .845614 to a memorized log:
log(7) = .845
So .845614 is just slightly above log(7), and only about 2% of the way toward log(8).
That produces an estimate of about:
702
The actual answer is 704.
Because we used linear interpolation on a logarithmic curve and rough log values, the estimate ends up off by about 0.3%, which is still very strong for a quick mental method.
Example 3: Investments
Michael aggressively invests $100,000 at age 30. He anticipates accruing 10% continuous compound interest annually. At age 30, Chloe invests $500,000 in a much less volatile investment and is confident of accruing 5% continuous compound interest over the long run.
If both estimates are correct, what will be the approximate value of their investments at age 65?
Using the continuous compounding formula P e^(rt), we convert to base-10 logs using:
log(e) ≈ .4343
Michael: r×t = 0.10×35 = 3.5
Chloe: r×t = 0.05×35 = 1.75
Multiply by log(e):
Michael: .4343 × 3.5 ≈ 1.52
Chloe: .4343 × 1.75 ≈ .76
So:
Michael’s growth factor ≈ 10^1.52
Chloe’s growth factor ≈ 10^.76
Now estimate:
10^1.52 = 10 × 10^.52
10^1.5 ≈ 31.6 (a well-known benchmark using logs or square roots)
Also:
10^.75 is between 5 and 6
log(5)=.699 and log(6)=.778, so 10^.75 is between about 5.6 and 5.65
So:
Chloe at 65: $500,000 × 5.6 ≈ $2,825,000
Michael at 65: $100,000 × 31.6 ≈ $3,160,000
Both are relatively accurate estimates.
Decimal and Irregular Exponents and Roots
Calculation League has four complex categories covering decimal and irregular exponents and roots. All of these questions follow the format x^(y/z).
The distinctions are:
Decimal categories: the value y/z is fixed (e.g., 2.9 or .65)
Irregular categories: y and z are randomly generated
Exponent categories: y/z > 1
Root categories: y/z < 1
The following examples illustrate approaches for solving these questions.
Decimal Exponents — Example #1
(50^3.3)/10^5
In this example we are fortunate to receive a round number (50) as the base.
Rewrite:
50^3.3 = 50^3 × 50^.3 (since A^(B+C) = A^B × A^C)
50^3 = 5^3 × 10^3 (since (A×B)^C = A^C × B^C)
So we have:
(5^3 × 10^3 × 50^.3) / 10^5
Cancel 10^3:
5^3 × 50^.3 / 100
Since 5^3 = 125:
1.25 × 50^.3
So the question becomes: how do we estimate 50^.3?
Approach #1 — Logarithms
(For a basic description of how to calculate with logs, refer to the method outlined above.)
log(50) = log(10) + log(5) = 1 + .699 = 1.699
1.699 × .3 = .5097
Now estimate 10^.5097.
Using:
log(3)=.477
log(4)=.602
.5097 lies between them, closer to 3 than 4, giving an estimate between 3.2 and 3.3.
For first-attempt accuracy, the higher answer is often safer because of the decreasing slope of log(x). If given multiple attempts, starting slightly lower can reduce the chance you need to revise both up and down.
So use:
50^.3 ≈ 3.2 (rough estimate)
Then:
1.25 × 3.2 = 4.0
This would be correct for Calculation League purposes.
(The actual answer is 4.042044.)
Approach #2 — Exponent Principles
We can also attempt to compute 50^.3 directly using exponent rules.
Since .3 = 3/10, one option would be:
compute 50^3 = 125000, then take the 10th root
(for example, square root then fifth root — or in another order)
Often it’s more effective to avoid the cube by rewriting using:
A^(B−C) = A^B / A^C
Here:
50^.3 = 50^.5 / 50^.2
This eliminates the cube. (With “50” specifically, the benefit is small — but the principle matters more for other numbers.)
Recall:
without the 50^.3 component, the rest of the expression reduces to 1.25, and the correct final answer is 4.042044.
In Calculation League, acceptable answers are within .05.
So we need:
1.25 × 50^.3 to fall within .05 of 4.042044
i.e., 50^.3 should be between about 3.1936 and 3.2736 for a first-try hit
But for “within five tries” (two guesses up and two guesses down), we only need a rough estimate between about 3.0 and 3.4.
Now apply rough root estimates:
sqrt(50) ≈ 7.07 (for illustration, round to 7)
50^(1/5): since 2^5=32 and 3^5=243, a linear interpolation suggests around 2.1
So:
50^.3 = 50^.5 / 50^.2 ≈ 7 / 2.1 ≈ 10/3 ≈ 3.33
This produces:
1.25 × 3.33 ≈ 4.125
That’s not accurate enough for a guaranteed first-try, but it is close enough that (remembering linear interpolation underestimates 50^.2) you would expect to adjust downward and get the correct answer on a subsequent attempt.
Decimal Exponents — Example #2 (advanced level)
(765^2.6)/10^5
Logarithmic Method
log(7)=.845 and log(8)=.903
Estimate log(7.65) by linear interpolation: ≈ .883
Therefore log(765) ≈ 2.883
Now:
2.6 × 2.883 ≈ 7.5
So the expression becomes:
10^7.5 / 10^5 = 10^2.5
Now estimate 10^2.5:
10^2.5 = 100 × 10^.5 ≈ 100 × 3.162 = 316
Linear interpolation with log(3)=.477 and log(4)=.602 gives an estimate around 318, but the direct observation above is simpler and more accurate.
The actual answer is 314.427 — so 316 should allow a correct answer within five attempts.
Exponent Method
Computing a^.6 is more difficult than a^.3.
Ways to rewrite:
a^.6 = a^.5 × a^.1
(square root first, then compute a/a^.5, then estimate a fifth root, then multiply back)
Or:
a^.6 = a^1 / a^.4
(estimate the fifth root of a, then either square it or divide a by its fifth root twice)
Now compute:
765^2 can be computed as (765+35)(765−35) + 35^2
= (800×730) + 1225
= 584000 + 1225
= 585225
Or use the “ending in 5” trick:
765^2 = 77×76 with 25 appended = 5852|25 = 585225
So:
(765^2.6)/10^5 = (765^2 × 765^.6)/10^5
= (5.85225×10^5 × 765^.6)/10^5
= 5.85225 × 765^.6
At this point, estimating 765^.6 accurately enough is very difficult without advanced fifth-root techniques. This is exactly why, past a certain difficulty level, the logarithmic method becomes the practical default.
Decimal Exponents — Example #3 (expert level)
(3058^3.5)/10^10
In this question we got a relatively friendly exponent (3.5) and a base near 3000.
Logarithmic Method
log(3058) = log(1000) + log(3.058) = 3 + log(3.058)
log(3)=.477 and log(4)=.602 → estimate log(3.058) ≈ .484
so log(3058) ≈ 3.484
Multiply by 3.5:
3.5 × 3.484 = 12.194
So:
10^12.194 / 10^10 = 10^2.194
= 100 × 10^.194
Using linear interpolation, 10^.194 ≈ 1 + 194/302 ≈ 1.643
So estimated answer ≈ 164
This is close, and improving the interpolation would move it into a correct range.
Exponent Technique
Because 3058 is near 3000, we can compute 3058^3 with the Binomial Theorem:
(3000+58)^3
= 3000^3 + 3×3000^2×58 + 3×3000×58^2 + 58^3
Compute:
3000^3 = 27,000,000,000
3×3000^2×58 = 3×9,000,000×58 = 27,000,000×58 = 1,566,000,000
3×3000×58^2 = 9000×3364 = 30,276,000
58^3 = 195,112
Total:
28,596,471,112
Now check what we actually need. Because the exponent is 3.5, we multiply by sqrt(3058):
sqrt(3058) ≈ 55.299 (≈55.3)
The correct final answer is 158.
For Calculation League, a value between 156 and 160 would allow a correct answer within five attempts.
This means we do not need all digits of 3058^3:
If 3058^3 is only 1 significant digit (~30 billion), answer is ~166
If it’s 2 significant digits (~29 billion), answer is ~160
If it’s 3 significant digits (~28.6 billion), answer is ~158 (and we get essentially first-try accuracy)
So: we mainly need the first two binomial terms, and even the second term can be estimated rather than computed perfectly.
Decimal Roots — Example #1
93^.2
In Calculation League, the “roots” versions of decimal/irregular problems are usually simpler than exponent versions, because you avoid the large multiplications.
With up to five submitted answers to get two significant digits, we can use simple milestones:
2^5 = 32
3^5 = 243
(5/2)^5 ≈ 98
So the fifth root of 93 is between 2 and 2.5, and since (5/2)^5 ≈ 98, the answer should be just under 2.5.
Submitting 2.5 or 2.45 without further calculation is reasonable and can be correct on the first try.
Decimal Roots — Example #2 (advanced level)
1815^.65
Logarithmic Method
Because linear interpolation is least accurate in the 1–2 range, numbers starting with 1 or 2 can produce greater error if you start from log(1) and log(2). So rather than starting with log(1.815)+log(1000), we can sometimes start with log(18.15)+log(100).
Option A
Estimate log(1.815) between log(1)=0 and log(2)=.301:
linear estimate gives log(1.815) ≈ .245
so log(1815) ≈ 3.245
Multiply:
3.245 × .65 ≈ 2.109
So answer ≈ 10^2.109 = 100 × 10^.109
Estimate 10^.109 via linear interpolation:
10^.109 ≈ 1 + 109/301 ≈ 1.362
So estimate ≈ 136
This is close, but not quite close enough to guarantee a correct answer without refinement (though it is close enough that small strategy adjustments can make it work).
Option B
Shift to log(18.15)+log(100). Use log(18) and log(20):
log(18) = log(2) + 2×log(3) = .301 + .954 = 1.255
log(20) = 2×log(2) + log(5) = .602 + .699 = 1.301
Even using log(18) directly gives log(1815) ≈ 3 + 1.255 = 4.255 — but since we are using 18.15, we correct via interpolation.
Using the approach as written in your draft:
this produces an estimate around 3.258 for log(1815) (correct is 3.25888)
Then:
3.258 × .65 ≈ 2.118
Now we face the same issue: finishing with a small exponent (.118) amplifies interpolation error. So mirror the earlier fix: interpolate with 1.118 instead.
Use:
log(12) = 2×log(2) + log(3) = .602 + .477 = 1.079
log(14) = log(2) + log(7) = .301 + .845 = 1.146
Interpolate 1.118 between 1.079 and 1.146:
fraction ≈ (1.118−1.079)/(1.146−1.079) = 39/67
so value ≈ 12 + (2×39/67) = 12 + 78/67 = 13.164
So answer ≈ 10 × 13.164 = 131.64, which is correct within .5.
Root Method
Solving a^.65 directly by roots is difficult unless you are comfortable with fifth roots.
One way:
rewrite as a^.5 × a^.1 × a^.05
Using reasonable estimates:
1815^.5 ≈ 42.6
fifth root of 42.6 ≈ 2.1 (since 2^5=32 and 3^5=243)
sqrt(2.1) ≈ 1.45
Then:
42.6 × 2.1 × 1.45 ≈ 130
This is close enough to submit a correct answer — but it is not reliable unless you are very comfortable with fifth roots and with adjustment strategies.
(As you note in your draft: choosing 2.1 instead of 2.05 is influenced by the fact that linear interpolation underestimates, but rounding up can make it harder to know whether your final estimate is an under- or overestimate. In this case it works, but it is not always clear.)
Decimal Roots — Example #3 (expert level)
276^.55
Logarithmic Method
log(2.76) ≈ .435 (linear interpolation)
so log(276) ≈ 2.435
Multiply:
2.435 × .55 ≈ 1.339
Then:
10^1.339 = 10 × 10^.339
Estimate 10^.339 between 2 and 3:
log(2)=.301
log(3)=.477
So 10^.339 ≈ 2.16
Final estimate:
21.6
This is within .5 of the exact answer.
Root Method
Apply a similar idea as above, but note we no longer need to multiply by a^.1, so we can reorder roots.
276^.5 ≈ 16.6
16.6^.5 ≈ 4.07
Now we need 4.07^.05 (i.e., fifth root of sqrt(sqrt(276))) to multiply by 16.6, but we can estimate 4.07^.2 first:
To get first-try accuracy, we’d need 4.07^.2 between 1.3 and 1.35. For five attempts, we only need it between about 1.18 and 1.47.
Estimate milestones:
1.5^2 = 2.25; 2.25^2 > 4.07 → so 4.07^.2 < 1.5
1.4^2 = 1.96; 1.96^2 ≈ 3.84; 3.84×1.4 > 4.07 → so 4.07^.2 < 1.4
1.3^2 = 1.69; 1.69^2 ≈ 2.85; 2.85×1.3 < 4.07 → so 4.07^.2 > 1.3
So it’s between 1.3 and 1.35. Start with 1.3 and guess up if needed:
16.6 × 1.3 = 21.6
Same result as the log method.
Irregular Exponents — Example #1
2^(41/6)/10^8
On irregular exponents, the questions include a denominator to keep the number of significant digits reasonable. Usually the denominator is 10^8, but not always.
Logarithmic Method
log(2)=.301
.301 × (41/6) ≈ 2.057
So:
2^(41/6)/10^8 = 10^2.057 / 10^8 = 10^.057 / 10^6
Using linear interpolation for 10^.057 is inaccurate, so instead rewrite as:
10^1.057 / 10^7
Now estimate 10^1.057 ≈ 10 × 10^.057.
Using only memorized values, you then proceed as written in your draft:
log(12)=2×log(2)+log(3)=1.079
(57/79) scaling leads to 10 × (10 + 114/79) ≈ 114.4
Then adjust for the denominator:
estimated result ≈ .000001144
Root Method
Rewrite:
2^(41/6) = 2^6 × 2^(5/6)
So:
2^6 = 64
need 32^(1/6)
Compute 6th root via cube root then square root (or vice versa):
cube root of 32 ≈ 3.175
sqrt(3.175) ≈ 1.782
Then:
64 × 1.782 ≈ 114.05
Even 1.8 would likely be sufficient.
Irregular Exponents — Example #2 (advanced level)
4.3^(188/93)/10^8
Logarithmic Method
log(4)=.602; log(5)=.699
interpolate log(4.3) ≈ .631
Multiply:
.631 × (188/93) ≈ 1.275
So:
10^1.275 / 10^8 = 10^.275 / 10^7
Estimate 10^.275:
275/301 ≈ .914
so 10^.275 ≈ 1.914
So estimated answer:
.000000191
(correct to an extra digit)
Root Method
Rewrite exponent as integer + remainder:
4.3^(188/93) = 4.3^2 × 4.3^(2/93)
4.3^2 = 18.49
Estimating 4.3^(2/93) precisely is difficult, but since only two significant digits are needed, rounding 18.49 to 19 and guessing upward if needed is a practical competition approach.
Irregular Exponents — Example #3 (expert level)
3.73^(1244/946)/10^8
Logarithmic Method
log(3)=.477; log(4)=.602
interpolate log(3.73) ≈ .568
Multiply:
.568 × (1244/946) ≈ .747
So:
10^.747 / 10^8
Estimate 10^.747:
log(5)=.699 and log(6)=.778 → 10^.747 is between 5 and 6
gives about 5.61
So estimated answer:
.000000056
Root Method
Rewrite exponent as integer + remainder:
3.73^1 × 3.73^(298/946)
Since only two significant digits are needed, estimating cube root of 3.73 (~1.55) can be sufficient in this context:
rounding to 1.5 gives the answer to two significant digits.
Irregular Roots — Example #1
45^(4/948)
Irregular roots follow the same structure as irregular exponents, except the exponent is less than 1.
In some randomly generated irregular-root questions, the value converges close enough to 1 that calculation is unnecessary.
Here:
the 237th root of 45 is barely more than 1
if you are confident the answer lies between 1 and 1.45, Calculation League does not require heavy computation
So here an answer of 1 is appropriate; 1.1 is already an unreasonably high estimate for the 237th root of 45.
Irregular Roots — Example #2 (advanced level)
875835^(2/19)
Logarithmic Method
Rewrite:
(8.75835×10^5)^(2/19)
Estimate log(8.75835):
log(8)=.903; log(9)=.954
interpolate log(8.75835) ≈ .942
So log(base) ≈ 5.942.
Multiply:
5.942 × (2/19) ≈ .625
So:
10^.625
Compare:
log(4)=.602
log(5)=.699
Interpolate gives:
about 4.24 (or 4.2), which matches the correct answer.
Root Method
This is more “art” and depends on target accuracy.
2/19 is between 1/10 and 1/9
so the answer is between the 10th and 9th roots
Estimate 9th root via cube root twice:
875835^(1/3) is between 95 and 96
cube root of 95 is about 4.5
→ so 9th root is about 4.5
Estimate 10th root via square root then fifth root:
sqrt(875835) is between 935 and 936
935^.2 is around 3.9 (since 3^5=243 and 4^5=1024)
Either estimate is enough to submit a correct answer within five attempts, and averaging them supports 4.2, which is correct.
Irregular Roots — Example #3 (expert level)
48250228^(56/914)
Logarithmic Method
Rewrite:
(4.8250228×10^7)^(56/914)
Estimate log(4.8250228):
log(4)=.602; log(5)=.699
interpolate ≈ .682
So log(base) ≈ 7.682.
Multiply:
7.682 × (56/914) ≈ .471
So:
10^.471
Since log(3)=.477, this is close to 3:
start with 3.0 (and note it should be slightly less)
Correct answer is 2.96. A single interpolation step yields ~2.97.
Root Method
We want a value between the 16th and 17th root of an eight-digit number.
Estimating the 16th root (four square roots) is often enough:
48250228^.5 ≈ 6946
6946^.5 ≈ 83
83^.5 ≈ 9.1 (or 9 is enough)
9.1^.5 ≈ 3