Deep Roots
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