2004 MCWC

The 2004 Mental Calculation World Cup was the first version of the event. There were only two surprise task categories, and there were no reported results from those categories. The overall event was won by Robert Fountain (UK).

Surprise Task Categories (2004)

The 2004 MCWC surprise task category included the following two tasks:

• A × B × C, with three-digit numbers

• Ax = B × C, involving two-digit and four-digit numbers

Example 1

76 × 78 × 33 (Easy Level)

For the easy-level question, we start with an A × B × C problem using two-digit numbers.

You actually can apply cross-multiplication to this kind of question.

Cross-Multiplication Approach

We begin by working from right to left.

• 6×8×3=1446 × 8 × 3 = 1446×8×3=144
  Write 4, carry 14.

• ((6×7)+(7×8))×3)+(8×6×3)+14((6 × 7) + (7 × 8)) × 3) + (8 × 6 × 3) + 14((6×7)+(7×8))×3)+(8×6×3)+14
  → (98×3)+144+14=452(98 × 3) + 144 + 14 = 452(98×3)+144+14=452
  Write 2, carry 45.

• ((6×7)+(7×8))×3)+(7×7×3)+45=486((6 × 7) + (7 × 8)) × 3) + (7 × 7 × 3) + 45 = 486((6×7)+(7×8))×3)+(7×7×3)+45=486
  Write 6, carry 48.

• 7×7×3+48=1957 × 7 × 3 + 48 = 1957×7×3+48=195

Final answer: 195624

I would not recommend practicing this method, however. It is important to develop the visualization and versatility skills needed to approach this type of question without cross-multiplication. There are a variety of better options for approaching this problem.

Standard Method (Recommended)

The simplest approach is to use standard school-method multiplication.

My recommendation is to break apart the “33”, because this reduces the work to a single difficult multiplication. I also recommend leaving the most difficult number for last, because this allows you to either write down or begin giving the answer as you go, which reduces the burden of remembering digits.

Step 1: Multiply 33 × 76

We can write:

• 33×76=(30×76)+(3×76)33 × 76 = (30 × 76) + (3 × 76)33×76=(30×76)+(3×76)

Compute each part:

• 30×76=228030 × 76 = 228030×76=2280

• 3×76=2283 × 76 = 2283×76=228

Add:

• 2280+228=25082280 + 228 = 25082280+228=2508

Step 2: Multiply 2508 × 78

At this point, depending on the context, you can proceed left to right or right to left, using as many digits at a time as you like. Below are several valid approaches.

Left to Right (2 × 1 Multiplications)

• 78×2=15678 × 2 = 15678×2=156
  Write 1, keep 56.

• 78×5=39078 × 5 = 39078×5=390; add 560 → 950
  Write 9, keep 50.

• 78×0=078 × 0 = 078×0=0
  Write 5.

• 78×8=62478 × 8 = 62478×8=624
  Write 624.

Because the tens digit of 2508 is zero, and because
(a) the hundreds digit is 5, and
(b) the ones digit of 78 is even,

this calculation reduces to two simple steps:

• 2.5×78=1952.5 × 78 = 1952.5×78=195

• 8×78=6248 × 78 = 6248×78=624

So the answer is 195624.

Left to Right (2 × 2 Multiplications)

• 78×25=195078 × 25 = 195078×25=1950
  Write 19, keep 50.

• 78×8=62478 × 8 = 62478×8=624; add 5000 → 5624

Right to Left (2 × 1 Multiplications)

• 78×8=62478 × 8 = 62478×8=624
  Write 4, carry 62.

• 78×0=078 × 0 = 078×0=0; add 62
  Write 2, carry 6.

• 78×5=39078 × 5 = 39078×5=390; add 6 → 396
  Write 6, carry 39.

• 78×2=15678 × 2 = 15678×2=156; add 39 → 195

Right to Left (2 × 2 Multiplications)

• 78×8=62478 × 8 = 62478×8=624
  Write 24, carry 6.

• 78×25=195078 × 25 = 195078×25=1950; add 6 → 1956

Alternative Methods

Of course, there are always many alternative ways to perform multiplications.

In this case, one alternative approach is the following.

Algebraic Decomposition

• 78×76=772−1278 × 76 = 77^2 - 1^278×76=772−12

Compute:

• 772=(80×74)+32=5920+977^2 = (80 × 74) + 3^2 = 5920 + 9772=(80×74)+32=5920+9

Therefore:

• 78×76=592878 × 76 = 592878×76=5928

Now multiply:

• 5928×335928 × 335928×33

We factor 33 as 3 × 11.

• 5928×3=177845928 × 3 = 177845928×3=17784

Then:

• 17784×11=177840+17784=19562417784 × 11 = 177840 + 17784 = 19562417784×11=177840+17784=195624

Example 2

(4117 × 1197) ÷ 210 (Advanced Level)

This question is of the form (A×B)÷C(A × B) ÷ C(A×B)÷C. We will need five significant digits for the answer.

The numbers here are large enough that it is worth briefly checking whether the problem can be simplified through factorization. This is especially true when the divisor is 210, which can immediately be factored as:

• 210=2×3×5×7210 = 2 × 3 × 5 × 7210=2×3×5×7

Checking the last digits of 4117 and 1197 shows that neither is divisible by 2 or 5. A quick factorization check does reveal, however, that 1197 is divisible by both 3 and 7.

As a result, we can divide the numerator and denominator by 21, obtaining:

• (4117×57)÷10(4117 × 57) ÷ 10(4117×57)÷10

• or equivalently, 4117×5.74117 × 5.74117×5.7

At this point, the problem has been reduced to the same type of calculation used in Example #1: a 4 × 2 multiplication.

Final Calculation

We can proceed as follows:

• 4117×5=205854117 × 5 = 205854117×5=20585 (this can be performed quickly)

• 4117×0.74117 × 0.74117×0.7 is slightly more difficult and equals 2882 (rounded to the nearest integer)

Add:

• 20585+2882=2346720585 + 2882 = 2346720585+2882=23467

Example 3

(3702 × 8227) ÷ 407 (Expert Level)

In this expert-level question, we again have a similar structure, but factorization does not simplify the problem.

Instead of factorization, we can use ratio balancing, particularly if we do not want to perform the full multiplication from left to right.

Ratio Balancing

It is useful to observe that:

• (x÷400)×(400÷407)=x÷407(x ÷ 400) × (400 ÷ 407) = x ÷ 407(x÷400)×(400÷407)=x÷407

This allows us to temporarily replace 407 with 400, and then correct the result afterward.

We can choose either 3702 or 8227 to divide by 400. I recommend using 3702 because:

• (a) the division produces a number less than 10, and

• (b) the ones digit is even, resulting in one fewer non-zero digit.

Compute:

• 3702÷400=9.2553702 ÷ 400 = 9.2553702÷400=9.255

Multiplying 9.255 × 8227

This transformation makes it much easier to apply standard methods.

Break the calculation into parts:

• 9×82279 × 82279×8227, plus

• (8227÷4)(8227 ÷ 4)(8227÷4) (this corresponds to 0.25), plus

• (8227÷2000)(8227 ÷ 2000)(8227÷2000)

Compute each:

• 9×8227=740439 × 8227 = 740439×8227=74043

• 8227÷4=20578227 ÷ 4 = 20578227÷4=2057 (rounded)
  Add immediately to obtain 76100.

• 8227÷200=418227 ÷ 200 = 418227÷200=41 (rounded)

So the result is:

• 76141

Correcting the Approximation

At this point, we could multiply by 400 and then divide by 407. This is not especially burdensome, as it is equivalent to performing a 5 × 1 multiplication followed by a 6 ÷ 3 division.

Alternatively, we can note that:

• 400÷407400 ÷ 407400÷407 is very close to 57÷5857 ÷ 5857÷58
  (since 399÷406399 ÷ 406399÷406 is exactly 57÷5857 ÷ 5857÷58)

Multiplying by 57÷5857 ÷ 5857÷58 is the same as subtracting 1/58 of the number.

Compute:

• 76141÷58≈131376141 ÷ 58 ≈ 131376141÷58≈1313

Subtract:

• 76141−1313=7482876141 − 1313 = 7482876141−1313=74828

Adjusting for the “Nearly”

To account for the small difference between 400÷407400 ÷ 407400÷407 and 399÷406399 ÷ 406399÷406, we use the identity:

• (x−y)(x+y)=x2−y2(x − y)(x + y) = x^2 − y^2(x−y)(x+y)=x2−y2

We rewrite:

• (400÷407)÷(399÷406)=(400×406)÷(399×407)(400 ÷ 407) ÷ (399 ÷ 406) = (400 × 406) ÷ (399 × 407)(400÷407)÷(399÷406)=(400×406)÷(399×407)

Compute:

• 400×406=4032−32400 × 406 = 403^2 − 3^2400×406=4032−32

• 399×407=4032−42399 × 407 = 403^2 − 4^2399×407=4032−42

Therefore, the correction term is:

• (42−32)÷4032=7÷165000(4^2 − 3^2) ÷ 403^2 = 7 ÷ 165000(42−32)÷4032=7÷165000
  which is approximately 1/23500 of the answer.

Since 1/23500 of 74828 is about 3, we add 3.

Final answer: 74831

Note: In Calculation League formats with five attempts, performing this final correction step is unnecessary.