2008 MCWC

The 2008 Mental Calculation World Cup was the third edition of the event. As in the previous competitions, there were only two surprise task categories, and no reported results from those categories. The overall event was won by Albert Coto (Spain).

Surprise Task Categories (2008)

The 2008 MCWC surprise task category included the following two task types:

• A × (B − C), where

  • A is a four-digit number,

  • B and C are three-digit numbers, and

  • B is greater than C.

• Finding the average of a list of durations of time, written in the format:
  X hours, Y minutes, Z seconds

Notes on Calculation League Modifications

In Calculation League, the first task format has been modified so that B is not necessarily greater than C.

The second task format has been modified more substantially. Instead of time averages, Calculation League uses a structurally similar arithmetic task of the form:

((a×b)+(c×d)+e)/f

Example 1

1333 × (748 − 337)

(Easy Level)

This is a relatively straightforward four-digit × three-digit multiplication problem, although the three-digit number is not initially visible (even though it is easy to calculate).

With the specific numbers used here, factoring turns out to be more efficient than either standard multiplication or cross-multiplication.

We begin by evaluating the subtraction:

• The difference between 748 and 337 is 411.

Now factor:

• 411 can be quickly factored into 3 × 137.

Compute:

• 1333×3=39991333 × 3 = 39991333×3=3999

As a result, the original question reduces to:

• 3999×1373999 × 1373999×137

This can be computed as:

• 4000×137=5480004000 × 137 = 5480004000×137=548000

• Subtract one copy of 137

So:

• 548000−137=547863548000 − 137 = 547863548000−137=547863

Final answer: 547863

Example 2

((8982 × 20) + (11 × 620) + 33) ÷ 2

(Advanced Level)

In this second question, the number generation was rather fortunate.

• The divisor is only 2, and

• Both products contain a factor that ends in zero.

As a result, it makes sense to divide by 2 before computing the products. Doing so gives:

(8982×10)+(11×310)+16.5(8982 × 10) + (11 × 310) + 16.5(8982×10)+(11×310)+16.5

Compute each part:

• 8982×10=898208982 × 10 = 898208982×10=89820

Now we are left with one relatively simple 2 × 2 multiplication and some additions:

• 11×310=341011 × 310 = 341011×310=3410

Add everything together:

• 89820+3410+16.5=93246.589820 + 3410 + 16.5 = 93246.589820+3410+16.5=93246.5

Final answer: 93246.5

Example 3

((334 × 86) + (79 × 299) + 7493) ÷ 82

(Expert Level)

For the expert-level task, the numbers are more difficult, and the answer is required to the nearest integer.

If you are comfortable calculating the entire numerator first and then dividing by the denominator, that approach will work fine. Otherwise, we can obtain a good estimate using ratio balancing.

Estimation via Ratio Balancing

We begin by adjusting each term relative to the denominator 82.

First term: 334 × (86 / 82)

This can be rewritten as:

(334×82/82)+(334×4/82)(334 × 82 / 82) + (334 × 4 / 82)(334×82/82)+(334×4/82)

Which simplifies to:

• 334+334/20.5334 + 334 / 20.5334+334/20.5

Since 334/20.5≈16334 / 20.5 ≈ 16334/20.5≈16, we obtain:

• 334 + 16 = 350

(In this step, we rounded down.)

Second term: 299 × (79 / 82)

This can be rewritten as:

(299×79/79)−(299×3/82)(299 × 79 / 79) − (299 × 3 / 82)(299×79/79)−(299×3/82)

Which simplifies to:

• 299−299/27.3299 − 299 / 27.3299−299/27.3

Since 299/27.3≈11299 / 27.3 ≈ 11299/27.3≈11, we obtain:

• 299 − 11 = 288

(In this step, we rounded up.)

Third term: 7493 ÷ 82

This division can be performed directly:

• 7493÷82=91.387493 ÷ 82 = 91.387493÷82=91.38

Accuracy Considerations

To guarantee correctness to the nearest integer, calculations would normally need to be accurate to the nearest 0.1.

In Calculation League, however, when the nearest integer is required, answers are accepted if they are within 0.5 of the correct value.

In this problem:

• We rounded down in the first estimation step.

• We rounded up in the second estimation step.

Because the rounding errors go in opposite directions, the combined error after the first two steps must be less than 0.5. As a result, by calculating the third term exactly, we can be confident that the final answer will fall within the acceptable range and be correct on the first submission.

Furthermore, because Calculation League allows up to five submissions per question, it is actually possible to compute one fewer significant digit in steps 1 and 2 and still guarantee correctness within five attempts. If we know whether our estimate after step 2 is an overestimate or an underestimate, we can still reach the correct answer reliably.