2014 MCWC

The 2014 Mental Calculation World Cup was the sixth edition of the event. German competitor Andreas Berger finished first overall in surprise tasks. Granth Thakkar (India) was the overall champion.

Marc Jornet Sanz finished second overall, Chie Ishikawa (Japan) finished third overall, Hua Wei Chan finished ninth, and Wenzel Gruss finished tenth overall (and seventh in surprise tasks).

Surprise Task Categories (2014)

The 2014 MCWC surprise task category included the following five task types:

• (a/b) × a decimal,
  where decimals generally had two or three significant digits

• Calculating km/hr,
  with kilometers given to three or four significant digits and minutes given to three significant digits

• Exact division by 7/3,
  with the answer always being four digits

• (a × b) − (c × d),
  with all numbers being five digits

• a^(1/2) + a^(1/3) + a^(1/4),
  one question, with as many significant digits correct as possible

Notes on Calculation League Modifications

Calculation League omits:

• the km/hr question, and

• the exact division question (since division already exists as a standard task).

Example 1

13181^(1/2) + 13181^(1/3) + 13181^(1/4)

(Easy Level)

For this question, we must calculate each individual term.

Estimating each root:

• 13181^(1/2) is 115 (or 114.8)

• 13181^(1/3) is 24 (or 23.6)

• 13181^(1/4) is 11 (or 10.7)

If we simply round each value to the nearest integer, we obtain:

• 115 + 24 + 11 = 150

However, the correct answer is 149.

If we keep track of the direction of rounding, we can arrive at the correct answer on the second attempt. Otherwise, we can use the rule for addition/subtraction that:

If we round in one direction three more times than the other direction, then we should adjust the answer one step away from the direction of rounding.

In this case, that means adjusting the estimate from 150 down to 149.

Logarithmic Estimation Alternative

In some situations — depending on the number of terms and the required accuracy — using logarithms can be appropriate. This works because we only need to estimate the logarithm of a single base (here, 13181), and then apply a series of simple divisions.

We estimate:

• log(13181) = 4 + log(1.3181)

Linear interpolation is most inaccurate between 1 and 2, so we estimate:

• 4 + log(1) + (0.301 × 0.3181) → 4.1

Now divide 4.1 by 2, 3, and 4, and sum the results:

• 4.1 / 2 = 2.05

• 4.1 / 3 = 1.37

• 4.1 / 4 = 1.025

Adding:

• 2.05 + 1.37 + 1.025

We then apply linear interpolation back to obtain estimates of:

• 13181^(1/2) → 117

• 13181^(1/3) → 24

• 13181^(1/4) → 11

Obviously, the above is not accurate enough in this case. For problems involving more terms or deeper roots, however, this strategy may be beneficial.

Example 2

(19125 × 21095) − (74169 × 32868)

(Advanced Level)

This problem requires performing two large multiplications in parallel, while finding the difference at each step. It is important to determine whether the final answer is positive or negative. Here, the answer is clearly negative, so it makes sense to start with the right-hand part.

Proceeding step by step:

• (9 × 8) − (5 × 5) → 47
  Type 7, carry 4.

• 4 + (9 × 6) + (6 × 8) − (5 × 9) − (5 × 2) → 51
  Type 1, carry 5.

• 5 + (9 × 8) + (6 × 6) + (1 × 8) − (2 × 9) − (1 × 5) → 98
  Type 8, carry 9.

• 9 + (9 × 2) + (6 × 8) + (1 × 6) + (4 × 8) − (5 × 1) − (1 × 9) − (9 × 5) → 54
  Type 4, carry 5.

• 5 + (9 × 3) + (6 × 2) + (1 × 8) + (4 × 6) + (7 × 8) − (5 × 2) − (2 × 1) − (9 × 9) − (1 × 5) → 34
  Type 4, carry 3.

• 3 + (6 × 3) + (1 × 2) + (4 × 8) + (7 × 6) − (2 × 2) − (1 × 1) − (1 × 9) → 83
  Type 3, carry 8.

• 8 + (1 × 3) + (4 × 2) + (7 × 8) − (1 × 2) − (9 × 1) → 64
  Type 4, carry 6.

• 6 + (4 × 3) + (7 × 2) − (9 × 2) − (1 × 1) → 13
  Type 3, carry 1.

• 1 + (7 × 3) − (1 × 2) → 20
  Type 20.

Of course, this question can be approached using many multiplication strategies. However, the additional visualization required here makes cross-multiplication substantially more useful than a standard A×B approach.

Example 3

(925 / 591) × 93239

(Expert Level)

Correct answer: 145932

For an inexact (a × b) / c question, we must decide whether to begin with multiplication or division. In this case, we need six significant digits.

We do not need to perform the division as written. The suggestion here is to “save” the number that is easier to multiply — almost always the smaller number (here, 925).

Choosing the Order of Operations

If we start with 925 / 591, the division must be computed to:

• 1.5651 to be within 5 of the answer, or

• 1.56514 to be correct to the nearest integer.

This leaves a 6×5 multiplication, with the six-digit number being unseen.

If instead we start with 93239 / 591, the division must be computed to:

• 157.76 to be within 5, or

• 157.765 to reach the nearest integer
  (this actually rounds to the wrong integer, but is within 0.2 of the correct answer).

The division requires the same accuracy, but the multiplication is only 6×3 instead of 6×5, which is a substantial difference in difficulty.

Exploiting Structure in 925

The number 925 is easier to work with than it first appears. Multiplying by a number ending in 25 simplifies many products.

More importantly, we can rewrite:

• 925 = (37 / 4) × 100

To compute 157.765 × 925, we proceed as follows:

1. Multiply by 100
   → 15776.5 × 9.25

2. Multiply by 9
   → 15776.5 × 9 = 141988.5

3. Add 15776.5 / 4
   → 141988.5 + 3944.125 = 145932.625

This result is within 0.2 of the correct answer.

Comparison of Effort

To multiply by 925, we effectively:

• shifted the decimal two places,

• performed one 6×1 multiplication and one 5×1 division, and

• added a six-digit number to a four-digit number
  (the decimal component being unnecessary here).

By comparison:

Standard / traditional / soroban methods require:

• six 3×1 multiplications (or three 6×1 multiplications), and

• adding 20 non-zero digits (instead of 10).

Cross-multiplication requires:

• 18 one-digit multiplications, and

• adding 34 non-zero digits.

Based on this, using basic arithmetic principles — here rewriting 925 as (37/4)×100 — is:

• substantially more effective than cross-multiplication, and

• both more effective and less burdensome to visualization than traditional or soroban approaches.