Maths Olympiad Prep

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Combinatorics Difficulty 5.1 AIME, harder Prove it Brazil

Elaine uses each of the digits 11 to 88 and writes down two 44-digit numbers.

a. If the sum of these numbers is the largest possible, what is their sum?

b. If the sum of these numbers is the least possible, what is their minimum value?

Solution

a. The largest sum is obtained when the largest digits are assigned to the leftmost positions, so it is equal to (8+7)1000+(6+5)100+(4+3)10+2+1=16373(8+7) \cdot 1000 + (6+5) \cdot 100 + (4+3) \cdot 10 + 2 + 1 = 16373.

b. The smallest sum is obtained when the smallest digits are assigned to the leftmost positions. So 11 and 22 are in the thousands, 33 and 44 are in the hundreds, 55 and 66 are in the tens and 77 and 88 are in the units. The biggest of the two numbers is the one beginning with 22. The minimum of its difference is obtained by making the digits in the smallest number bigger and the digits in the biggest number smaller, so it is 23571468=8892357 - 1468 = 889.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.