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Difficulty: 5/102022 IOQM 2022 (Q19)

Consider string of 1’s. We wish to place some + signs in between so that the sum is 1000. For instance, if = 190, one may put + signs so as to get 11 ninety times and 1 ten times, and get the sum 1000. If is the number of positive integers for which it is possible to place + signs so as to get the sum 1000, then find the sum of the digits of a.

Guide / Hint

Hint 1: Start by analyzing the initial conditions and setting up the basic equations.  1000 = 1a1 + 11a2 + 11a3 + …….

Hint 2: Look for algebraic properties, symmetry, or geometric theorems to simplify. where a1, a2, a3, …… are non-negative integers for all ai, when i > 3, ai = 0.

Hint 3: Proceed with the final algebraic steps to solve the system. => 1000 solve for the final valuep + 11q + r.

Solution

Step 1:  1000 = 1a1 + 11a2 + 11a3 + ……

Step 2: where a1, a2, a3, …… are non-negative integers for all ai, when i > 3, ai = 0

Step 3: => 1000 = 111p + 11q + r

Step 4: If = 0, then there are 91 possibilities for q, r.

Step 5: For = 1, there are 81 possibilities for q, r.

Step 6: For = 2, there are 71 possibilities for q, r.

Step 7: For = 3, there are 61 possibilities for q, r.

Step 8: For = 4, there are 51 possibilities for q, r.

Step 9: …………………………………………………

Step 10: …………………………………………………

Step 11: …………………………………………………

Step 12: For = 9, there are 1 possibility for q, r.

Step 13: Hence, total number of possibilities = 460

Step 14: Sum of digits of 460 is 10.

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