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.
Hint 1: Start by analyzing the initial conditions and setting up the basic equations. 1000 = 1a1 + 11a2 + 11a3 + …….
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.
Step 1: 1000 = 1a1 + 11a2 + 11a3 + ……
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.
Ready to track your progress and master these topics?
Create a free account