What is the least number that must be subtracted from 1294 so that, when the result is divided by 9, 11 and 13, the remainder in each case is 6?

Introduction / Context:
This number theory problem checks your understanding of common remainders and the use of the least common multiple for several divisors. You are given a starting number and asked to adjust it so that dividing by 9, 11 and 13 always leaves the same remainder 6. The goal is to find the smallest possible subtraction to achieve this uniform remainder condition.

Given Data / Assumptions:

• Original number = 1294.
• Divisors: 9, 11 and 13.
• Desired remainder when divided by each divisor = 6.
• Let the new number be N = 1294 - k, where k is the least number to subtract.
• N leaves remainder 6 when divided by 9, 11 and 13.

Concept / Approach:
If N leaves remainder 6 when divided by each of 9, 11 and 13, then N - 6 is divisible by all three of these numbers. Therefore, N - 6 must be a multiple of the least common multiple of 9, 11 and 13. Once we compute this least common multiple, we look for multiples of it that are close to 1294 and then choose the one that gives the largest N less than or equal to 1294. The difference between 1294 and that N gives the required k.

Step-by-Step Solution:
Let N = 1294 - k. We want N ≡ 6 (mod 9), N ≡ 6 (mod 11), N ≡ 6 (mod 13). This means N - 6 is divisible by 9, 11 and 13. Compute least common multiple of 9, 11 and 13. Since they are pairwise coprime, LCM = 9 * 11 * 13 = 1287. So N - 6 must equal 1287 * t for some integer t. We need N as large as possible but not greater than 1294, so try t = 1. Then N - 6 = 1287, so N = 1287 + 6 = 1293. Now 1293 ≤ 1294, which is valid. Therefore k = 1294 - 1293 = 1.

Verification / Alternative check:
Check that 1293 leaves remainder 6 with each divisor. 1293 ÷ 9 = 143 with remainder 6, 1293 ÷ 11 = 117 with remainder 6, and 1293 ÷ 13 = 99 with remainder 6. All three checks confirm that 1293 satisfies the condition, so subtracting 1 from 1294 is both necessary and sufficient.

Why Other Options Are Wrong:
2, 3, 4 and 5: Subtracting any of these values gives numbers that do not simultaneously leave remainder 6 when divided by 9, 11 and 13. They fail at least one of the divisibility checks for N - 6.

Common Pitfalls:
Trying to test each option only against one divisor and not checking all three. Confusing common remainder problems with problems that require remainder zero, that is pure divisibility. Forgetting to account for the remainder shift, that is working with N instead of N - 6 when applying the least common multiple idea.

Final Answer:
The least number that must be subtracted is 1