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

Introduction / Context:We want to adjust a given number downward so that the resulting remainder is congruent to 6 modulo 9, 11, and 13 simultaneously. This is a modular arithmetic problem relying on least common multiples (LCMs).

Given Data / Assumptions:

• Original number: 1294
• Desired property: R ≡ 6 (mod 9, 11, 13)
• We seek the least nonnegative subtraction x where R = 1294 − x.

Concept / Approach:If R ≡ 6 modulo each of 9, 11, and 13, then R − 6 is divisible by LCM(9, 11, 13). Compute 1294 − 6 and reduce modulo that LCM to find the smallest x so that 1294 − x − 6 is exactly that multiple.

Step-by-Step Solution:LCM(9, 11, 13) = 9*11*13 = 1287 (since they are pairwise coprime).Compute 1294 − 6 = 1288.We need x such that 1288 − x is a multiple of 1287 ⇒ choose x = 1 to get 1287.Thus the least number to subtract is 1.

Verification / Alternative check:R = 1294 − 1 = 1293. Then 1293 − 6 = 1287 which is divisible by 9, 11, and 13. Hence 1293 leaves remainder 6 when divided by each of 9, 11, and 13.

Why Other Options Are Wrong:0, 2, 3, and 5 do not reduce 1294 sufficiently to land exactly one LCM step down from 1294 − 6. They fail at least one modulus check.

Common Pitfalls:Forgetting to subtract the target remainder first; computing gcd instead of LCM; arithmetic slips with 1287.

Final Answer:1