What least number must be subtracted from 1936 so that the remainder, when divided by 9, 10, and 15, leaves the same remainder 7 in each case?

Introduction / Context:We need to reduce 1936 to a number that is congruent to 7 modulo 9, 10, and 15. This again uses the idea that if a number leaves the same remainder with several moduli, then subtracting that remainder yields a multiple of their LCM.

Given Data / Assumptions:

• Original number: 1936
• Desired remainder: 7 for each divisor 9, 10, and 15
• Find the least nonnegative subtraction x so that R = 1936 − x meets the condition.

Concept / Approach:If R ≡ 7 (mod 9, 10, 15), then R − 7 is divisible by LCM(9, 10, 15). Compute that LCM, reduce 1936 − 7 modulo it, and choose the smallest subtraction to hit an exact multiple.

Step-by-Step Solution:LCM(9, 10, 15) = 90 (since 9 = 3^2, 10 = 2*5, 15 = 3*5 ⇒ LCM = 2*3^2*5 = 90).Compute 1936 − 7 = 1929.Find x such that 1929 − x is a multiple of 90. 1929 mod 90 = 39.Therefore x = 39 gives 1929 − 39 = 1890 which is 90*21.Hence the least number to subtract is 39.

Verification / Alternative check:R = 1936 − 39 = 1897. Then R − 7 = 1890 which is divisible by 9, 10, and 15. Thus R leaves remainder 7 in each division.

Why Other Options Are Wrong:46, 53, and 44 correspond to incorrect remainders modulo 90; 36 misses the exact multiple by 3.

Common Pitfalls:Not subtracting the target remainder first; computing LCM improperly; or miscalculating remainders modulo 90.

Final Answer:39