Find the unique number between 900 and 1000 that leaves a remainder of 23 when divided by both 38 and 57.

Introduction / Context:This problem asks for a number within a range satisfying two simultaneous remainder conditions. It is a straightforward application of the Chinese Remainder Theorem when moduli are not coprime but share a common factor appropriately, or by using the least common multiple (LCM) if the same remainder is required.

Given Data / Assumptions:

• N ≡ 23 (mod 38)
• N ≡ 23 (mod 57)
• 900 ≤ N ≤ 1000

Concept / Approach:If the desired remainder is the same for both moduli, then N − 23 must be a multiple of LCM(38, 57). Compute that LCM and then list candidates within the range by adding multiples of the LCM to 23.

Step-by-Step Solution:38 = 2*19, 57 = 3*19 ⇒ LCM(38, 57) = 2*3*19 = 114.General solution: N = 23 + 114t for integer t.Find t with 900 ≤ 23 + 114t ≤ 1000. Try t = 8 ⇒ N = 23 + 912 = 935 (in range). t = 9 ⇒ N = 1049 (exceeds 1000).Therefore the required number is 935.

Verification / Alternative check:935 ÷ 38 leaves 23 (since 38*24 = 912; 935 − 912 = 23). 935 ÷ 57 leaves 23 (since 57*16 = 912; 935 − 912 = 23). Both conditions hold.

Why Other Options Are Wrong:945, 925, 955, and 965 are not of the form 23 + 114t and fail at least one modulus.

Common Pitfalls:Using gcd instead of LCM; not ensuring the same remainder approach; or missing the range restriction when enumerating solutions.

Final Answer:935