Two-digit number with digit sum 10 — Reversing the digits decreases the number by 36. Decide which statements are correct: I) The number is divisible by a composite number. II) The number is a multiple of a prime number.

Introduction / Context:
This is the same structural setup as a classic digit problem: a + b = 10 for a two-digit number 10a + b, and reversing to 10b + a reduces the value by 36. From the digits you can identify the number and then assess statements about divisibility by composite and prime numbers.

Given Data / Assumptions:

• Digits a (tens) and b (units) satisfy a + b = 10.
• (10a + b) - (10b + a) = 36 → 9(a - b) = 36 → a - b = 4.
• We evaluate two statements after finding the number.

Concept / Approach:
Solve the linear system to find (a, b). Then check whether the number is divisible by some composite and whether it is a multiple of a prime. Every integer greater than 1 is a multiple of at least one prime (by unique factorization), but primes are not divisible by any composite > 1.

Step-by-Step Solution:
Solve a + b = 10 and a - b = 4 → adding gives 2a = 14 → a = 7, so b = 3.Number = 73; reversed 37; indeed 73 - 37 = 36.73 is prime → not divisible by any composite number > 1.However, 73 is trivially a multiple of a prime (itself): 73 = 73 * 1.

Verification / Alternative check:
Confirm digit sum condition and difference condition; a primality check for 73 (no small divisors) confirms the conclusions about statements I and II.

Why Other Options Are Wrong:

• Only I / Both / Neither: Statement I is false; Statement II is true. Hence “Only II”.

Common Pitfalls:
Misinterpreting “multiple of a prime” as implying non-prime; overlooking that primes are multiples of themselves.

Final Answer:
Only II