If 103 A + A is a perfect square, then the minimum number of digits in A is

Concept: A whole number is a perfect square only when every prime in its factorization occurs to an even power. So if a fixed multiplier P is combined with an unknown A, any prime that appears to an odd power in P must also appear to an odd power in A — otherwise that prime's total exponent in P × A stays odd, and the product cannot be a perfect square.

Application: Apply this to 103A + A.

1. 103A + A = 1000A + A = 1001A — so the perfect-square condition applies to 1001A.

2. Factor 1001 = 7 × 11 × 13, three distinct primes, each to power 1 (odd).

3. For 1001A to be a perfect square, A must contain a factor of 7, a factor of 11, and a factor of 13 — this pairs up all three odd exponents. Any further prime factor in A must itself appear to an even power.

4. The smallest A meeting this requirement (no extra square factor) is A = 7 × 11 × 13 = 1001.

5. 1001 has 4 digits.

Cross-check: 1001 × 1001 = 10012 = 1002001, a perfect square, confirming A = 1001 works. No smaller A can work: a 2-digit or 3-digit number cannot be divisible by 7, 11, and 13 all at once, since their product already reaches 1001, a 4-digit number — and dropping any one of the three primes leaves an odd exponent in 1001A, so the product would fail to be a perfect square.

Result: The minimum number of digits in A is 4.