Question:medium

'abcd' is a four-digit number. It has only two factors excluding 1 and itself. In addition, the first two digits form a perfect square and the next two digits form a number which is one more than a perfect square. Which of the following could be the number?

Show Hint

The number must have exactly 4 factors in total (so it is \(p \times q\) or \(p^3\)); check the digit-square condition first to shortlist, then factorise.
Updated On: Jul 20, 2026
  • 1626
  • 1665
  • 2565
  • 2582
  • 3682
Show Solution

The Correct Option is D

Solution and Explanation

A cleaner route is to factorise all five options first, since only semiprimes (product of two distinct primes) or perfect cubes of a prime can have exactly two factors besides 1 and itself.

$1626 = 2 \times 3 \times 271$ (three distinct primes, 8 factors) - rejected.
$1665 = 3^2 \times 5 \times 37$ (12 factors) - rejected.
$2565 = 3^3 \times 5 \times 19$ (16 factors) - rejected.
$3682 = 2 \times 7 \times 263$ (8 factors) - rejected.
$2582 = 2 \times 1291$. Testing 1291 for primality by trial division up to $35$: not divisible by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, or 31, so it is prime. So $2582$ is a product of exactly two distinct primes, giving exactly 4 total factors - exactly two besides 1 and itself.

Only this option survives the factor test. Now checking the digit rule as a confirmation: the first two digits give $25 = 5^2$, a perfect square, and the last two digits give $82 = 9^2+1$, one more than a perfect square. Both conditions hold together only for this number.

\[\boxed{abcd = 2582}\]
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