If x and y are two-digit prime numbers such that y is obtained from x by interchanging its digits and x − y = 36, then what is the value of xy ?

Correct — B, 2701. Write x as 10a + b, so that y, the number with its digits interchanged, is 10b + a. The difference x − y is then 9(a − b), and setting that equal to 36 gives a − b = 4 immediately — the difference of a two-digit number and its reverse is always a multiple of 9, which is the fastest way into this item. Now cut the search down. Both x and y are primes above 10, so both are odd, which means both digits must be odd. Odd digits differing by 4 leave only three candidates: 51 and 15, 73 and 37, 95 and 59. Of these, 51 is 3 × 17 and 95 is 5 × 19, so only 73 and 37 are both prime. Hence x = 73, y = 37, and the product is 73 × 37. Compute it as 73 × 40 − 73 × 3 = 2920 − 219 = 2701.

• (a)1611 — Its factorisation is 3 × 3 × 179, which contains a three-digit prime and no pair of two-digit ones. A quick check also disposes of it: 1611 is divisible by 3 because its digits sum to 9, and no product of two primes above 10 can be.
• (c)4031 — This factorises as 29 × 139. One of the factors runs to three digits, so the number cannot be the product of two two-digit primes at all, let alone a digit-reversed pair — 29 reversed is 92, which is even.
• (d)5603 — Its factorisation is 13 × 431, which again includes a three-digit prime. Even the tempting near-miss of 95 × 59 gives 5605 rather than 5603, so no reading of the digit pairs reaches this figure.

For any two-digit number the value is 10a + b and its reverse is 10b + a, so the difference is 9(a − b) and the sum is 11(a + b). Those two identities settle a very large family of digit-reversal questions on sight — a stated difference must be a multiple of 9 and immediately fixes the gap between the digits, while a stated sum must be a multiple of 11 and fixes their total. The three-digit version works the same way, with the difference always a multiple of 99.

The parity argument is what makes the search short enough to do in the head. Every prime above 2 is odd, so the units digit of a two-digit prime must be odd, and since y is x with the digits swapped, both digits of x have to be odd. Combined with a difference of 4, the field narrows to three pairs before any primality test is attempted. A pair of primes that remain prime when their digits are reversed, such as 37 and 73, are called emirps, and the small ones — 13 and 31, 17 and 71, 37 and 73, 79 and 97 — are worth carrying, since examiners return to them.

• A two-digit number minus its reverse always equals 9 times the difference of its digits.
• A two-digit number plus its reverse always equals 11 times the sum of its digits.
• Every prime above 2 is odd, so both digits of x must be odd if x and its reverse are both prime.
• The only two-digit prime pair whose reversal differs by 36 is 73 and 37.
• 73 × 37 = 2701; primes that stay prime when reversed, such as 13 and 31 or 79 and 97, are known as emirps.

• Testing every two-digit number instead of using the multiple-of-9 identity to fix the digit gap first.
• Forgetting that both digits must be odd, which triples the amount of primality testing required.
• Slipping on 73 × 37; splitting it as 73 × 40 minus 73 × 3 is safer than a long multiplication.

Asked as a number-system item where one place-value identity plus a parity observation replaces what would otherwise be an exhaustive search.

• practice — not a real PYQ

  The difference between a two-digit number and the number obtained by reversing its digits is 45. The difference between the two digits is

  • (a)3
  • (b)4
  • (c)5
  • (d)9

  Answer(c) 5 — the difference is always 9 times the gap between the digits, and 45 divided by 9 is 5.

• practice — not a real PYQ

  The sum of a two-digit number and the number formed by reversing its digits is always divisible by

  • (a)9
  • (b)10
  • (c)11
  • (d)13

  Answer(c) 11 — the sum equals 11 times the sum of the digits, just as the difference equals 9 times their difference.