The sum of two numbers is 168. If their HCF is 14, how many such pairs of numbers are possible?

Correct — B, 2. If the highest common factor is 14, both numbers are multiples of 14, so write them as 14a and 14b where a and b share no common factor greater than 1. Their sum is 14(a + b) = 168, so a + b = 12. Now list the pairs of positive integers adding to 12 and keep only the coprime ones: 1 and 11 are coprime; 2 and 10 share 2; 3 and 9 share 3; 4 and 8 share 4; 5 and 7 are coprime; 6 and 6 share 6. Two pairs survive. They give the numbers 14 and 154, whose highest common factor is 14, and 70 and 98, whose highest common factor is also 14. So exactly two such pairs exist.

• (a)1 — Undercounts by missing one of the two coprime splits of 12. Both 1 + 11 and 5 + 7 qualify, and each yields a genuine pair with highest common factor 14.
• (c)3 — Comes from counting a pair whose parts are not coprime — most often 3 and 9, which gives 42 and 126. Those two do add to 168, but their highest common factor is 42, not 14.
• (d)0 — The construction plainly works: 70 + 98 = 168 and their highest common factor is 14. Since 14 divides 168, at least one such pair must exist.

Two numbers with highest common factor h are always h times a coprime pair. So a question that fixes the sum and the highest common factor becomes a question about coprime pairs summing to the reduced total. That reduction is the standard method, and it converts an open search into a short list.

Nearly every wrong answer here comes from skipping the coprimality test. Dividing 168 by 14 gives 12, and there are six unordered pairs adding to 12, but four of them share a factor and would push the highest common factor above 14. Note also that any common factor of a and b divides a + b = 12, so only the divisors of 12 need checking — which is why the surviving pairs are the ones containing no shared divisor of 12. Sum and highest common factor questions also carry a hidden gate: h must divide the sum, and 14 does divide 168, giving 12.

• If the highest common factor of two numbers is h, they can be written as ha and hb with a and b coprime.
• Here 14(a + b) = 168, so a + b = 12.
• The coprime pairs summing to 12 are (1, 11) and (5, 7); (2,10), (3,9), (4,8) and (6,6) all share a factor.
• The two number pairs are 14 and 154, and 70 and 98.
• For such a question to have any solution, the highest common factor must divide the given sum.

• Counting every split of 12 instead of only the coprime ones.
• Counting (a, b) and (b, a) as different pairs.
• Forgetting to check that the given highest common factor divides the given sum.

A counting item on number theory, where the whole method is dividing out the common factor and testing coprimality.

No directly related past PYQ was found.

• practice — not a real PYQ

  The sum of two numbers is 216 and their highest common factor is 27. How many such pairs are possible?

  • (a)1
  • (b)2
  • (c)3
  • (d)4

  Answer(b) 2 — dividing by 27 gives a + b = 8, and the coprime pairs are (1, 7) and (3, 5).

• practice — not a real PYQ

  If the highest common factor of two numbers is 12 and their product is 1728, their lowest common multiple is

  • (a)12
  • (b)72
  • (c)144
  • (d)288

  Answer(c) 144 — the product of two numbers equals the product of their HCF and LCM, so the LCM is 1728 divided by 12.