For any three natural numbers a, b and c, if (bc + 1)/(abc + a + c) = 11/43, then the value of abc is

Correct — A, 30. The denominator hides a factorisation. abc + a + c is a(bc + 1) + c, and the numerator is bc + 1, so turning the equation upside down gives 43/11 = a + c/(bc + 1). Now c is at most bc, so c/(bc + 1) is a proper fraction, smaller than 1. Since 43/11 is 3 + 10/11, the whole-number part must be a = 3 and the fractional part must be c/(bc + 1) = 10/11. Cross-multiplying, 11c = 10bc + 10, that is c(11 − 10b) = 10. For b = 1 this reads c = 10; for any b of 2 or more the bracket is negative and c cannot be a natural number. So a = 3, b = 1, c = 10 and abc = 30. Checking forwards: bc + 1 = 11 and abc + a + c = 30 + 3 + 10 = 43.

• (b)45 — No triple of natural numbers gives abc = 45 here. The equation pins a, b and c individually, not just their product, and the only solution is 3, 1 and 10.
• (c)60 — A plausible-looking multiple of the numbers in sight, but substituting any factorisation of 60 into (bc + 1)/(abc + a + c) fails to produce 11/43. With abc = 60 and the equation's own constraint a = 3, one needs bc = 20 and c = 10, which contradicts bc + 1 = 11.
• (d)75 — Fails the same test. Once the continued-fraction step fixes a = 3 and bc + 1 = 11, the product abc is 3 x 10 = 30 and nothing else is available.

When an equation in several unknowns has only whole-number solutions available, the useful move is almost always to isolate a whole number and a proper fraction on the same side. Here inverting the fraction produces 43/11 = a + c/(bc + 1), and because a proper fraction cannot supply any part of the integer 3, the split is forced: the integer part is a and the fractional part is everything else. This is the first step of a continued-fraction expansion, and it is the standard tool for equations of this shape.

The instinct in the hall is to cross-multiply into 43(bc + 1) = 11(abc + a + c) and start hunting. That is one equation in three unknowns and it goes nowhere without the factorisation. Spotting that abc + a + c = a(bc + 1) + c is the whole item; everything after it is arithmetic. A second habit worth building is the reverse check — plug 3, 1 and 10 back in and confirm 11/43 before committing, which takes five seconds and rules out any slip in the algebra.

• abc + a + c factorises as a(bc + 1) + c, which shares the numerator bc + 1.
• Inverting the given equation gives 43/11 = a + c/(bc + 1).
• Because c is at most bc, the term c/(bc + 1) is always a proper fraction, so the integer part must be a.
• 43/11 = 3 + 10/11, so a = 3 and c(11 − 10b) = 10, which admits only b = 1 and c = 10.
• The solution is a = 3, b = 1, c = 10, so abc = 30 and bc + 1 = 11, abc + a + c = 43.

• Cross-multiplying at once into one equation with three unknowns.
• Guessing a factorisation of an offered product instead of solving for a, b and c.
• Missing that c/(bc + 1) is always less than 1, which is what forces the split.

A number-theory item disguised as an algebraic fraction; the marks are for spotting the factorisation, not for the arithmetic.

No directly related past PYQ was found.

• practice — not a real PYQ

  If a and b are natural numbers and (b + 1)/(ab + a + 1) = 2/9, what is the value of ab?

  • (a)2
  • (b)3
  • (c)4
  • (d)6

  Answer(c) 4 — the denominator is a(b + 1) + 1, so inverting gives 9/2 = a + 1/(b + 1) = 4 + 1/2, hence a = 4 and b = 1.

• practice — not a real PYQ

  The fraction 30/7 written as an integer plus a proper fraction is

  • (a)4 + 1/7
  • (b)4 + 2/7
  • (c)3 + 9/7
  • (d)5 − 5/7

  Answer(b) 4 + 2/7 — the integer part is 4 and the remainder over 7 is 2/7.