The value at the tenth place of a three-digit number is twice the value at the unit's place. The value at the hundredth place is 75% of the value at the unit's place. How many such numbers are possible?

Correct — B, 1. Call the units digit u. The tens digit is 2u and the hundreds digit is 0.75u, and every digit has to be a whole number from 0 to 9, with the hundreds digit at least 1 for the number to have three digits. Two constraints bite. For 0.75u to be a whole number, u must be a multiple of 4, so u is 0, 4 or 8. For 2u to be a single digit, u can be at most 4. That leaves u = 0 and u = 4. With u = 0 every digit is 0, which is not a three-digit number. With u = 4 the tens digit is 8 and the hundreds digit is 3, giving 384. Exactly one number satisfies the description.

• (a)0 — 384 satisfies every clause: its tens digit 8 is twice its units digit 4, and its hundreds digit 3 is three quarters of 4. So at least one such number exists.
• (c)2 — Requires a second solution, and the two constraints leave no room for one. u = 8 would give a tens digit of 16, and u = 0 gives 000, which is not a three-digit number.
• (d)3 — Three solutions would need three units digits that are multiples of 4 and no greater than 4, and only 0 and 4 qualify, of which 0 fails.

Digit problems are searches over a very small set, and the productive method is to express every digit in terms of one of them and then apply the two standing constraints: each digit is a whole number between 0 and 9, and the leading digit cannot be 0. Percentages inside such a question are really divisibility conditions in disguise — 75 per cent of a digit is whole only when that digit is a multiple of 4.

Anchor on the units digit, since both other digits are defined from it, and the search collapses to ten cases before the constraints cut it to one. The phrase 'the value at the hundredth place' means the digit standing in the hundreds column, not that digit multiplied by a hundred — read as place value the condition would demand 100h = 0.75u, which no digit satisfies, and the item would have no answer at all. The wording is loose but the intended reading is the only one that produces a number.

• With units digit u, the tens digit is 2u and the hundreds digit is 0.75u.
• 0.75u is a whole number only when u is a multiple of 4, so u is 0, 4 or 8.
• 2u must stay below 10, which caps u at 4.
• u = 0 gives 000, not a three-digit number, so the only solution is u = 4 and the number is 384.
• The hundreds digit of a three-digit number cannot be zero, which is what rules out the u = 0 case.

• Allowing a units digit of 8, which pushes the tens digit past 9.
• Counting 000 as a three-digit number.
• Reading 'value at the hundredth place' as the digit times one hundred, which makes the item unsolvable.

A digit-counting item where two constraints pull in opposite directions and leave a single number.

• practice — not a real PYQ

  In a three-digit number, the tens digit is three times the units digit and the hundreds digit equals the units digit. How many such numbers are possible?

  • (a)1
  • (b)2
  • (c)3
  • (d)4

  Answer(c) 3 — the units digit can be 1, 2 or 3, giving 131, 262 and 393; 0 fails because the hundreds digit would vanish.

• practice — not a real PYQ

  For 75 per cent of a single digit to be a whole number, that digit must be

  • (a)even
  • (b)a multiple of 3
  • (c)a multiple of 4
  • (d)a multiple of 5

  Answer(c) a multiple of 4 — three quarters of the digit is whole only when the digit is divisible by 4.