CSAT Solved Papers/ 2022/Q18

2022 CSAT — Q18

Quant Counting & combinatorics 2.5 marks Medium

How many 33-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 55?

  1. A 8
  2. B 12 Answer
  3. C 16
  4. D 24

Worked rationale

Divisible by 55 means the units digit is 00 or 55. But every digit must be odd, so the units digit must be 55 (it cannot be 00).

With the units digit fixed at 55, the hundreds and tens digits are odd, distinct, and different from 55 — drawn from {1,3,7,9}\{1, 3, 7, 9\} (four digits):

hundreds: 4 choices,tens: 3 choices⇒4×3=12.\text{hundreds: } 4 \text{ choices},\quad \text{tens: } 3 \text{ choices} \Rightarrow 4 \times 3 = 12.

Answer: (b) 12.

Why the other options miss

  • A
    over-restricted the digits: clamps the remaining digits too tightly (e.g. treats one position as fixed), undercounting.
  • C
    forgot the no-repeat rule: counts 4×44 \times 4, letting the tens digit repeat the hundreds digit.
  • D
    let a barred units digit in: allows units ∈{0,5}\in \{0,5\}, or reuses 55, double-counting the placement.

Specialist insight

Two constraints collide and the divisor wins: “divisible by 55” wants units ∈{0,5}\in\{0,5\}, “all odd” forbids 00, so units =5= 5 is forced — that is the whole insight. Then it is a clean 4×34\times 3 over the odd digits {1,3,7,9}\{1,3,7,9\}. Lock the most restrictive position first; the rest is permutation arithmetic.

The trap, in one line

Odd + divisible-by-55 forces units =5=5; then 4×3=124\times 3 = 12 over {1,3,7,9}\{1,3,7,9\} ⇒\Rightarrow (b).

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