CSAT Solved Papers/ 2025/Q20

2025 CSAT — Q20

Quant Number theory 2.5 marks Easy

What is the unit digit in the multiplication of 1×3×5×7×9×…×9991 \times 3 \times 5 \times 7 \times 9 \times \ldots \times 999?

  1. A 1
  2. B 3
  3. C 5 Answer
  4. D 9

Worked rationale

The product runs over all odd numbers from 11 to 999999. Two facts pin the unit digit without any heavy multiplication:

  1. The product contains 55 (and 15,25,…15, 25, \dots), so it is divisible by 55 — its unit digit is 00 or 55.
  2. Every factor is odd, so the product is odd — its unit digit cannot be 00.

A number that is divisible by 55 and is odd must end in 55.

Answer: (c) 5.

Why the other options miss

  • A
    wrong formula: chases a 44-cycle of unit digits (as for a single power of one base) and lands on 11, ignoring the factor of 55 that fixes the answer.
  • B
    an arithmetic slip: multiplies the first few unit digits (1⋅3⋅5⋅7→51\cdot3\cdot5\cdot7 \to 5, then mis-tracks) and stops at a wrong residue.
  • D
    solved the wrong question: reads off the unit digit of the last factor (999999), not of the product.

Specialist insight

The decisive observation is a parity-and-divisibility sandwich: “contains a 55” forces the last digit into {0,5}\{0,5\}, and “all factors odd” eliminates 00. That two-line argument sidesteps the unit-digit cycle entirely. The general rule worth internalising: any product that includes a multiple of 55 and stays odd ends in 55; if it also includes an even number, it ends in 00. Recognising which of these regimes you are in is a sub-five-second decision and a frequent CSAT freebie — do not start cycling powers of 33 and 77.

The trap, in one line

The product is odd and divisible by 55 — so it must end in 55; don't run a unit-digit cycle or read off the last factor (999→9999 \to 9).

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