2023 CSAT — Q7

Quant Number theory 2.5 marks Medium

What is the remainder when $85 \times 87 \times 89 \times 91 \times 95 \times 96$ is divided by $100$ ?

A 0 Answer
B 1
C 2
D 4

Worked rationale

The remainder mod $100$ is just the last two digits, so reduce the running product mod $100$ at every step — never multiply the whole thing out.

85 \times 87 = 7395 \equiv 95.

95 \times 89 = 8455 \equiv 55.

55 \times 91 = 5005 \equiv 05.

05 \times 95 = 475 \equiv 75.

75 \times 96 = 7200 \equiv 00.

The remainder is $0$ .

Shortcut (the real time-saver): a multiple of $4$ times a multiple of $25$ is a multiple of $100$ . Here $96 = 4 \times 24$ supplies the factor $4$ , and $85 \times 95$ supplies two factors of $5$ (i.e. $25$ ), so the product is divisible by $4 \times 25 = 100$ . Remainder $0$ .

Answer: (a) 0.

Why the other options miss

B

an arithmetic slip: a single mis-multiplication in the running product, landing on a wrong two-digit tail.
C

wrong formula: confuses the last-digit pattern (a mod- $10$ idea) with the mod- $100$ question, reading off a units digit instead.
D

solved the wrong question: notices the factor $4$ from $96$ but misses that $85$ and $95$ together carry the matching $25$ , so stops at “divisible by $4$ ” with a wrong tail.

Specialist insight

Two equally fast routes: (i) reduce mod $100$ after every multiplication so no number ever exceeds four digits; or (ii) spot the $4 \times 25 = 100$ factorisation — $96$ gives a $4$ , the two numbers ending in $5$ give a $25$ — proving divisibility by $100$ in one line. The trap is reaching for the units-digit cycle, which answers a different (mod $10$ ) question.

The trap, in one line

Mod $100$ is the last two digits: reduce after each step (or spot $4\mid 96$ and $25\mid 85\cdot 95$ ), giving remainder $0$ .

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