2021 CSAT — Q6

Quant Number theory 2.5 marks Medium

The number $3798125P369$ is divisible by $7$ . What is the value of the digit $P$ ?

A 1
B 6 Answer
C 7
D 9

Worked rationale

Work modulo $7$ using the place-value weights $10^k \bmod 7$ , which cycle with period $6$ :

10^0,10^1,\dots \equiv 1,3,2,6,4,5,\;1,3,2,6,4 \pmod 7.

Read the digits of $3798125P369$ from the units place up: $9,6,3,P,5,2,1,8,9,7,3$ with weights $1,3,2,6,4,5,1,3,2,6,4$ :

9(1)+6(3)+3(2)+P(6)+5(4)+2(5)+1(1)+8(3)+9(2)+7(6)+3(4)

=9+18+6+6P+20+10+1+24+18+42+12 = 160 + 6P.

Now $160 \equiv 6 \pmod 7$ , so we need

6 + 6P \equiv 0 \pmod 7 \;\Rightarrow\; 6(1+P)\equiv 0 \;\Rightarrow\; 1+P \equiv 0 \pmod 7 \;\Rightarrow\; P \equiv 6.

The only digit is $P=6$ .

Answer: (b) 6.

Why the other options miss

A

an arithmetic slip: a single weight or carry mis-added in the mod- $7$ sum, landing on the wrong residue.
C

solved the wrong question: $7$ is not even a valid digit-by-digit fix here; chosen because the divisor is $7$ (a lure), not because the arithmetic forces it.
D

an arithmetic slip: drops or mis-weights one of the higher place values, shifting the required $P$ .

Specialist insight

For divisibility by $7$ there is no one-line digit-sum rule, so the scoring method is the place-value cycle $1,3,2,6,4,5$ (period $6$ ). Tabulate weights against digits from the units up, sum once, and solve a single congruence in $P$ . The structure $6(1+P)\equiv 0$ is clean: since $\gcd(6,7)=1$ , it collapses to $P\equiv 6$ . Resist the urge to do trial long-division on each option under the clock — the congruence is faster and less error-prone.

The trap, in one line

Weighted sum $\equiv 160+6P\equiv 6+6P\pmod7$ ; force $\equiv 0 \Rightarrow P\equiv 6 \Rightarrow$ (b).

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