CSAT Solved Papers/ 2021/Q79

2021 CSAT — Q79

Quant Number theory 2.5 marks Hard

When a certain number is multiplied by $7$ , the product entirely comprises ones only $(1111\ldots)$ . What is the smallest such number?

A 15713
B 15723
C 15783
D 15873 Answer

Worked rationale

We need the smallest $N$ with $7N$ equal to a repunit $R_k = \underbrace{11\ldots1}_{k}$ . So $7 \mid R_k$ , and we want the smallest such repunit.

Repunits and $7$ : $R_1=1, R_2=11, R_3=111, R_4=1111, R_5=11111, R_6 = 111111$ . Checking divisibility by $7$ , the first repunit divisible by $7$ is $R_6 = 111111$ (the multiplicative order of $10$ mod $7$ is $6$ ). Then

N = \frac{111111}{7} = 15873.

Check: $15873 \times 7 = 111111$ ✓.

Answer: (d) 15873.

Why the other options miss

A

an arithmetic slip: $15713\times 7 = 109991$ , not a repunit — a slip in the division.
B

an arithmetic slip: $15723\times 7 = 110061$ , not all ones.
C

an arithmetic slip: $15783\times 7 = 110481$ , not all ones; a near-miss digit error.

Specialist insight

The product is a repunit, so the task is “smallest repunit divisible by $7$ ” — and because $\text{ord}_7(10) = 6$ , that is $R_6 = 111111$ , giving $N = 111111/7 = 15873$ . Don’t grind the four options through $\times 7$ blindly; recognise the repunit structure and divide $111111$ by $7$ directly (long division gives $15873$ cleanly). The distractors are all near $15873$ , betting on a division slip.

The trap, in one line

Smallest repunit divisible by $7$ is $R_6 = 111111$ ; $111111/7 = 15873 \Rightarrow$ (d).

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