CSAT Solved Papers/ 2023/Q27

2023 CSAT — Q27

Quant Number theory 2.5 marks Hard

A number $N$ is formed by writing $9$ for $99$ times. What is the remainder if $N$ is divided by $13$ ?

A 11 Answer
B 9
C 7
D 1

Worked rationale

A string of $99$ nines is $N = \underbrace{99\cdots9}_{99} = 10^{99} - 1$ . So work out $10^{99} \bmod 13$ , then subtract $1$ .

Find the order of $10$ mod $13$ (the powers cycle):

10^1 \equiv 10,\quad 10^2 \equiv 9,\quad 10^3 \equiv 12,\quad 10^4 \equiv 3,\quad 10^5 \equiv 4,\quad 10^6 \equiv 1 \pmod{13}.

The order is $6$ . Reduce the exponent: $99 = 6\times 16 + 3$ , so $99 \equiv 3 \pmod 6$ , giving

10^{99} \equiv 10^{3} \equiv 12 \pmod{13}.

Therefore

N = 10^{99} - 1 \equiv 12 - 1 = 11 \pmod{13}.

Answer: (a) 11.

Why the other options miss

B

an arithmetic slip: reduces the exponent to $\equiv 2 \pmod 6$ (a slip in $99 \bmod 6$ ), giving $10^2 - 1 \equiv 8$ , or mis-reads the cycle table to land on $9$ .
C

wrong formula: forgets the " $-1$ " (computes a power residue, not $10^{99}-1$ ) or uses a wrong order for $10$ .
D

a remainder-rule slip: assumes $10^{99} \equiv 1$ by mis-applying Fermat ( $10^{12}\equiv 1$ , but $99 \not\equiv 0 \pmod{6}$ ), wrongly concluding $N \equiv 0$ or echoing a $1$ .

Specialist insight

The unlock is rewriting a repunit-of-nines as $10^{k} - 1$ — then it is a clean order-reduction. The order of $10$ mod $13$ is $6$ (not $12$ , even though Fermat guarantees $10^{12}\equiv 1$ ), so reduce $99$ mod $6$ , not mod $12$ . The mandatory final step is subtracting $1$ ; dropping it is the most common −1/3 leak. Build the small cycle table once and read off $10^3 \equiv 12$ .

The trap, in one line

$N = 10^{99}-1$ ; $\mathrm{ord}_{13}(10)=6$ , $99\equiv 3 \Rightarrow 10^{99}\equiv 12$ , so $N\equiv 12-1 = 11 \Rightarrow$ (a).

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