CSAT Solved Papers/ 2023/Q17

2023 CSAT — Q17

Quant Number theory 2.5 marks Easy

For any choices of values of $X, Y$ and $Z$ , the $6$ -digit number of the form $XYZXYZ$ is divisible by:

A 7 and 11 only
B 11 and 13 only
C 7 and 13 only
D 7, 11 and 13 Answer

Worked rationale

Write the $6$ -digit repeat-block number as the $3$ -digit number $\overline{XYZ}$ times a structural factor:

\overline{XYZXYZ} = \overline{XYZ} \times 1001.

Indeed $\overline{XYZXYZ} = \overline{XYZ}\cdot 1000 + \overline{XYZ} = \overline{XYZ}\cdot 1001$ . Now factor

1001 = 7 \times 11 \times 13.

So for every choice of $X, Y, Z$ , the number $XYZXYZ$ is divisible by $7$ , $11$ and $13$ simultaneously.

Answer: (d) 7, 11 and 13.

Why the other options miss

A

mishandled a repeated factor: factors $1001$ incompletely (e.g. stops at $7 \times 143$ without splitting $143 = 11 \times 13$ ), dropping the $13$ .
B

mishandled a repeated factor: partial factorisation $1001 = 11 \times 91$ left as $11 \times 13 \times 7$ but mis-records, dropping the $7$ .
C

an arithmetic slip: writes $1001 = 7 \times 13 \times 11$ but omits the $11$ in the final read-off.

Specialist insight

$1001 = 7 \times 11 \times 13$ is one of the most-tested constants in CSAT/aptitude number theory — worth committing to memory alongside $1001 = 7\cdot 11\cdot 13$ and $111 = 3\cdot 37$ . The instant you see a $3$ -digit block repeated, the answer is “divisible by all of $7, 11, 13$ .” The trap options exist only to catch an incomplete factorisation of $1001$ ; there is no actual case-work.

The trap, in one line

$XYZXYZ = XYZ \times 1001$ and $1001 = 7\times 11\times 13$ , so it is divisible by all three $\Rightarrow$ (d).

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