CAT 2024 Slot 3 · Quant · Question 6

Remainders of the powers of 10 modulo 13 repeat in a cycle, so the first task is to find its length.

Working through, {10}^{1}\equiv 10, {10}^{2}\equiv 100\equiv 9, {10}^{3}\equiv 90\equiv 12, {10}^{4}\equiv 120\equiv 3, {10}^{5}\equiv 30\equiv 4 and {10}^{6}\equiv 40\equiv 1 (\mod 13).

The remainder returns to 1 at the sixth power, so the cycle length is 6.

Now reduce the exponent modulo 6: 68=6\times 11+2, so 68\equiv 2 (\mod 6).

Therefore {10}^{68}={\left({10}^{6}\right)}^{11}\times{10}^{2}\equiv{1}^{11}\times{10}^{2}(\mod 13).

That leaves {10}^{2}=100, and 100=7\times 13+9.

The remainder is 9, the second option.

Shortcut. {10}^{6}-1=999999 is divisible by 13, since 999999=999\times 1001 and 1001=7\times 11\times 13. That single fact gives the cycle length of 6 without listing any powers, leaving only {10}^{2}\mod 13 to compute.