Question

${10}^x$ when divided by 13 leaves a remainder = -1. If x is a natural number less than 100, then how many solutions are possible for x?

• A. 33
• B. 21
• C. 16
• D. 17

Answer 1

The correct option is D 17

The first power of 10 for which a division by 13 gives a remainder -1 is x=3. The next power of 10 will be ${10}^9$. Hence it falls in an AP with first term as 3 and last term as 99 with a common difference of 6.

Number of terms $=\frac{(99-3)}{6}+1=\frac{96}{6}+1=16+1=17$ terms.