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UPSC Prelims 2024CSATBasic NumeracyNumber System - Divisibility

Q54. 32^5 + 2^27 is divisible by

A. 3
B. 7
C. 10✓ Correct
D. 11

Detailed Solution

✓ Correct Answer: Option C

The given expression is 32^5 + 2^27. First, rewrite 32 as a power of 2: 32 = 2^5. So, 32^5 = (2^5)^5 = 2^(5*5) = 2^25. Now, the expression becomes 2^25 + 2^27. Factor out the common term, which is 2^25: 2^25 + 2^27 = 2^25 + (2^25 * 2^2) = 2^25 (1 + 2^2) = 2^25 (1 + 4) = 2^25 * 5.

To check for divisibility by 10, we need a factor of 10 (which is 2 * 5). We can rewrite 2^25 as 2^24 * 2: = (2^24 * 2) * 5 = 2^24 * (2 * 5) = 2^24 * 10. Since the expression can be written as 2^24 * 10, it is clearly divisible by 10. This question tests basic exponent rules and divisibility properties.

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