JIPMAT 2025 (QA) - 2^122 + 4^62 + 8^42 + 4^64 + 2^130 is divisible by which one of the following integers? | PYQs + Solutions

JIPMAT 2025

Number System

Divisibility Rules

Hard

$2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}$ is divisible by which one of the following integers?

✅ Correct Option: 4

Let's simplify :

2^{122} + 4^{62} + 8^{42} + 4^{64} + 2^{130}

= 2^{122} + 2^{2 \times{62}} + 2^{3 \times{42}} + 2^{2 \times {64}} + 2^{130}

= 2^{122} + 2^{124} + 2^{126} + 2^{128} + 2^{130}

= 2^{122} + 2^{122} \times 2^2 + 2^{122} \times 2^4 + 2^{122} \times 2^6 +2^{122} \times 2^8

= 2^{122} (1 + 4 + 16 + 64 + 256)

= 2^{122} (341)

341

is divisible by

11

, hence, entire expression will be divisible by 11.

Hence, the correct answer is Option 4.

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