$2^{122}+4^{62}+8^{42}+4^{64}+2^{130}$ is divisible by which one of the following integers?
Solution
✅ Correct Option: 4
Write every term as a power of 2: $4^{62}=2^{124}$, $8^{42}=2^{126}$ and $4^{64}=2^{128}$. So, $2^{122}+4^{62}+8^{42}+4^{64}+2^{130}$ $=2^{122}\left(1+2^2+2^4+2^6+2^8\right)$ $=2^{122}(1+4+16+64+256)$ $=2^{122}\times341$. Now, $341=11\times31$. Therefore, the entire expression is divisible by 11.
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