2^122+4^62+8^42+4^64+2^130 is divisible by which one of the following integers? Option 1: 3 Option 2: 5 Option 3: 7 Option 4: 11

Answer: 11. Solution: 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(1+2^2+2^4+2^6+2^8) =2^122(1+4+16+64+256) =2^122x341. Now, 341=11x31. Therefore, the entire expression is divisible by 11.

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

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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