Q1:jipmat 2025QA › Divisibility RulesMediumQA · MCQ

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

• A3
• B5
• C7
• D11

Pick an option to attempt

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The Setup: This expression looks like a final boss battle, but it's really just base-2 math in disguise. We need to convert all terms to base $2$ and factor out the smallest common power. Step 1: Convert bases $4$ and $8$ to base $2$. $4^{62}=(2^2)^{62}=2^{124}$ $8^{42}=(2^3)^{42}=2^{126}$ $4^{64}=(2^2)^{64}=2^{128}$ Step 2: Rewrite the entire expression. $$E=2^{122}+2^{124}+2^{126}+2^{128}+2^{130}$$ Step 3: Factor out the lowest power, which is $2^{122}$. $$E=2^{122}(1+2^2+2^4+2^6+2^8)$$ $$E=2^{122}(1+4+16+64+256)$$ $$E=2^{122}(341)$$ Step 4: Check the divisibility of $341$ among the given options. Let's test $11$ using the alternating sum rule: $3-4+1=0$, which is completely divisible by $11$. Since $341=11 \times 31$, the whole expression is a multiple of $11$. Final Answer: 11