IPMAT Indore 2024 · Quantitative Ability · Question 33

Step 1 — put everything over a common exponent of 100. This is the whole technique.

{2}^{300}={\left({2}^{3}\right)}^{100}={8}^{100}

{3}^{200}={\left({3}^{2}\right)}^{100}={9}^{100}

{4}^{100}={4}^{100}

Step 2 — compare the first three. With the same exponent, the larger base wins:

{4}^{100}<{8}^{100}<{9}^{100}

Step 3 — handle the sum. Bound it above:

{2}^{100}+{3}^{100}<{3}^{100}+{3}^{100}=2\cdot{3}^{100}

and since {3}^{200}={3}^{100}\cdot{3}^{100} with {3}^{100} astronomically larger than 2,

{2}^{100}+{3}^{100}<2\cdot{3}^{100}⋘{3}^{100}\cdot{3}^{100}={3}^{200}

So the sum is nowhere near the largest.

Answer: Option 1 — {3}^{200}

Shortcut. Force a common exponent, then read off the base. \gcd \left( 300,200,100 \right)=100, so rewrite as {8}^{100},{9}^{100},{4}^{100} — and {9}^{100}={3}^{200} wins immediately.

Dismiss the sum on sight: a sum of two 100th powers can never compete with a product of two of them. More generally, adding exponentials is negligible beside multiplying them — the sum is bounded by twice the larger term, while {3}^{200} is {3}^{100} times it.