IPMAT Indore Indices — PYPs

The Setup: We need to identify the greatest number among the expressions $2^{300}$, $3^{200}$, $4^{100}$, and $2^{100}+3^{100}$.

Step 1: Normalize the exponents to a common power for direct comparison.
We can rewrite each single-term expression by factoring out a power of $100$ in the exponent:
$2^{300} = (2^3)^{100} = 8^{100}$
$3^{200} = (3^2)^{100} = 9^{100}$
$4^{100} = 4^{100}$

Step 2: Evaluate the additive term against the largest single base.
Compare the addition term $2^{100} + 3^{100}$ to the largest normalized term $9^{100}$:
Since $2^{100} < 3^{100}$, their sum satisfies $2^{100} + 3^{100} < 3^{100} + 3^{100} = 2 \times 3^{100}$.
Clearly, $2 \times 3^{100}$ is vastly smaller than $9^{100}$ (which is $3^{200}$).

Step 3: Conclude the greatest term.
Among the expressions $8^{100}$, $9^{100}$, and $4^{100}$, the term with the largest base is $9^{100}$, which corresponds back to $3^{200}$.

Final Answer: $3^{200}$

The Setup: A power-scaling comparison. Computing these directly is hopeless, so rewrite every term with the same exponent - then only the bases need comparing.
Step 1: Normalise to the exponent 100. Each exponent is a multiple of 100, so pull that out:

$$2^{300}=\left(2^3\right)^{100}=8^{100}, \qquad 3^{200}=\left(3^2\right)^{100}=9^{100}, \qquad 4^{100}=4^{100}$$

Step 2: Compare the three pure powers. With a common positive exponent, $t\mapsto t^{100}$ is increasing on positive bases, so the ordering of the bases carries straight over:

$$9^{100}>8^{100}>4^{100} \implies 3^{200}>2^{300}>4^{100}$$

Step 3: Dispose of the sum. The remaining candidate $2^{100}+3^{100}$ is not a pure power, so bound it. Since $2^{100}<3^{100}$:

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

Compare that generous overestimate against the leader:

$$\frac{3^{200}}{3^{100}}=3^{100} \quad\text{which is vastly larger than }2$$

so $2\cdot 3^{100}<3^{100}\cdot 3^{100}=3^{200}$. The sum loses even after being inflated, and the gap is not close - $3^{200}$ exceeds it by a factor of roughly $\frac{3^{100}}{2}$.
Step 4: Conclude. The full ordering is $3^{200}>2^{300}>4^{100}>2^{100}+3^{100}$, and the greatest is $3^{200}$.
The lesson: adding two large powers is worth far less than multiplying the exponent. $2^{100}+3^{100}$ is barely bigger than its larger half, while squaring $3^{100}$ multiplies it by itself.

Final Answer: $3^{200}$