# Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"

## Problem

What is the leftmost digit of the product $$\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }}?$$

## Solution

We notice that $$16000\cdots \times 25000\cdots = 16 \times 25 \times 10^{198} = 400 * 10^{198}$$ In addition, we notice that $$16200\cdots \times 25300\cdots = 162 \times 253 \times 10^{194} = 40986 \times 10^{194}$$

Since $$16000\cdots \times 25000\cdots < \underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }} < 16200\cdots \times 25300\cdots$$

We conclude that the leftmost digit must be $\boxed{4}$.

## Solution 2

By multiplying out $16 \cdot 25$, $161 \cdot 252$, and $1616 \cdot 2525$, we notice that the first $2$ digits don't change even when we continue to add more digits. With this observation, we can conclude that the first digit of the product is $\boxed{4}$.

~Mathdreams

## Solution 3

Remove factors of $16$ and $25$ to get $\left(\underbrace{101010101 \cdots}_{\text{50 0s and 50 1s}} \right)^2 \cdot 400$. Recall by Pascal's triangle that $11=121$, $101=10201$, so the leftmost digit is guaranteed to be $1$. Now, multiplying by our scale factor the answer is $\boxed{4}$ $\linebreak$ ~Geometry285