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 $\boxed{4}$ .

~Bradygho

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 $4$ .

~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

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2021 JMPSC Accuracy Problems/Problem 14

Contents

Problem

Solution

Solution 2

Solution 3