2021 JMPSC Accuracy Problems/Problem 14

Revision as of 11:38, 11 July 2021 by Mathdreams (talk | contribs) (Solution 2)

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

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