View source for 2021 JMPSC Accuracy Problems/Problem 14

==Problem==
What is the leftmost digit of the product <cmath>\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }}?</cmath>

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

Since 
<cmath>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</cmath>

We conclude that the leftmost digit must be <math>\boxed{4}</math>.

~Bradygho

== Solution 2 ==

By multiplying out <math>16 \cdot 25</math>, <math>161 \cdot 252</math>, and <math>1616 \cdot 2525</math>, we notice that the first <math>2</math> 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 <math>4</math>.

~Mathdreams