Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"
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==Solution== | ==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>. | ||
Revision as of 22:52, 10 July 2021
Problem
What is the leftmost digit of the product ![\[\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }}?\]](http://latex.artofproblemsolving.com/7/c/c/7cca56c840db1d2457346e67fff42c3701c000fc.png)
Solution
We notice that
![\[16000\cdots \times 25000\cdots = 16 \times 25 \times 10^{198} = 400 * 10^{198}\]](http://latex.artofproblemsolving.com/0/f/5/0f5beb98db8fd0f0bc4753f731bf456b8b7530b9.png)
In addition, we notice that
![\[16200\cdots \times 25300\cdots = 162 \times 253 \times 10^{194} = 40986 \times 10^{194}\]](http://latex.artofproblemsolving.com/5/f/9/5f96a5d9f1c9fceafd162bec5e821f757dd3b7af.png)
Since
![\[16000\cdots \times 25000\cdots < \underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }} < 16200\cdots \times 25300\cdots\]](http://latex.artofproblemsolving.com/b/4/9/b49bd73af4a65ca8e0dc4ae360e80fd92d8aa181.png)
We conclude that the leftmost digit must be 
.
