Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"
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~Bradygho | ~Bradygho | ||
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| + | == Solution 2 == | ||
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| + | 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>. | ||
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| + | ~ Mathdreams | ||
Revision as of 10:38, 11 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 
.
~Bradygho
Solution 2
By multiplying out 
, 
, and 
, we notice that the first 
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 
.
~ Mathdreams
