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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 14"

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==Solution 2==
 
==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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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>\boxed{4}</math>.
  
 
~Mathdreams
 
~Mathdreams

Revision as of 11:34, 11 July 2021

Problem

What is the leftmost digit of the product \[\underbrace{161616 \cdots 16}_{100 \text{ digits }} \times \underbrace{252525 \cdots 25}_{100 \text{ digits }}?\]

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 $\boxed{4}$.

~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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