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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
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<math>\linebreak</math>
 
<math>\linebreak</math>
 
~Geometry285
 
~Geometry285
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==See also==
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 16:25, 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


See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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