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

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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>. | + | 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== | ||

+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||

+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||

+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||

+ | {{JMPSC Notice}} |

## Latest revision as of 17:25, 11 July 2021

## Problem

What is the leftmost digit of the product

## Solution

We notice that In addition, we notice that

Since

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

## Solution 3

Remove factors of and to get . Recall by Pascal's triangle that , , so the leftmost digit is guaranteed to be . Now, multiplying by our scale factor the answer is ~Geometry285

## See also

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