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Difference between revisions of "2000 PMWC Problems/Problem T1"

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==Solution==
 
==Solution==
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If there is always one candy left over when the candies are evenly distributed to <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, or <math>9</math> children, then there must also be exactly one candy left over when the candies are distributed to <math>LCM(5, 6, 7, 8, 9)</math> children. Since <math>LCM(5, 6, 7, 8, 9) = 2^3 * 3^2 * 5^1 * 7^1 = 2520</math>, the total number of candies can be expressed as <math>2520n + 1</math>, where <math>n</math> is any nonnegative integer. However, we know that there are between <math>4000</math> and <math>6000</math> candies, so the only viable value for <math>n</math> is <math>2</math>, meaning the total number of candies is <math>2520(2) + 1 = 5040 + 1 = 5041</math>.
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The problem then asks us to find the largest number of candies below <math>4000</math> such that the candies are evenly distributed and no candies are left over. This is equivalent to asking what the largest factor of <math>5401</math> is that is less than <math>4000</math>. Since <math>5401 = 71^2</math>, the only factors of <math>5401</math> are <math>1</math>, <math>71</math>, and <math>5401</math>. Thus the answer is <math>\boxed{71}</math>.
  
 
==See Also==
 
==See Also==
 
Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems
 
Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems

Revision as of 11:13, 23 December 2019

Problem

A box contains $4000$ to $6000$ candies. When the candies are evenly distributed to $5$, $6$, $7$, $8$, or $9$ children, there is always one candy left. If the candies are packaged into small bags each having the same number of candies, what is the largest number of candies below $4000$ in each bag so that no candies are left?

Solution

If there is always one candy left over when the candies are evenly distributed to $5$, $6$, $7$, $8$, or $9$ children, then there must also be exactly one candy left over when the candies are distributed to $LCM(5, 6, 7, 8, 9)$ children. Since $LCM(5, 6, 7, 8, 9) = 2^3 * 3^2 * 5^1 * 7^1 = 2520$, the total number of candies can be expressed as $2520n + 1$, where $n$ is any nonnegative integer. However, we know that there are between $4000$ and $6000$ candies, so the only viable value for $n$ is $2$, meaning the total number of candies is $2520(2) + 1 = 5040 + 1 = 5041$.

The problem then asks us to find the largest number of candies below $4000$ such that the candies are evenly distributed and no candies are left over. This is equivalent to asking what the largest factor of $5401$ is that is less than $4000$. Since $5401 = 71^2$, the only factors of $5401$ are $1$, $71$, and $5401$. Thus the answer is $\boxed{71}$.

See Also

Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems

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