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Difference between revisions of "1957 AHSME Problems/Problem 32"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(E) }30}</math>.
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To begin, we look for a maximum value that this common divisor could have by looking at the smallest terms. <math>1^5-1=0</math>, but this gives us no information because all positive integers divide zero. However, <math>2^5-2=30</math>, so our answer cannot be greater than <math>30</math> (because <math>30</math> divided by an integer larger than <math>30</math> is never an integer). This fact eliminates options (B) and (D).
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Now, we will factorize <math>n^5-n</math> to discover some properties:
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\begin{align*}
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n^5-n &= n(n^4-1) \\
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&= n(n^2-1)(n^2+1) \\
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&= n(n-1)(n+1)(n^2+1)
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\end{align*}
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One of <math>n</math> and <math>n-1</math> must be even, so <math>2</math> divides <math>n^5-n</math> for all <math>n</math>. Likewise, one of <math>n-1</math>, <math>n</math>, and <math>n+1</math> must be divisible by <math>3</math>, so <math>3</math> must divide <math>n^5-n</math> as well. In the case of <math>5</math>, there are, in total, three cases:
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 +
Case 1: <math>5</math> divides either <math>n-1</math>, <math>n</math>, or <math>n+1</math>. Then, <math>5</math> clearly must divide <math>n^5-n</math>.
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Case 2: <math>5</math> divides <math>n+2</math>. In this case, <math>n+2 \equiv 0</math> [[modulo|<math>\mod</math>]] <math>5</math>, so <math>n \equiv 3 \mod 5</math>. Then, <math>n^2+1 \equiv 10 \equiv 0 \mod 5</math>, so <math>n^2-1</math> (and, consequently, <math>n^5-n</math>) is divisible by <math>5</math>.
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Case 3: <math>5</math> divides <math>n+3</math>. In this case, <math>n+3 \equiv 0 \mod 5</math>, so <math>n \equiv 2 \mod 5</math>. Thus, <math>n^2+1 \equiv 5 \equiv 0 \mod 5</math>, so <math>n^2-1</math> (and, consequently, <math>n^5-n</math>) is divisible by <math>5</math>.
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Thus, 5 divides <math>n^5-n</math> for all <math>n</math>.
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Because <math>2</math>, <math>3</math>, and <math>5</math> all divide <math>n^5-n</math> and are [[relatively prime]], <math>2 \cdot 3 \cdot 5 = 30</math> also divides <math>n^5-n</math>. Thus, because <math>30</math> is the largest of our remaining choices, we choose answer <math>\boxed{\textbf{(E) }30}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:57, 25 July 2024

Problem

The largest of the following integers which divides each of the numbers of the sequence $1^5 - 1,\, 2^5 - 2,\, 3^5 - 3,\, \cdots, n^5 - n, \cdots$ is:

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 120\qquad \textbf{(E)}\ 30$

Solution

To begin, we look for a maximum value that this common divisor could have by looking at the smallest terms. $1^5-1=0$, but this gives us no information because all positive integers divide zero. However, $2^5-2=30$, so our answer cannot be greater than $30$ (because $30$ divided by an integer larger than $30$ is never an integer). This fact eliminates options (B) and (D).

Now, we will factorize $n^5-n$ to discover some properties: \begin{align*} n^5-n &= n(n^4-1) \\ &= n(n^2-1)(n^2+1) \\ &= n(n-1)(n+1)(n^2+1) \end{align*} One of $n$ and $n-1$ must be even, so $2$ divides $n^5-n$ for all $n$. Likewise, one of $n-1$, $n$, and $n+1$ must be divisible by $3$, so $3$ must divide $n^5-n$ as well. In the case of $5$, there are, in total, three cases:

Case 1: $5$ divides either $n-1$, $n$, or $n+1$. Then, $5$ clearly must divide $n^5-n$.

Case 2: $5$ divides $n+2$. In this case, $n+2 \equiv 0$ $\mod$ $5$, so $n \equiv 3 \mod 5$. Then, $n^2+1 \equiv 10 \equiv 0 \mod 5$, so $n^2-1$ (and, consequently, $n^5-n$) is divisible by $5$.

Case 3: $5$ divides $n+3$. In this case, $n+3 \equiv 0 \mod 5$, so $n \equiv 2 \mod 5$. Thus, $n^2+1 \equiv 5 \equiv 0 \mod 5$, so $n^2-1$ (and, consequently, $n^5-n$) is divisible by $5$.

Thus, 5 divides $n^5-n$ for all $n$.

Because $2$, $3$, and $5$ all divide $n^5-n$ and are relatively prime, $2 \cdot 3 \cdot 5 = 30$ also divides $n^5-n$. Thus, because $30$ is the largest of our remaining choices, we choose answer $\boxed{\textbf{(E) }30}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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