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Difference between revisions of "1991 AHSME Problems/Problem 26"
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An <math>n</math>-digit positive integer is cute if its <math>n</math> digits are an arrangement of the set <math>\{1,2,...,n\}</math> and its first  
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== Problem ==
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<math>k</math> digits form an integer that is divisible by <math>k</math> , for  <math>k  = 1,2,...,n</math>. For example, <math>321</math> is a cute <math>3</math>-digit integer because <math>1</math> divides <math>3</math>, <math>2</math> divides <math>32</math>, and <math>3</math> divides <math>321</math>. Howmany cute <math>6</math>-digit integers are there?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>An <math>n</math>-digit positive integer is cute if its <math>n</math> digits are an arrangement of the set <math>\{1,2,...,n\}</math> and its first  
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<math>k</math> digits form an integer that is divisible by <math>k</math>, for  <math>k  = 1,2,...,n</math>. For example, <math>321</math> is a cute <math>3</math>-digit integer because <math>1</math> divides <math>3</math>, <math>2</math> divides <math>32</math>, and <math>3</math> divides <math>321</math>. How many cute <math>6</math>-digit integers are there?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
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<math>\text{(A) } 0\quad
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\text{(B) } 1\quad
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\text{(C) } 2\quad
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\text{(D) } 3\quad
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\text{(E) } 4</math>
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== Solution ==
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<math>\fbox{C}</math> Let the number be <math>abcdef</math>. We know <math>1</math> will always divide <math>a</math>. <math>5</math> must divide <math>abcde</math>, so <math>e</math> must be <math>5</math> or <math>0</math>, but we can only use the digits <math>1</math> to <math>6</math>, so <math>e = 5</math>. <math>4</math> must divide <math>abcd</math>, so it must divide <math>cd</math> (the test for divisibility by 4 is that the last two digits form a number divisible by 4), and so <math>cd</math> must be <math>12</math>, <math>16</math>, <math>24</math>, <math>32</math>, <math>36</math>, (not <math>52</math> or <math>56</math> as the <math>5</math> is already used), or <math>64</math>. We know <math>2</math> divides <math>ab</math> so <math>b</math> is even, <math>4</math> divides <math>abcd</math> so <math>d</math> is even, and <math>6</math> divides <math>abcdef</math> so <math>f</math> is even, and thus <math>b</math>, <math>d</math>, and <math>f</math> must be <math>2</math>, <math>4</math>, and <math>6</math> in some order, so <math>c</math> must be odd, so <math>cd</math> must be <math>12</math>, <math>16</math>, <math>32</math>, or <math>36</math>. Now <math>6</math> divides <math>abcdef</math> implies <math>3</math> divides <math>abcdef</math>, and we know <math>3</math> divides <math>abc</math>, so <math>3</math> must also divide <math>def</math>, so we need a multiple of <math>3</math> starting with <math>2</math> or <math>6</math>, using the remaining digits. If <math>cd</math> is <math>12</math>, then <math>def</math> must be <math>243</math>, <math>234</math>, <math>246</math>, or <math>264</math>, but none of these work as we need <math>e = 5</math>. If <math>cd</math> is <math>16</math>, <math>def</math> must be <math>624</math>, <math>642</math>, <math>645</math>, or <math>654</math>, so <math>654</math> works (it has <math>e=5</math>). If <math>cd</math> is <math>32</math>, similarly nothing works, and if <math>cd</math> is <math>36</math>, <math>651</math> or <math>654</math> works. So now we have, as possibilities, <math>ab1654</math>, <math>ab3651</math>, and <math>ab3654</math>. But clearly <math>ab3651</math> can't work, as we need <math>abcdef</math> to be divisible by <math>6</math>, so it must be even, so we eliminate this possibility. Now with <math>ab1654</math>, we need <math>b</math> to be even, so it must be <math>2</math>, giving <math>a21654</math>, and then <math>321654</math> works as <math>3</math> does divide <math>321</math>. With <math>ab3654</math>, we get <math>a23654</math> as <math>b</math> is even, and then <math>123654</math> works as <math>3</math> divides <math>123</math>. Hence the number of such numbers is <math>2</math>: <math>321654</math> and <math>123654</math>.
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== See also ==
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{{AHSME box|year=1991|num-b=25|num-a=27}} 
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[[Category: Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:40, 24 February 2018

Problem

An $n$-digit positive integer is cute if its $n$ digits are an arrangement of the set $\{1,2,...,n\}$ and its first $k$ digits form an integer that is divisible by $k$, for $k = 1,2,...,n$. For example, $321$ is a cute $3$-digit integer because $1$ divides $3$, $2$ divides $32$, and $3$ divides $321$. How many cute $6$-digit integers are there?

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$

Solution

$\fbox{C}$ Let the number be $abcdef$. We know $1$ will always divide $a$. $5$ must divide $abcde$, so $e$ must be $5$ or $0$, but we can only use the digits $1$ to $6$, so $e = 5$. $4$ must divide $abcd$, so it must divide $cd$ (the test for divisibility by 4 is that the last two digits form a number divisible by 4), and so $cd$ must be $12$, $16$, $24$, $32$, $36$, (not $52$ or $56$ as the $5$ is already used), or $64$. We know $2$ divides $ab$ so $b$ is even, $4$ divides $abcd$ so $d$ is even, and $6$ divides $abcdef$ so $f$ is even, and thus $b$, $d$, and $f$ must be $2$, $4$, and $6$ in some order, so $c$ must be odd, so $cd$ must be $12$, $16$, $32$, or $36$. Now $6$ divides $abcdef$ implies $3$ divides $abcdef$, and we know $3$ divides $abc$, so $3$ must also divide $def$, so we need a multiple of $3$ starting with $2$ or $6$, using the remaining digits. If $cd$ is $12$, then $def$ must be $243$, $234$, $246$, or $264$, but none of these work as we need $e = 5$. If $cd$ is $16$, $def$ must be $624$, $642$, $645$, or $654$, so $654$ works (it has $e=5$). If $cd$ is $32$, similarly nothing works, and if $cd$ is $36$, $651$ or $654$ works. So now we have, as possibilities, $ab1654$, $ab3651$, and $ab3654$. But clearly $ab3651$ can't work, as we need $abcdef$ to be divisible by $6$, so it must be even, so we eliminate this possibility. Now with $ab1654$, we need $b$ to be even, so it must be $2$, giving $a21654$, and then $321654$ works as $3$ does divide $321$. With $ab3654$, we get $a23654$ as $b$ is even, and then $123654$ works as $3$ divides $123$. Hence the number of such numbers is $2$: $321654$ and $123654$.

See also

1991 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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
All AHSME Problems and Solutions

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