1991 AHSME Problems/Problem 26

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 $123654$ .

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
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1991 AHSME Problems/Problem 26

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