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CEMC Gauss (Grade 8), 2022

Number theory Difficulty 3.8 AMC 10/12 Find the answer Canada

The number 385 is an example of a three-digit number for which
one of the digits is the sum of the other two digits. How many numbers
between 100 and 999 have this property?

Pick one

Solution

A palindrome greater than 10 000 and less than 100 000 is a
5-digit positive integer of the form $abcba$ , where $a, b$ and $c$ are digits and $a\neq0$ .

A positive integer is a multiple of 18 if it is a multiple of both 2 and
9 (and a positive integer that is a multiple of both 2 and 9 is a
multiple of 18).

A positive integer is a multiple of 2 if it is even, and thus the digit
$a$ is equal to $2,4,6$ or 8 (recall $a\neq0$ ).

A positive integer is a multiple of 9 exactly when the sum of its digits
is a multiple of 9, and thus $a+b+c+b+a$ or $2a+2b+c$ is a multiple of 9.

Next we consider four possible cases, one case for each of the possible
values of $a$ .

Case 1: $a=2$

When $a=2$ , we require that $2a+2b+c=4+2b+c$ be a multiple of 9.

Since $4+2b+c\geq4$ , then the
smallest possible multiple of 9 that $4+2b+c$ can equal is 9.

Since $b\leq 9$ and $c\leq9$ , then $4+2b+c$ is at most $4+2(9)+9=31$ .

Thus, $4+2b+c$ can equal $9, 18$ or 27, which gives $2b+c$ equal to $5, 14$ or 23 respectively.

Next, we determine the possible values of $b$ and $c$ so that $2b+c$ is equal to $5, 14$ or 23.

$2b+c=5$	$2b+c=14$	$2b+c=23$
$b=2, c=1$	$b=7, c=0$	$b=9, c=5$
$b=1, c=3$	$b=6, c=2$	$b=8, c=7$
$b=0, c=5$	$b=5, c=4$	$b=7, c=9$
	$b=4, c=6$
	$b=3, c=8$

Thus when $a=2$ , there are $3+5+3=11$ such palindromes.

Case 2: $a=4$

When $a=4$ , we require that $2a+2b+c=8+2b+c$ be a multiple of 9.

Since $8+2b+c\geq8$ , then the
smallest possible multiple of 9 that $8+2b+c$ can equal is 9.

Since $b\leq 9$ and $c\leq9$ , then $8+2b+c$ is at most $8+2(9)+9=35$ .

Thus, $8+2b+c$ can equal $9, 18$ or 27, which gives $2b+c$ equal to $1, 10$ or 19 respectively.

Next, we determine the possible values of $b$ and $c$ so that $2b+c$ is equal to $1, 10$ or 19.

$2b+c=1$	$2b+c=10$	$2b+c=19$
$b=0, c=1$	$b=5, c=0$	$b=9, c=1$
	$b=4, c=2$	$b=8, c=3$
	$b=3, c=4$	$b=7, c=5$
	$b=2, c=6$	$b=6, c=7$
	$b=1, c=8$	$b=5, c=9$

Thus when $a=4$ , there are $1+5+5=11$ such palindromes.

Case 3: $a=6$

When $a=6$ , we require that $2a+2b+c=12+2b+c$ be a multiple of 9.

Since $12+2b+c\geq12$ and $12+2b+c\leq12+2(9)+9=39$ , then $12+2b+c$ can equal 18, 27 or 36, which
gives $2b+c$ equal to 6, 15 or 24,
respectively.

$2b+c=6$	$2b+c=15$	$2b+c=24$
$b=3, c=0$	$b=7, c=1$	$b=9, c=6$
$b=2, c=2$	$b=6, c=3$	$b=8, c=8$
$b=1, c=4$	$b=5, c=5$
$b=0, c=6$	$b=4, c=7$
	$b=3, c=9$

Thus when $a=6$ , there are $4+5+2=11$ such palindromes.

Case 4: $a=8$

When $a=8$ , we require that $2a+2b+c=16+2b+c$ be a multiple of 9.

Since $16+2b+c\geq16$ and $16+2b+c\leq16+2(9)+9=43$ , then $16+2b+c$ can equal 18, 27 or 36, which
gives $2b+c$ equal to 2, 11 or 20,
respectively.

$2b+c=2$	$2b+c=11$	$2b+c=20$
$b=1, c=0$	$b=5, c=1$	$b=9, c=2$
$b=0, c=2$	$b=4, c=3$	$b=8, c=4$
	$b=3, c=5$	$b=7, c=6$
	$b=2, c=7$	$b=6, c=8$
	$b=1, c=9$

Thus when $a=8$ , there are $2+5+4=11$ such palindromes.

Therefore, the number of palindromes that are greater than 10 000 and
less than 100 000 and that are multiples of 18 is $11+11+11+11=44$ .

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