Problem of the week #94 - January 13th, 2014

[Jameson]

What is the smallest three digit palindrome divisible by 18? Can you solve this without a brute force examination of possibilities?

Note: a "palindrome" is a number (or word) that is the same backwards and forwards. For example, 121 and 777 are both palindromes.
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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) magneto
2) kaliprasad
3) MarkFL
4) mente oscura
5) Ackbach
6) Pranav
7) soroban
8) eddybob123

Solution (from Pranav):

Spoiler

Let the three digit number be $\overline{abc}$. Clearly, we have the condition that $a\neq 0$.

Since the number is a palindrome, we have $a=c$ i.e $\overline{aba}$. For the number to be divisible by $18$, it must be divisible by $2$ and $9$.

The number is divisible by $2$ if $a=2, 4, 6 \,\text{or}\, 8$ ($a=0$ not allowed). Also, the number is divisible if $a+b+a=2a+b$ is a multiple of $9$. Since we look for the smallest palindrome, let $a=2$. Clearly, $b=5$ for $2a+b$ to be a multiple of $9$. Hence, the smallest palindrome divisible by 18 is $\fbox{252}$.