Question

What is the smallest 6 digit palindrome that is divisible by 99

I've been trying for 3 hours and I still cant figure it out PLEASE help.

Answer 1

Hello,

Let's assume n the palindrome

`n=\overline{abccba}\\\\Since\ 99=9*11:\\\\the \ smallest\ \Longrightarrow \ a=1\\\\a+b+c+c+b+a=9*k\ ,\ k\in \mathbb{N}\\2*(a+b+c)=9*k\ \Longrightarrow \ k\in 2\mathbb{N} , \ : k=2\\\\1+b+c=9 \Longrightarrow \ b=0\ and\ c=8\\\\n=108801\\\\Proof:\\108801=9*12089 =11*9891\\Nota\ bene:(a+c+b)-(b+c+a)=11*p \Longrightarrow \ 0=11*p \Longrightarrow \ p=0`

Answer 2

• 108801

Explanation:

6-digit palindrome is the number n the form of:

• xyzzyx

This is divisible by 11 by default as the sum of the digits in odd placed is same as sum of the number in even places (remember the divisibility rule by 11):

• x + z + y = y + z + x

Now, in order to be divisible by 99, the number must be divisible by 11 and 9.

According to divisibility rule by 9 the sum of all digits must be divisible by 9. You can see In our case we need to have (the minimum):

• x + y + z = 9

The smallest number we could get is when x is minimum, y is minimum, so:

• x = 1, y = 0, then y = 8

The number we get is:

• 108801

Proof:

• 108801/99 = 1099