If there are N integers between 100 and 999, inclusive, have the property that some permutation of its digits is a multiple of 11 between 100 and 999. For example, both 121 and 211 have this property. Then find the value of (N-1)/ 5 .
In this question, do i have to write down all the numbers that satisfy this property and then count (but it would take a lot of time) or is there any easier way?
count the total number of multiples of $11$ between 100 to 999.
they are all 3 digits so each of them can be arranged in $3!$ ways.
but changing unit digit and 100th's place digit will give us another multiple of $11$ which is already counted, So there are $\frac{3!}{2}$ arrangements for each multiple.
But some multiple has $0$ in them, and $0$ cannot be in the 100th's place. Find all such multiples and subtract these cases.
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