Same Digit in First and Last Place Values
Let be an integer that has the same digit appearing in its first and last place values. The rest are zeroes.
Is there a value of that is divisible to all digits except for and ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
For n to be divisible by 8 , the last three digits must be a multiple of 8 . Thus n must end with 8 .
For n to be divisible by 9 , the sum of all the digits n has must be a multiple of 9 . But if we apply the statement in the first paragraph, the sum will always be 1 6 which is not a multiple of 9 .
Therefore, n can't be a number divisible by all digits except for 5 and 0