I am thinking of a fifty-digit number divisible by 239, of which, each digit is the same, except the ones digit. What is the ones digit?

We have that the 50-digit number is made up of:

aaaa...ak, with 49 a's and k is the 1's digit.

We could also write this as:

a*10^50 + a*10^49 ... + a*10 + k

a(10^50 + 10^49 + ... 10) + k

a(1111111...10) + k, with 49 1's


Now, we have to find a number that divides the term in paranthesis, then we take this new number and divide it by 239. If the remainder is a number from 0 to 9, that's the value of k.

A direct way to find the first number is to separate the parenthesis term in groups. Since there are 49 1's and 49 = 7*7, with 7 being prime, we can only split the parenthesis term in groups of 7 digits.

In this case, we have that the number found is:

(term)/x = (term)/1111111 = 11111110

If we now take x and divide by 239, we have 0 as remainder, which is in the 0 to 9 scope. We conclude that k must therefore be 0.

I hope it made sense to you guys, because I think it kind of just a little perhaps made possibly a little sense in my senseless mind. :P

Bruno.

Posted by Bruno on 2005-05-17 02:22:28