A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.
(In reply to
re: A similar problem but perhaps a bit trickier by Charlie)
Hmmm..the right answer, but I didn't understand the method!!! The method I like *only* works when dividing numbers by 11 and is a method one should learn!
If a number is divisible by 11, then, after putting +++ etc between its digits, the result should be a multiple of 11, whether positive or negative or 0, i.e. 22, 11, 0, 11, 22 are common answers. E.g. 132 IS divisible by 11, as 13+2=0. 1342 IS divisible by 11 as 13+42=0 but 6543 is not as 65+43 does not equal 0 or 11 or the other "multiples" of 11.
So, in this question, calling the unknown number 'x', we have:
56+43+87+9x+28+41. This evaluates to 7x. Now this must equal one of the multiples of 11.
7x = 11 ....gives x=18
7x = 0 ....gives x=7
7x = 11 ...gives x= 4
As 'x' is a single digit from 0 to 9, you can see the solution we require is x=7. Obviously, you could work out all the others, i.e. 7x=22 and 7x=33 but that would be pointless, as you can tell 'x' is getting smaller and smaller!! So, x=7 is the required solution.
Nice answer by you though, Charlie!
Remember that rule for division by 11, putting ++ in, very hand for maths challenges I've found!
Regards,
Kirk
P.s It *only* works on 11s though, perhaps your method works on any number!

Posted by Kirk
on 20031120 17:13:45 