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
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 1-3+2=0. 1342 IS divisible by 11 as 1-3+4-2=0 but 6543 is not as 6-5+4-3 does not equal 0 or 11 or the other "multiples" of 11.
So, in this question, calling the unknown number 'x', we have:
5-6+4-3+8-7+9-x+2-8+4-1. This evaluates to 7-x. Now this must equal one of the multiples of 11.
7-x = -11 ....gives x=18
7-x = 0 ....gives x=7
7-x = 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. 7-x=-22 and 7-x=-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!
P.s It *only* works on 11s though, perhaps your method works on any number!
Posted by Kirk
on 2003-11-20 17:13:45