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.

I get an answer of 8 using the following approach:

A 50-digit integer, N, with all digits equal to 1, can be obtained as follows:

N = (10^50)/9 - 1/9

This can be converted to a 50-digit integer, M, with all 1s except for the digit x in the 26th position, by addition:

M = N + x*10^25

where x is chosen to make M divisible by 13.

If we divide (10^50)/9 by 13 we get a repeating pattern of the six numbers 854700, beginning in the 48th digit to the left of the decimal. This means that the decimal part of the resulting number is 0.854700854700...

If we divide 1/9 by 13 we get this same repeating pattern as follows: 0.008547008547...

Taking the difference (i.e., calculating N/13), we get for the decimal part: 0.8461538462, with the last decimal rounded.

We want an x such that (x*10^25)/13 will have a decimal part that, when added to N/13, gives all zeros to the right of the decimal (i.e., makes M an integer). The integer 8 meets this condition, because 8/13 gives a repeating pattern of the six numbers 615384. Calculating (8*10^25)/13 leads to the the following decimal portion: 0.1538461538 to 10 decimal places. The leading 6 in the series ends up to the left of the decimal. Now we add the decimal portions of N/13 and (8*10^25)/13:

0.8461538462 + 0.1538461538 = 1.0000000000

Thus, M is divisible by 13 if the number 8 occupies the 26th digit.

Posted by NK on 2004-02-10 10:32:01