56th Kürschák 1955 Problem 2 How many five digit numbers are divisible by 3 and contain the digit 6?
Consider first the number of 5-digit numbers divisible by 3. The smallest is 10002 = 3334·3, the largest is 99999 = 33333·3, so there are 29999. Now consider the number that do not contain the digit 6. There are 8 choices for the 1st digit, 9 choices for each of the 2nd, 3rd and 4th digits. If the total of the first 4 digits is 0 mod 3, then the last digit must be 0, 3 or 9. If it is 1 mod 3, then the last digit must be 2, 5 or 8. If it is 2 mod 3, then the last digit must be 1, 4 or 7. So in all cases there are 3 choices for the last digit. Hence the total number is 8·9 3 3 = 17496. So the total no. which do contain a 6 is 12503. 56th Kürschák 1955 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03