Show that there are arbitrarily large numbers n such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides n.
Solution
3232 = 16 x 202 and 10000 = 16 x 625. So any number with 3232 as its last 4 digits is divisible by 16. So consider N = 22223232. Its sum of digits is 18, so it is divisible by 9. Hence it is divisible by 9.16 = 144. But any four digits have at most four 2s and at most two 3s, so the product of any four digits divides 144 and hence N. But now we can extend N by inserting an additional 9m 2s at the front. Its digit sum is increased by 18m, so it remains divisible by 144 and it is still divisible by the product of any four digits.
Alternative solution
The number 111111111 with nine 1s is divisible by 9. Hence the number with twenty-seven 1s which equals 111111111 x 1000000001000000001 is divisible by 27. So N, the number with twenty-seven 3s, is divisible by 34. Now the number with 27n 3s is divisible by N and hence by 34.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002