1,3

The first number must be some power of 2, the second the same power of 5. Such a pair is the only positive solution because neither factor could have both a 2 and a 5 in its prime factorization, or else it would be a multiple of 10 and would thus have a 0 as its last digit, which is ruled out. "We may wonder what powers of 10 are products of two integers without any zero digits. For large numbers, this is very unlikely because there will normally be 10% of zeros among many random digits... In fact, there seems to be only 11 possibilities... The probability is roughly (0.9)^n that the n-th power of 10 would yield a solution. So the expected number of solutions above the n-th power of 10 is someting like 10*(0.9)^n. Since we've actually checked that there's no other solution below n = 1500, we can be very confident that we've not missed anything..."

G. P. Michon, What two integers without zero digits have a product of 1000000000?

10^0 = 1 * 1

10^1 = 2 * 5

10^2 = 4 * 25

10^3 = 8 * 125

10^4 = 16 * 625

10^5 = 32 * 3125

10^6 = 64 * 15625

10^7 = 128 * 78125

10^9 = 512 * 1953125

10^18 = 262144 * 3814697265625

10^33 = 8589934592 * 116415321826934814453125

Sequence in context: A037405 A048333 A007496 this_sequence A082274 A029804 A084690

Adjacent sequences: A108939 A108940 A108941 this_sequence A108943 A108944 A108945

base,fini,nonn

Jonathan Vos Post (jvospost2(AT)yahoo.com), Jul 20 2005