a x b = 1,000,000

Here's an easy one I saw in a paper on Friday...

Choose 2 numbers which can be multiplied together to make 1,000,000. Neither number can contain the digit '0' (zero).

To make it a little more challenging - your answer must contain the least possible number of digits.

well 1000000 = 2^6*5^6 and as soon as we multiply a 2 by a five the answer ends in a zero.....

...hence least possible digits as the given solution is the only

guess i didn't read Ted S' answer then

Well, the only possible solution is 64 * 15,625. Since 1,000,000 = 10^6 = 2^6 * 5^6, it is readily seen that if any of the 2s and 5s are multiplied together, the number will end in 0. So, try 2^6 as one number and 5^6 as the other and hope that they don't have any 0s in them. Since they don't, the problem is solved.

how about 333333.3333333333333333333333333....*3 ....only 1 digit

Based on comments in other threads in this ng I'd say that this answer has infinitely many digits, but only one numeral

Hmm, the question didn't call for digits, it called for 2 numbers...

'Choose 2 numbers which can be multiplied together to make 1,000,000. Neither number can contain the digit '0' (zero)'

Given that I could only come up with 15625 x 64 and haven't seen another solution yet. Even the 2^6*5^6 doesn't work as it doesn't use 2 numbers, it uses 2 formulas, equastions, etc. what ever you want to call it, but 2^6 certainly isn't a number, although it will eventually lead to one

If you want to add the criteria that the numbers are positive integers, then there is no other solution, as has been argued here in other posts.

However, numbers are not necessarily positive integers. This was pointed out by a previous poster. The numbers -15625 and -64 also satisfy the condition. As do the numbers 7812.5 and 128. But this doesn't have any fewer digits than 15625 and 64 has. Allowing non-integer solutions gives plenty more pairs of numbers that when multiplied give 1,000,000, but none that have fewer digits than 15625 and 64 (when written in decimal notation). (This is easily seen by noting that for a number to have a finite decimal representation, the denominator of the fraction (in lowest terms) can only have 2 and 5 as non-unit factors.)

Brian

Alan O.D.:

Well, in that case, I can beat that: 'sixty-four times fifteen thousand six hundred twenty-five'.

Your last comment- the one in parnthesesgives me another 7-digit solution: 11x11 (base 999)

I did restrict myself to the standard 10 digits. Obviously }*} (base 1111) is a 6-digit answer, if we agree that } is the digit in base 1111 that is equal to 1000 in base 10. (two }s and the four digits in the description of the base)

Using alternate bases, that are actually used (such as 2,8,16,and 60) we also have: A^6*1, base 16 This has 5 digits (including the two in the base) If you say A^6*1, hexadecimal, there are technically 3 digits. Since you need two factors (by the problem description), you can't get rid of the *1. (with one factor, in a base greater than 1000,000 you can have a 1-digit solution

Anyone else with strange/ unorthodox solutions?

Thanks for your idea, mighty dragon. What about 1 * 11 (base 999999) which is one digit less? Now, how acceptable is changing base for the solution but not for the problem?