The product of some of the numbers 1 to 10 was divided by the product of the remaining numbers.
arithmetic/ratio.64.7 New
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Problem
The product of some of the numbers 1 to 10 was divided by the product of the remaining numbers. 1 was placed in the numerator and 2 in the denominator. The result was 64/7. Where should the remaining numbers 3 to 10 be placed? 1 x ... 64 ----------- = ---- 2 x ... 7 [ Editor: posted by Mark van Hoeij <hoeij@sci.kun.nl>. ]
Show solution
The top and bottom each have a unique prime factorization composed of
all the prime factors of the numbers that were multiplied there.
After canceling out all common factors, 2^6 is left on top and 7 on
the bottom.
Since the only number in 1..10 with a prime factor of 7 is 7, the only
way to have 7 remaining below is to put the 7 in the denominator.
Now look at all the numbers that could contribute to the 2^6 on top:
4, 6, 8, 10. Together these contribute 7 factors of 2. But you need
at least 2^7 to have 2^6 remaining after the 2 we already see on the
bottom is canceled; so all the even numbers (other than 2) go in the
numerator.
This leaves us with an extraneous factor of 3 and another of 5 on top.
The 5 can be canceled only by placing the number 5 in the denominator.
Now only the numbers 3 and 9 are left. The only place we can put the
9 without too many leftover factors of 3 is in the denominator, and
then the 3 goes into the numerator to cancel the last leftover factor.
1 x 4 x 6 x 8 x 10 x 3
---------------------- = 64/7.
2 x 7 x 5 x 9
-- David A. Karr <karr@cs.cornell.edu>