So you think you can do math question II (difficulty 3/5):
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shouldn't you be doing your homework on your own?
Feb 17, 2008
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#3
Easy: 5 * 3 * 1 * -1 * -3 = 45, so a=1, b=3, c=5, d=7, e=9, so a+b+c+d+e = 25
Ok, good job! Next winner omnipaul!
savy_swede wrote: shouldn't you be doing your homework on your own?
This is part of a math contest,not my homework. I take statistics
Nilesh wrote: Ok, good job! Next winner omnipaul!
This isn't a general proof, it's an example where the sum is 25. A correct answer would show that the answer is always 25.
To do this, note that 45 = 3*3*5, in prime factorization. Therefore, each number (6-x) must be one of +/- 1,3,5,9,15. Because at most two numbers can be +/- 1 (since we have distinctness), we must have at least three non-unit factors. But the prime factorization is the factorization with the maximal number of factors, therefore, the three numbers we are given must be the three prime factors (in absolute value). Therefore, we must have both +/- 3, since there are two threes in the prime factorization, and the remaining factor, +/- 5, must be positive (since we have to have an even number of negatives).
Therefore, any time 45 is written as the product of five distinct integers, the only possibility is 45 = 1*3*5*(-1)*(-3). Therefore, the only solution we have (up to permutation) is a,b,c,d,e = 5,3,1,7,9. The sum of these numbers is 25.
Let a, b, c, d, and e be distinct (different) integers such that
(6-a)(6-b)(6-c)(6-d)(6-e) = 45. What is the sum of a, b, c, d, and e?
This one isn't that bad. Again, first one to submit the correct answer wins.