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Let $a_1,a_2,.....a_{169} $ represent any arbitrary permutation of the number $1,2,3....169$. Then the product $(a_1-1)(a_2-2)......(a_{169} - 169)$ is

  1. Odd only for some permutation, not all
  2. Always even, whatever be the permutation
  3. Always odd, whatever be the permutation
  4. Even only for some permutation, not all

I have no idea how to solve this problem .

Aakash Kumar
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    Is "even" a possible choice? – Hagen von Eitzen Jul 09 '16 at 16:15
  • @HagenvonEitzen yes – Aakash Kumar Jul 09 '16 at 16:16
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    Is "an integer" a possible choice? Is "difficult and I don't want to do it" a possible choice? Obviously (1-1)(2-2)....(169 - 169) = 0 but (2-1)....(169-168)(169-1) $\ne$ 0$ so there is not single answer so ... what are we expected to answer? Everything we can? Anything that's true? Anything that's definite? Nothing that is indefinite? – fleablood Jul 09 '16 at 16:21
  • @fleablood I have added the choices – Aakash Kumar Jul 09 '16 at 16:24
  • Hagen van Eitzens answer is pretty good. If so much as one single $a_i - i$ is even then the product is even. To be odd every single $a_i - i$ must be odd. Obviously some $a_i-i$ can be even. So the question boils down to: is it possible to make a permutation where all the $a_i - i$ are odd. As there are more odd $i$ and $a_i$ then even there must be at least one $a_i$ and $i$ both odd. I find it odd that there is no option 5) more than one are even and more than one are odd. If 1 and 4 were supposed to be "permutations" than 1 and 4 are equivalent. – fleablood Jul 09 '16 at 16:32
  • So, does it have to be positive, negative, non-positive, non-negative? – fleablood Jul 09 '16 at 16:35
  • What's the difference between option 1 and option 4? Hint: When is the product of two integers odd? – Henrik supports the community Jul 09 '16 at 18:12

1 Answers1

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There are more odd than even indices, hence some $a_{\text{odd}}$ must be odd, thus making the product even.

Hagen von Eitzen
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