<p>this is one of the toughest problems i have come across yet
a,b,c,d,e,f,g,h,i are 9 numbers from 1-9.
in no specific order.a,d,e,f are even numbers
b,c,g,h,i are odd numbers
NOTE:-it is not neccesarry for c to be greater than b
now(^means raised to the power to)
a<em>c</em>i*d=b^f + e^f + g^f=h^f + i^f +d^f</p>
<p>find all the numbers from a to i</p>
<p>this sort of thing is terribly boring.</p>
<p>i agree with Ben</p>
<p>neaah...
i like such types of questions...
and you don't have to know calc to solve it
which makes it more democratic and fair lol</p>
<p>Wow this question is lame. =</p>
<p>yeah i know lol
its from a local mensa column that runs every monday in an afternoon newspaper here</p>
<p>The humility on this thread is ridiculous.</p>
<p>I think the humility on these boards is ridiculous...</p>
<p>lol it looks like no one can crack the puzzle.240 reads and 7 posts lol.nothing in connection with the puzzle.
PS:-even i cant crack it hence the posting</p>
<p>are they unique numbers?</p>
<p>odd...my brute force solution says there is no answer...I'll get back to you on that...</p>
<p>please recheck which ones are even and odd (not that I doubt you...I just want to check before running a larger brute-force program...)</p>
<p>tanonev, i tried brute force also. it didnt seem to give me an answer even without considering the evens and odds.</p>
<p>The point isn't about humility or lack thereof. Where's the potential for real mathematical insight in this problem, or even for a single-stroke, clever approach? Mathematicians don't do puzzles for the sake of knowing what numbers fit in some silly contrived set of equations -- we do them to discover new worlds of previously unseen mathematical beauty. When you are resorting to brute-force crunchery, it means that somehow nothing deep is coming out of this right now. Plus, I have a very strong hunch that this is the sort of problem that won't ever yield anything broad and nontrivial that isn't already known.</p>
<p>Summary: puzzles are good if they create new math or open your mind in a new way. Puzzles in Mensa columns are typically just silly toys with neither virtue (I defy anyone to show me I am wrong about this one). Is that really such a bad thing to say?</p>
<p>I was wondering if there was a clever application of Jensen's inequality here somewhere, but when someone said it was a mensa question I knew right away it's a math problem with no point or non-trivial solution.</p>
<p>Maybe I'm not a math whiz, but I thought the problem was interesting. Since a<em>c</em>i<em>d is the product of two evens and two odds, it must have a four as a factor and either 8 or 12. On the other hand, the even number to the f power also has to have a factor of 4. So the odd number powers of f must sum to a number that is divisible by four. For f=2 or f=4 (the logical choices since numbers raised to the 6 or 8 power get too big for a</em>c<em>i</em>d) there turn out to be very few possible choices that add up to a number divisible by four.</p>
<p>I didn't solve the problem either, it would be great to see if there is a solution, but IMO it was interesting to think along these lines.</p>
<p>"Maybe I'm not a math whiz, but I thought the problem was interesting. Since a<em>c</em>i<em>d is the product of two evens and two odds, it must have a four as a factor and either 8 or 12. On the other hand, the even number to the f power also has to have a factor of 4. So the odd number powers of f must sum to a number that is divisible by four. For f=2 or f=4 (the logical choices since numbers raised to the 6 or 8 power get too big for a</em>c<em>i</em>d) there turn out to be very few possible choices that add up to a number divisible by four."</p>
<p>Thanks ADad, that's just what I needed to prove that it can't be done, EVEN IF repeats are allowed, AND EVEN IF we extend them to be ALL COUNTING NUMBERS.</p>
<p>(a<em>c</em>i<em>d) modulo 4 = 0, no matter what
therefore (b^f + e^f + g^f) modulo 4 must equal 0
e is even and f is even, so e^f modulo 4 = 0
Proof: e = 2</em>A, f = 2<em>B,
e^f = (2</em>A)^(2<em>B) = 2^(2</em>B)<em>A^(2</em>B) = 4<em>(2^B)</em>(A^(2<em>B))
therefore (b^f + e^f + g^f) modulo 4 = (b^f + g^f) modulo 4
but b is odd and f is even, so b^f modulo 4 = 1
Proof: 1^(anything) = 1, and 1 modulo 4 = 1
3^2 = 9 = 1 + 8
5^2 = 25 = 9 + 16
7^2 = 49 = 25 + 24
...
(2n+1)^2 = (2n-1)^2 + 8</em>n for all counting numbers n
by similar reasoning, since g is odd and f is even, g^f modulo 4 = 1
therefore (b^f + g^f) modulo 4 = 2
but we said before that (b^f + g^f) modulo 4 must equal 0
Therefore we have a contradiction
Therefore THERE IS NO SOLUTION</p>
<p>Wow! </p>
<p>Fascinating, elegant, and very well done!</p>
<p>Congratulations! And thanks. :)</p>
<p>thanks...I can't believe I didn't think of that yesterday...lol</p>
<p>When you're given a property, always change it.</p>
<p>i.e.
even numbers = 2k
odd numbers = 2k+1
divisible by 3 = n(n+1)(n+2)</p>
<p>it simplifies the question so much</p>