Math Problems

[eggmanneo]

I've been studying for the GREs. I just got a 780 on the math part of a practice exam, so hopefully I carry this success over to the actual test. Here's a problem that was quite hard but interesting. I don't know why I'm posting this, I doubt any of my friends would be interested, but here it is

Comments

[eggmanneo]

The number abcde is really 10^{4}a + 10^{3}b + 10^{2}c + 10^{1}d + 10^{0}e so the difference becomes...

10^4 (a-e) + 10^3 (b-d) + 10^2 (c-c) + 10 (d-b) + (e-a)

= (10^4 - 1)(a-e) + (10^3 - 10)(b-d)

= 9999(a-e) + 990(b-d)
So the answer is E.9

Solution is from this message board
http://www.physicsforums.com/showthread.php?t=182509

Casey's Opinion--Math is Evil

[anonymous]

My thoughts:

1) What the heck?! and...

2) If I see a problem like this on the GRE, I will laugh, then cry, then start filling out my application to work at McDonald's.

[jgranhill]

You can logically deduce this pretty well without the use of an analytic method too. Clearly, an integer with an odd number in the front and an even number in the first digit will reflect and cause the difference to be odd. Therefore, the five-digit reflection cannot be necessarily divisible by two or four. It is easy to consider a counter-example to five, so the only viable solutions are six and nine. However, in order for something to be divisible by six, it must be divisible by three and two, and it's already been shown that two is not a viable candidate, so only nine remains (by process of elimination).

Also, it is clear that any five digit permutation will result in a number whose difference in reflection will be divisible by three, so you require a number whose sole divisor requires a divisibility by three, therefore nine.

Obviously, the analytic solution gives the most surity in the situation, but not give the obvious intuition about the digit relations and powers of ten, this requires less intuition.