9768: #36, #50 and #61

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

9768: #36, #50 and #61

Postby YKM » Wed Sep 07, 2011 5:27 am

#36, For each real number x, let u(x) be the mean of the numbers, 4, 9, 7, 5 and x; and let n(x) be the median of these numbers. For how many values of x is u(x) = n(x)?

Ans: D, Three.
Can anyone show me which three?

#50, How many continuous real-valued functions f are there with domain [-1, 1] such that (f(x))^2 = x^2 for each x in [-1, 1]?

Ans: D, Four.
Can anyone show me which four?

#61, What is the greatest integer that divides (p^4) - 1 for every prime number p greater than 5?

Ans: E, 240.
I can deduce the answer by using calculator, but no other idea by hand.

Thank you very much.

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: 9768: #36, #50 and #61

Postby blitzer6266 » Wed Sep 07, 2011 11:13 am

36. x= 0, 6.25, 10

50.
f(x) = x^2
f(x) = -x^2
f(x) = x^2 for x </= 0 and -x^2 for x>0
f(x) = -x^2 for x </= 0 and x^2 for x>0

61. This uses some basic number theory.

p^4 - 1 = (p^2-1)(p^2+1)

Every odd prime is congruent to 1 mod 8. Thus p^2-1 is divisible by 8 and p^2-1 is clearly divisible by 2. Therefore p^4-1 will always be divisible by 2^4.

Since p^2 is not divisible by 3, either p^2-1 or p^2+1 is divisible by 3. Therefore, p^4-1 is always divisible by 3.

If p is not divisible by 5, then p^2 is either congruent to 1 or -1 mod 5. Thus, either p^2-1 or p^2+1 is divisible by 5.

Putting it all together, if p>5 where p is prime, 2^4*3*5= 240 divides p^4- 1

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: 9768: #36, #50 and #61

Postby blitzer6266 » Wed Sep 07, 2011 11:24 am

Thus p^2-1 is divisible by 8 and *p^2+1* is clearly divisible by 2

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: 9768: #36, #50 and #61

Postby blitzer6266 » Wed Sep 07, 2011 12:14 pm

And for 50, replace all of the x^2 with abs(x). My bad

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: 9768: #36, #50 and #61

Postby YKM » Thu Sep 08, 2011 4:50 am

thank you very much

aherring
Posts: 21
Joined: Tue Oct 02, 2012 11:45 pm

Re: 9768: #36, #50 and #61

Postby aherring » Thu Oct 03, 2013 4:59 pm

I suspect I am missing something, or perhaps reading the problem incorrectly... But in what sense do the functions +/- x and +/- abs(x) have domain [-1,1]? It seems to me that for each of the four functions the domain is all of R. Any ideas? Much appreciated!

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: 9768: #36, #50 and #61

Postby DDswife » Thu Aug 14, 2014 5:32 pm

29 is an odd prime and it's not 1 mod 8

I think that you meant p^2 is 1 mod 8

There is another way to prove that p^2-1 is a multiple of 8 for any odd. It doesn't evenneed to be a prime number.

(p+1)(p-1) is the product of 2 consecutive even numbers. But one of themmust be a multiple of 4
Last edited by DDswife on Thu Aug 14, 2014 5:36 pm, edited 1 time in total.

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: 9768: #36, #50 and #61

Postby DDswife » Thu Aug 14, 2014 5:33 pm

Their domain is R, yes. But you can restrict it as mch as you want.




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: No registered users and 4 guests