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9768: #36, #50 and #61

Posted: Wed Sep 07, 2011 5:27 am
by YKM
#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.

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

Posted: Wed Sep 07, 2011 11:13 am
by blitzer6266
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

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

Posted: Wed Sep 07, 2011 11:24 am
by blitzer6266
Thus p^2-1 is divisible by 8 and *p^2+1* is clearly divisible by 2

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

Posted: Wed Sep 07, 2011 12:14 pm
by blitzer6266
And for 50, replace all of the x^2 with abs(x). My bad

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

Posted: Thu Sep 08, 2011 4:50 am
by YKM
thank you very much

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

Posted: Thu Oct 03, 2013 4:59 pm
by aherring
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!

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

Posted: Thu Aug 14, 2014 5:32 pm
by DDswife
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

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

Posted: Thu Aug 14, 2014 5:33 pm
by DDswife
Their domain is R, yes. But you can restrict it as mch as you want.