9768: #36, #50 and #61
9768: #36, #50 and #61
#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 realvalued 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.
Ans: D, Three.
Can anyone show me which three?
#50, How many continuous realvalued 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.

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: 9768: #36, #50 and #61
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^21)(p^2+1)
Every odd prime is congruent to 1 mod 8. Thus p^21 is divisible by 8 and p^21 is clearly divisible by 2. Therefore p^41 will always be divisible by 2^4.
Since p^2 is not divisible by 3, either p^21 or p^2+1 is divisible by 3. Therefore, p^41 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^21 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
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^21)(p^2+1)
Every odd prime is congruent to 1 mod 8. Thus p^21 is divisible by 8 and p^21 is clearly divisible by 2. Therefore p^41 will always be divisible by 2^4.
Since p^2 is not divisible by 3, either p^21 or p^2+1 is divisible by 3. Therefore, p^41 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^21 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

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: 9768: #36, #50 and #61
Thus p^21 is divisible by 8 and *p^2+1* is clearly divisible by 2

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: 9768: #36, #50 and #61
And for 50, replace all of the x^2 with abs(x). My bad
Re: 9768: #36, #50 and #61
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
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^21 is a multiple of 8 for any odd. It doesn't evenneed to be a prime number.
(p+1)(p1) is the product of 2 consecutive even numbers. But one of themmust be a multiple of 4
I think that you meant p^2 is 1 mod 8
There is another way to prove that p^21 is a multiple of 8 for any odd. It doesn't evenneed to be a prime number.
(p+1)(p1) 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.
Re: 9768: #36, #50 and #61
Their domain is R, yes. But you can restrict it as mch as you want.