Number theory problem

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mgmt_gmat: Master | Next Rank: 500 Posts; Posts: 221; Joined: Mon Dec 22, 2008 6:24 pm; Thanked: 2 times

by mgmt_gmat » Thu Feb 11, 2010 5:32 am

70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.

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Source: Beat The GMAT — Problem Solving | Thu Feb 11, 2010 5:32 am

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

by dtweah » Thu Feb 11, 2010 6:00 am

mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.

4(x+Y) =100
We know only one number multiplied by 4 gives positive 100. It is 25. So if both X and Y must sum to 25. If we consider the set of positive integers and zero, then the max each can take is 25. If only positive integers then the max is 24 and min is 1. If we consider all real numbers, then there is NO MAX or MIN since we can stretch x to an infinite number of decimal points and reduce Y by the same amout to infinity. Under this, negative numbers are also possible X could be 50 and y -25 and we still get hundred. Botched question.

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mohit11: Master | Next Rank: 500 Posts; Posts: 293; Joined: Thu Jan 14, 2010 9:08 amFollowed by: 5 members; Location: India; Thanked: 36 times; GMAT Score:730

by mohit11 » Thu Feb 11, 2010 7:27 am

I had the following approach to solve this question.

X^4 + Y^4 = 100

and 4^4 = 256

There fore answer has to be below 4, also, since we have to maximize X, we need to minimize Y, and 3^4 = 81 .. however we are not dealing with integers, so we can also consider 3.1 , 3.2 etc.

So the answer is B - between 3 and 6

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Thu Feb 11, 2010 7:34 am

mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.

@mgmt_gmat will u please make it clear whether it is X^4 + y^4 =100 or x4+y4=100
If it is X^4 +Y^4 =100
since X^4 +Y^4 =100
now 1^4 =1
2^4 =16
3^4 = 81
4^4 = 256

Here u can see X is going to lie between 3 and 4 for max value of X.
So B is correct

In case question is X4 +Y4 =100 dtweah approach is very correct.

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Thu Feb 11, 2010 8:22 am

To maximize 'x', minimize 'y' as far as possible. Best value of 'y' is 0

x^4 + y^4 = 100

Sub. y=0,

x^4 = 100

3^4 is 81. So, x should be little greater than 3. Best choice is B (3 to 6)

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komal: Legendary Member; Posts: 777; Joined: Fri Jan 01, 2010 4:02 amFollowed by: 47 members; Location: Mumbai, India; Thanked: 117 times

by komal » Tue Feb 16, 2010 9:24 am

mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.

Greatest possible value of X when Y^4 is minimum.
The possible minimum value of Y^4 is 0
X^4=100 ---> X^2=10 --> X> 3 AND X<4
B is the answer.

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LalaB: Master | Next Rank: 500 Posts; Posts: 425; Joined: Wed Dec 08, 2010 9:00 amFollowed by: 7 members; Thanked: 56 times; GMAT Score:690

by LalaB » Tue Dec 13, 2011 9:44 am

to maximize one variable, we need to minimize another one.so, let y=0
then x^4=10^2 x=sqrt10 =3.16
[spoiler]B[/spoiler] is the answ

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