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f(x) ... ?

Expert replies
Source: — Problem Solving |

Re: f(x) ... ?

by dtweah » Sat Apr 11, 2009 8:07 am
feeet wrote:Hi,

I have some problems when formular such as f(x)=... is involved. see attached two examples.

Please help!

Thanks
f(x)=f(1-x)
start with each fuction in the aswer choice and substitute (1-x) where ever you see x.

So f(x)=x^2 (1-x)^2
f(1-x) = (1-x)^2 (1-(1-x))^2
= (1-x)^2 x^2
So f(x)=f(1-x).
In a problem like this you just have to evaluate each answer choice and eliminate. My hunch is that in such question where they expect you to try each answer choice, the correct answer is usually located at the bottom so I would start from E or D. Take this with a grain of salt as I am not a GMAT expert.

For the second part you have to put h(n) in plain English.
h(n=1)= 2
h(n=2) 2 x 4
h(n=3) 2 x 4 x 8
h(n=4) 2 x 4 x 8 x 10

h(n=100) 2 x 4 x8 x 10 x 12 …….100
For all you know, h(100) must end with 2 zero’s and
If you add 1 to a number that ends with 2 zero the sum
must end with 01
h(100) +1 = xxxxxxxxxxxxxxxxx01
If a prime factor divides this number evenly, then the quotient
Must be such that when we multiply the quotient by that prime factor, we get a 01 ending. Examine all primes that are candidates. 2 is out b/c number is odd: (3 5,7,11,13, 17, 19, 23, 29, 31, 37, 41)
The strongest candidate is a xxxxxxxxxxxxx07(placing 0 before the last digit to mimimize as much as possible) in case of 13, 23, but you get 21. and you know 0+2>0. For 37, try xxxxxxxxxxxxxxx03 and you get the same outcome as above. 11 is a little tricky:
xxxxxxxxxxxxxxxxx01
X 11
_____________________
xxxxxxxxxxxxxxxxx01
xxxxxxxxxxxxxxxx01
_____________________
xxxxxxxxxxxxxxxxp11, which fails. 19 we have xxxxxxxxxxxxx09 but we get 81. 8+0>0. Continue and you see that the prime factor we are looking for has to be > 40.
Hope this gives you an idea what a question like this is getting at.
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Re: f(x) ... ?

by Ian Stewart » Sat Apr 11, 2009 1:10 pm
dtweah wrote: For the second part you have to put h(n) in plain English.
h(n=1)= 2
h(n=2) 2 x 4
h(n=3) 2 x 4 x 8
h(n=4) 2 x 4 x 8 x 10

h(n=100) 2 x 4 x8 x 10 x 12 …….100
For all you know, h(100) must end with 2 zero’s and
If you add 1 to a number that ends with 2 zero the sum
must end with 01
h(100) +1 = xxxxxxxxxxxxxxxxx01
If a prime factor divides this number evenly, then the quotient
Must be such that when we multiply the quotient by that prime factor, we get a 01 ending. Examine all primes that are candidates. 2 is out b/c number is odd: (3 5,7,11,13, 17, 19, 23, 29, 31, 37, 41)
The strongest candidate is a xxxxxxxxxxxxx07(placing 0 before the last digit to mimimize as much as possible) in case of 13, 23, but you get 21. and you know 0+2>0. For 37, try xxxxxxxxxxxxxxx03 and you get the same outcome as above. 11 is a little tricky:
xxxxxxxxxxxxxxxxx01
X 11
_____________________
xxxxxxxxxxxxxxxxx01
xxxxxxxxxxxxxxxx01
_____________________
xxxxxxxxxxxxxxxxp11, which fails. 19 we have xxxxxxxxxxxxx09 but we get 81. 8+0>0. Continue and you see that the prime factor we are looking for has to be > 40.
Hope this gives you an idea what a question like this is getting at.
It is not because h(100) + 1 ends in '01' that we know that its smallest prime factor is greater than 40. There are many numbers ending in '01' which are divisible by primes less than 40. For example:

201 ends in '01', but is divisible by 3;
301 ends in '01' but is divisible by 7;
1001 ends in '01' but is divisible by 7, 11 and 13;

and 901 is divisible by 17, 1501 is divisible by 19, 2001 is divisible by 23 and 29, and so on. The only primes you can rule out as divisors of h(100) + 1 by looking at its last two digits are 2 and 5.

Fortunately, we don't need to test primes exhaustively to solve the problem. I've posted a few solutions on this forum, as have others; one is here:

www.beatthegmat.com/gmat-prep-q-t14339.html
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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