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by sourabh33 » Thu Jun 02, 2011 4:12 am
The function g(x) is defined for integers x such that if x is even, g(x) = x/2 and if x is odd, g(x) = x + 5. Given that g(g(g(g(g(x))))) = 19, how many possible values for x would satisfy this equation?

A. 1
B. 5
C. 7
D. 8
E. 11
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by cans » Thu Jun 02, 2011 4:40 am
2:42 sec
method: if g(x) is odd, it means x was even. (because if x is odd, then g(x) will be x+5, odd+odd= even and thus not possible). and thus it was divided by 2. So multiply g(x) by 2 to get x.
And if g(x) is even, either x is odd or even.. 2 options
thus g(g(g(g(g(x)))))=19
odd thus, g(g(g(g(x)))) has only one value=38
as its even, g(g(g(x))) was either 74(=38*2) or 33(38-3)
if 74, then again g(g(x)) can have 2 values. (148 or 69) and if 33, then only 1 value = 66
value of g(x) = 296,143,64,132,61
value of x = 296*2,291,286,128,59,264,127,122
IMO D
While finding the answer i didn't calculate all values
Just remember if odd, then we get even. and if even, then we get both even & odd
as 19 is odd,
1st g() = even
2nd g()=even/odd
3rd g()=even/odd/even
4th g()=even/odd/even/even/odd
5th g()=even/odd/odd/even/odd/even/odd/even
Total 8 :)
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by sourabh33 » Thu Jun 02, 2011 5:57 am
2:42!!!! is a brilliant effort

your A & E are exactly as OA & OE !!

Except, the possible values of x will be
608, 299, 294, 284, 137, 264, 127, 122

And
as 19 is odd,
1st g() = even
2nd g()=even/odd
3rd g()=even/odd/even
4th g()=even/odd/even/even/odd
5th g()=even/odd/odd/even/odd/even/odd/even
the
5th g()=even/odd/even/even/odd/even/odd/even
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by vikram4689 » Thu Jun 02, 2011 7:14 am
Nice Method Canes....thanks
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by cans » Thu Jun 02, 2011 7:36 am
sourabh33 wrote:2:42!!!! is a brilliant effort

your A & E are exactly as OA & OE !!

Except, the possible values of x will be
608, 299, 294, 284, 137, 264, 127, 122

And
as 19 is odd,
1st g() = even
2nd g()=even/odd
3rd g()=even/odd/even
4th g()=even/odd/even/even/odd
5th g()=even/odd/odd/even/odd/even/odd/even
the
5th g()=even/odd/even/even/odd/even/odd/even
Thanks for the correction. :)
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Cans!!
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