Re: HOT Competition: A function g is defined, for all positive integers x
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26 Aug 2020, 10:13
IMO ANSWER : C
We have,
g(x)=4x−5,x=eveng(x)=4x−5,x=even
g(x)=5x−15,x=odd,x>3g(x)=5x−15,x=odd,x>3
We can conclude from the above equation that,
If x=even , g(x)=odd.
If x=odd , g(x)=even.
Need to find : p=?
Statement (1) : g(g(p))+g(p)=195,
Case I : p=even,
g(4p-5)+4p-5=195
(5(4p-5)-15)+4p-5=195 , since (4p-5) is an odd integer.
24p-45=195
p=10
Case II : p=odd,
g(5p-15)+5p-15=195
(4(5p-15)-5)+5p-15=195 , since (4p-5) is an odd integer.
25p-80=195
p=11
Hence, Statement (1) is not sufficient.
Statement (2) : g(g(g(g(p)))) is even,
If g(g(g(g(p)))) is even => g(g(g(p))) is odd => g(g(p)) is even => g(p) is odd => p is even. But we cannot deduce the value of p
Hence, Statement (2) is not sufficient.
Combining both we get, p=10.
Hence Both statements together are sufficient