NP Guide (3rd Ed) - Divisibility & Primes p.106

[hellokimkim36]

I have a question regarding In Action Problem #25 on p.106 of the number properties (3rd Edition).

If x,y, and z are integers, is x even?
1) 10^x = (4^y)*(5^z)

--> I got (2^x)*(5^x) = (2^2y)*(5^z)
from this, I can infer that 2^x = 2^2y and 5^x = 5^z since 2 and 5 are primes.

Thus --> x=2y which i know x is for sure even;
BUT: -->x = z which i DO NOT know anything about z.

Therefore I think A is insufficient because x can be both even (2y) or odd (since z can be odd or even). However, the answer is that A is sufficient.

Please explain why A is sufficient.

Thanks.

[Ben Ku]

If x,y, and z are integers, is x even?
1) 10^x = (4^y)*(5^z)

--> I got (2^x)*(5^x) = (2^2y)*(5^z)
from this, I can infer that 2^x = 2^2y and 5^x = 5^z since 2 and 5 are primes.

Thus --> x=2y which i know x is for sure even;
BUT: -->x = z which i DO NOT know anything about z.

Therefore I think A is insufficient because x can be both even (2y) or odd (since z can be odd or even). However, the answer is that A is sufficient.

I think that x = 2y (x is even) and x = z (z is an integer) do NOT contradict each other. Basically x must fit BOTH conclusions, so x must be even.

[hellokimkim36]

Thank you for your response. I agree with your explanation.

[Ben Ku]

Glad it helped!

[ankitp]

I have a question though I disagree with the solution.

Statement A by itself is NOT SUFFICIENT, x could = 0, which is not odd.

You need statement 1 and 2 to be sufficient , as statement 2 guarantees x != -

[Ben Ku]

I have a question though I disagree with the solution.

Statement A by itself is NOT SUFFICIENT, x could = 0, which is not odd.

You need statement 1 and 2 to be sufficient , as statement 2 guarantees x != -

Can you expand on your thoughts? I don't really understand your question. The problem asks whether x is even. As we worked out above, we figured out that in statement (1), x must be even. You're right, x could be 0; this fits what we said about statement (1) and does not present a counterexample.

[sudhir.18n]

Hi ,
Reviving a very old thread..
Was just going through this problem on NPG 4 Ed.
Fairly easy one ; but the nly doubt I have is '
Do we have to assume in exponential questions that the integers are +ve? Bcoz the question stem says "If x, y, and z are integers, is x even?" cant X, Y , Z be negative?

Thanks
Sudhir

[jnelson0612]

Hi ,
Reviving a very old thread..
Was just going through this problem on NPG 4 Ed.
Fairly easy one ; but the nly doubt I have is '
Do we have to assume in exponential questions that the integers are +ve? Bcoz the question stem says "If x, y, and z are integers, is x even?" cant X, Y , Z be negative?

Thanks
Sudhir

x, y, and z could possibly be negative, but the issue is whether x is even. We know that x=2y, so even if y=-1 and x=-2, x would still be even. Negative numbers can absolutely be distinguished as even and odd.