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If a, b, and c are all integers, is ab + bc + ca + a^2 odd?

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If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   16 Feb 2021, 01:15
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If a, b, and c are all integers, is \(ab + bc + ca + a^2\) odd?

(1) a is odd.
(2) (b + c) is odd.

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If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   Updated on: 16 Feb 2021, 08:52
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The terms can be rearranged and written as \(a^2 + a(b + c) + bc\)


Statement 1: a is odd

We do not know about b and c, if they are odd or even.

For eg, if b and c are both even, we have O + O(E + E) + E = O + O *E + E = O + E + E = O + E = Odd

if b and c are both Odd, we have O + O(O + O) + O = O + O *E + O = O + E + O = Even


Therefore Statement 1 Alone is Insufficient. Answer options could be B, C or E



Statement 2: b + c is Odd

This means that one of them has to be odd and the other even, so bc = Even


If a is Odd, then we get O + (O * O) + E = O + O + E = O + O = Even

If a is even, then we get E + (E * O) + E = E + E + E = Even


Therefore Statement 2 Alone is Sufficient.



Option B

Arun Kumar

Originally posted by CrackverbalGMAT on 16 Feb 2021, 06:34.
Last edited by CrackverbalGMAT on 16 Feb 2021, 08:52, edited 1 time in total.
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If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   16 Feb 2021, 07:18
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Let us simplify this expression as a(a+b+c)+bc. We have to find whether this expression is odd or even.

Let us denote odd as O and even as E.

Statement 1:a is odd.

applying this to the expression O(O+b+c)+ bc

Whether bc is odd or even gives different results for the expression. We don't know whether b and c are odd or even. INSUFFICIENT.

Statement 2: (b+c) is odd.

This means bc is even.

Applying this,a(a+O)+E.

Assume a is odd.

O(O+O)+E gives O*E+E which makes the result Even.

Assume a is even

E(E+O)+E gives E*O+E which makes the result Even.

SUFFICIENT.

Ans: B

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Re: If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   16 Feb 2021, 12:48
If a, b, and c are all integers, is ab+bc+ca+a^2 odd?

Stat1: a is odd; so expression = even/odd (depends on b) + even/odd (depends on b & c) + even/odd (depends on b & c) + even/odd (depends on c) + odd; we need b and c. Not sufficient.

Stat2: (b + c) is odd; so either b is odd and c is even or vice versa.
Case1: if b = even and c= odd, expression = even + even + even/odd (depends on a) + even/odd (depends on a); if a is even, then expr = even and if a= odd, expr. = even
Case2: if b = odd and c= even, expression = even/odd (depends on a) + even + even + even/odd (depends on a); if a is even, then expr = even and if a= odd, expr. = even. Sufficient

So, I think B. :)
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Re: If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   17 Feb 2021, 00:43
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If a, b, and c are all integers, is ab+bc+ca+a^2 odd?


ab+bc+ca+a^2
= b(a+c) + a(a+c)
(a+b)(a+c)

(1) a is odd
No info. about b and C

Not sufficient

(2) (b + c) is odd
So, b is even and c is odd or vice-versa.

Let a be odd

(a+b)(a+c)
(odd + even)(odd + odd)
odd × even
even

Let a be even

(a+b)(a+c)
(even + even)(even + odd)
even × odd
even

So, ab+bc+ca+a^2 is not odd.
Sufficient

Option B

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Re: If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink] New post   14 Oct 2023, 07:46
in first statement value of bc is not given but for any value odd or even answers comes out to be even. so it should be enough right?
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Re: If a, b, and c are all integers, is ab + bc + ca + a^2 odd? [#permalink]
14 Oct 2023, 07:46
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Bunuel
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