GO Classes CS/DA 2025 | Weekly Quiz 2 | Fundamental Course | Question: 7

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GO Classes asked Mar 8, 2023 • edited Mar 9 by shadymademe

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commented Mar 8, 2023

a.b.c will always be necessarily even.

There will be 3 cases possible :

$a$ $is$ $even$ $,$ $b$ $is$ $even$ . $In$ $this$ $case$ $c$ $will$ $also$ $be$ $even$ . $So$ $,$ $a$$.$$b$$.$$c$ $will$ $be$ $even$
$a$ $is$ $even$ $,$ $b$ $is$ $odd$ . $In$ $this$ $case$ $c$ $will$ $be$ $even$ $because$ $a^2$ $+$ $b^2$ $will$ $be$ $(even)^2$ $+$ $(odd)^2$ $which$ $will$ $even$. $So$ , $a.b.c$ $will$ $be$ ($even$ $.$ $odd$ $.$ $even$) $which$ $will$ $be$ $even$.
$a$ $is$ $odd$ $,$ $b$ $is$ $even$ . $In$ $this$ $case$ $c$ $will$ $be$ $even$ $because$ $a^2$ $+$ $b^2$ $will$ $be$ $(odd)^2$ $+$ $(even)^2$ $which$ $will$ $even$. $So$ , $a.b.c$ $will$ $be$ ($odd$ $.$ $even$ $.$ $even$) $which$ $will$ $be$ $even$.

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@chokostar

a.b.c will always be necessarily even.

There will be 3 cases possible :

$a$ $is$ $even$ $,$ $b$ $is$ $even$ . $In$ $this$ $case$ $c$ $will$ $also$ $be$ $even$ . $So$ $,$ $a$$.$$b$$.$$c$ $will$ $be$ $even$

$a$ $is$ $even$ $,$ $b$ $is$ $odd$ . $In$ $this$ $case$ $c$ $will$ $be$ $even$ $because$ $a^2$ $+$ $b^2$ $will$ $be$ $(even)^2$ $+$ $(odd)^2$ $which$ $will$ $even$. $So$ , $a.b.c$ $will$ $be$ ($even$ $.$ $odd$ $.$ $even$) $which$ $will$ $be$ $even$.

$a$ $is$ $odd$ $,$ $b$ $is$ $even$ . $In$ $this$ $case$ $c$ $will$ $be$ $even$ $because$ $a^2$ $+$ $b^2$ $will$ $be$ $(odd)^2$ $+$ $(even)^2$ $which$ $will$ $even$. $So$ , $a.b.c$ $will$ $be$ ($odd$ $.$ $even$ $.$ $even$) $which$ $will$ $be$ $even$. — Harsh Saini_1, Mar 8, 2023
However now I am confused with option (a) now because all a,b,c need not be even.

@chokostar

In Option A, $(abc) $ is the Multiplication of $a,b,c.$ Similarly, $(bc) = b \times c.$

Since, at least one of $a,b$ must be even, So, $a \times b \times c = even.$ — Deepak Poonia, Mar 8, 2023
If a, b, c are three integers and $a^ 2 +b^ 2 =c^{2}$ then a or b is even.

and any even number multiplied with any other number results as even. So, in abc either a or b is even which implies product abc is even , and same will be the case for product ab.

That is why option (a) and (c) will be always true .( note : only for a, b, c $\epsilon$ $\mathbb{Z}$ ) — Priyanshu Dubey, Aug 13, 2023

GO Classes · Answer 1 · 2023-03-08T02:38:41+0000

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