Every Number Serves A Purpose

15 is not divisible by 6.
But for an integer A A , the product 15 × A 15\times A is divisible by 6.

Is it true that A A must be divisible by 6?

Yes, it is true No, it is not true

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ronald Chén
Nov 6, 2016

No. For example, A might be 2. The reason is that the number 3, which is one of the prime factors of the number 6, also belongs to the decomposition of the number 15. Thus we can only be sure that A is even.

Maybe this problem should be extended by replacing 15 and 6 with variables so that you cannot find the number 2.

William Nathanael Supriadi - 4 years, 7 months ago
Viki Zeta
Nov 7, 2016

6 15 A 15 A = 6 m + 0 5 A = 2 m 2 5 A 2 A 6|15A \\ \implies 15A = 6m + 0 \\ \implies 5A = 2m \\ \implies 2 | 5A \\ \implies 2 | A

Or

15 A 0 m o d 6 5 A 0 m o d 2 A 0 m o d 2 15A \equiv 0 ~ \mathrm{mod}~6 \\ \implies 5A \equiv 0 ~ \mathrm{mod}~2\\ \implies A \equiv 0 ~ \mathrm{mod} 2

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...