If $a, b \in \mathbb{Z}$ and $a≡b(mod3)$ and $a≡b(mod5)$, then $a≡b(mod15)$

Question

I don't know how to approach this question.

All I know is that:

$a≡b(mod 3)$ $\rightarrow $ $a = 3k + b$, $ m \in \mathbb{Z}$

$a≡b(mod 5)$ $\rightarrow $ $a = 5m + b$, $ n \in \mathbb{Z}$

I don't know where to go from here...

$3k = 5m$, so $3 | m$ and $5 | k$. Thus $a = 15l +b$ – Jorge Dec 10 '21 at 07:12 — Jorge, Dec 10 '21 at 07:12
See CCRT in the linked dupe. – Bill Dubuque Dec 10 '21 at 08:12 — Bill Dubuque, Dec 10 '21 at 08:12

score 0 · Accepted Answer · answered Dec 10 '21 at 07:45

0

Based on the given conditions we can say that a-b is a multiple of both 3 and 5 and thereby it is a multiple of 15 too.

a = 3k+b
a = 5m+b
3k+b = 5m+b
3k = 5m = x

x is divisible by both 3 and 5 and hence by 15.

answered Dec 10 '21 at 07:45

Anuneet Anand

1 Answers1