Let

. Then

. By convention, every non-zero integer

divides 0, so

.
Suppose this relation holds for

, i.e.

. We then hope to show it must also hold for

.
You have

We assumed that

, and it's clear that

because

is a multiple of 3. This means the remainder upon divides

must be 0, and therefore the relation holds for

. This proves the statement.
Answer:
$32.50
Step-by-step explanation:
He needs $395, cuz he already had $5, and 15×20=300. subtract the 5 from 300.
Answer:
yes
Step-by-step explanation:
can i have brainliest
2^4 x 3 is a prime factor of 48