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Division without Reminder

Yankel

Active member
Jan 27, 2012
398
Hello,

I got a very basic question...

A number n is dividable by 15 and 18. Can I assume from that that it is dividable by 27?

(dividable - you can divide it by 15 and get no reminder).

If it is dividable by 15, it is by 3 and 5. If by 18, it is dividable by 3 and 6, which means 3 and 2.

Can I say that since not every number that is dividable by 3 is also dividable by 9, this number is not dividable by 27, not necessarily anyway ?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
The least common multiple of 15 and 18 is 90, which is divisible by both 15 and 18. but not 27. However, 270 is also divisible by 15, 18 ... and 27.

So, one cannot assume the number n is divisible by 27, yet that doesn’t mean there is no value of n divisible by 27.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Hello,

I got a very basic question...

A number n is dividable by 15 and 18. Can I assume from that that it is dividable by 27?

(dividable - you can divide it by 15 and get no reminder).

If it is dividable by 15, it is by 3 and 5. If by 18, it is dividable by 3 and 6, which means 3 and 2.

Can I say that since not every number that is dividable by 3 is also dividable by 9, this number is not dividable by 27, not necessarily anyway ?
If a number is divisible by 15, it is divisible by 3 and 5.

If a number is divisible by 18, it is divisible by 2, 3 and 3.

So if the number is divisible by both 15 and 18, it is divisible by 2, 3, 3 and 5.

There is not any multiplicative combination of 2, 3, 3 and 5 to give 27. So no, we can not assume that the number is divisible by 27.