MHB Is a Number Divisible by 15 and 18 Also Divisible by 27?

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A number divisible by both 15 and 18 is not necessarily divisible by 27. While it is true that being divisible by 15 implies divisibility by 3 and 5, and being divisible by 18 implies divisibility by 2 and 3, the combination does not guarantee divisibility by 27. The least common multiple of 15 and 18 is 90, which is not divisible by 27. Therefore, one cannot assume that a number divisible by 15 and 18 is also divisible by 27.
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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 ?
 
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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.
 
Yankel said:
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.
 
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