I guess the question should be like 27, 54 or both. Pick some numbers to solve:
3,6 and 9 are positive consecutive multiples of 3. They are divisible by 27 (3*9) and by 54 (6*9). Hence both.

- Deepak

jnbimmer wrote:If r,s,and t are consecutive positive multiples of 3, is rst divisible by 27, 54, or both?

Because they are three multiples of 3, their product will definitely be able to divide 27, which is just 3^3. Consecutive multiples of three rotate in evenness and oddness; because they are consecutive, and because there are three of them, at least one of them is even, and so their product will definitely be able to divide 54, whose prime factorization is just 2*3^3.

Let s=r+3 and t=r+6

Then rst = r(r+3)(r+6)

Since r,s & t are each multiples of 3, the resulting number rst has three 3s as one of its factors. Therefore rst is divisible by 27 since 27=3x3x3.

Also rst must be even as at least one of rst is even. Therefore rst is divisible by 54 since an even number contains the factor 2 that is also present in 54.

Rule: A number is divisible by 3 if the sum of the digits is divisble by 3.

27: 2+7 = 9
54: 5+4 = 9

If r,s,and t are consecutive positive multiples of 3, rst is divisible by both.

nittanylion530 wrote:Rule: A number is divisible by 3 if the sum of the digits is divisble by 3.

27: 2+7 = 9
54: 5+4 = 9

If r,s,and t are consecutive positive multiples of 3, rst is divisible by both.

Hi there,

actually, using the rule the way you did doesn't quite lead us to the answer: It is a given that 27 and 54 are divisible by 3...We need to know whether rst is divisible by 27 and 54.

Let's think about 27 first. What would a number have to do in order to divide by 27 successfully? Well, because 27 is "made up of" three 3s (ie, its prime factorization is 3^3), a number would have to have three 3s in it in order to be successfully divided by 27. Well, because r, s, and t are all multiples of 3, we know rst has three 3s in it, and so of course rst can be divided by 27.

54 is just 2*27. So, in order to successfully divide by 54, a number would, in addition to having three 3s, also have to have a 2. Because these are consecutive multiples of 3, we know that at least one of them is even, and so rst has at least one 2 in it. So, rst would also be a multiple of 54.