Answer:
The 3 ses that we found are:
{4, 5, 6}
{64, 65, 66}
{124, 125, 126}
Explanation:
3 consecutive numbers are written as:
n, (n + 1), (n + 2)
We know that the first one is a multiple of 4.
Then n is a multiple of 4:
n = k*4
where k is an integer.
The second one is a multiple of 5, then:
(n + 1) = j*5
where j is an integer.
The last one is a multiple of 6.
(n + 2) = p*6
where p is an integer.
We want to find 3 sets.
The first one is trivial:
n = 4
n + 1 = 5
n + 2 = 6
4 is a multiple of 4
5 is a multiple of 5
6 is a multiple of 6.
Now let's find a set that is not trivial.
First, remember that all the multiples of 5 end with a 0 or a 5,
Then we can look for a value n, that is multiple of 4, and that has a last units digit equal to 4 (then the next one will have a units digit and will be multiple of 5)
For example, 4*6 = 24
24 is a multiple of 4.
The next number is 24 + 1 = 25, which is multiple of 5.
The next number is 25 + 1 = 26, which is not multiple of 6.
So let's try again.
4*11 = 44, is a multiple of 4.
The next number is 44 + 1 = 45
45 is a multiple of 5.
The next number is 45 + 1 = 46, which is not multiple of 6
So we need to try with another set.
4*16 = 64, is a multiple of 4.
The next number is 64 + 1 = 65, which is multiple of 5.
The next number is 66, which is multiple of 6:
6*11 = 66
Then the set:
n = 64
n + 1 = 65
n + 2 = 66
Is a possible set.
Now we can keep trying this, we can see that the next set has the numbers:
n = 4*31 = 124 is a multiple of 4.
The next number, is 124 + 1 = 125, which is a multiple of 5.
The next number is 125 + 1 = 126, which is multiple of 6
6*21 = 126
Then the set:
n = 124
n + 1 = 125
n + 2 = 126
Is another possible set.
The 3 ses that we found are:
{4, 5, 6}
{64, 65, 66}
{124, 125, 126}