Yes. Here's why...

What we need to do is find five numbers out of which a combination of three can not be picked whose sum is divisible by three.

The easiest way to see whether or not that's possible is to look at all possible numbers as sets, grouped by their offsets from multiples of three. That gives us three sets:

a(x) = 3x + 0 = {0, 3, 6, 9, 12, 15, 18, 21, 24 ... }

b(x) = 3x + 1 = {1, 4, 7, 10, 13, 16, 19, 22, 25, ... }

c(x) = 3x + 2 = {2, 5, 8, 11, 14, 17, 20, 23, 26, ...}

There are two important things to note here:

1) First, any three numbers selected from one of those sets will add up to a multiple of three. This can be demonstrated very easily. Let's take set C. We'll pick three random numbers out of the selection, calling them x, y, and z. Their sum then would be:

3x + 2 + 3y + 2 + 3z + 2

= 3x + 3y + 3z + 6

= 3(x + y + z + 2)

which means that all possible selections will be a multiple of three. Now let's try that with set B:

3x + 1 + 3y + 1 + 3z + 1

= 3(x + y + z + 1)

again, all answers are multiples of three.

This is most obvious with set A, where the results would be expressed simply as:

3x + 3y + 3z

= 3(x + y + z)

This means that in order for our set of five numbers to meet the conditions we want, no more than two can be picked out of any of those three sets.

2) The second important note to look at is that if we pick a random number out of each of those sets, and add them together, they too will add up to a multiple of three. Here's the proof, again with our three random selections of (x, y, z):

a(x) + b(y) + c(z)

= 3x + 0 + 3y + 1 + 3z + 2

= 3x + 3y + 3z + 3

= 3(x + y + z + 1)

This means that we can't pick a number out of all three sets. Otherwise, a sum that's divisible by three can be found.

Now consider these facts together:

- we have three sets that include every possible number that we can select
- we can pick at most two numbers out of each of those sets
- we can pick numbers out of at most two of those sets

These conditions can not be met if we want to pick five numbers. We can find four that meet this condition (a pair out of any two of the sets), but if we want to pick a fifth one, it must either come from the third set, breaking our limit of two sets, or from one of the ones we've already picked from, breaking our limit of two per set.

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If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.

4 is a multiple of three numbers.

If the sum of their digits is a multiple of three, the whole number is a multiple of three.

no because first of all you should have said differentpositive whole #s because now i can say 1*1*1=1 which is not a multiple of 3but assuming they are different it is still a no because 2*1*5=10 which is not a multiple of 3

You group the numbers 11,12,1,2.Then you group the numbers 10,9,3,4.last you group the numbers 8,7,6,5.

Commas are used to separate large numbers into groups of three digits. Each group is called a period.

There are no such whole numbers. The sum of three consecutive whole numbers must be a multiple of 3; as 68 is not a multiple of 3 (68 = 3 × 22 2/3) it cannot be the sum of three whole numbers.

For this to be possible with whole numbers, 175 has to be a multiple of three. It's not, it isn't.

A multiple of 3 Also a multiple of 6.

3,6,9,12,15,18,21,24,27,30,33,36,39,42,46,49,52,55,58,61,64,67,80,83,86,89,92, 95,98 and 101 They are all the numbers I know. XD

These do exist.

128

Calculate the least common multiple of the three numbers. Any multiple of that has all those numbers as factors.

EVERY three consecutive numbers add to a multiple of 3: Proof: numbers are n, n + 1 and n + 2. The total is 3n + 3 or 3(n + 1) This means that for any three consecutive numbers, the total is 3 times the middle number.

Period

idont know

The answer is 9 × 6 × 7

I'll guess those are three-digit numbers. The way to find if a number is a multiple of 3 is to total its digits. If that total is a multiple of 3, the whole number is a multiple of 3. 285 and 126, yes. 770, 176, 410, 452, 650, no.

It means write down three pairs of numbers that are relatively prime like 4 and 9, 5 and 6, 13 and 25

That doesn't work. The number has to be divisible by three. Any three consecutive numbers add up to a multiple of three.

10

The LCM of the given three numbers is 60

Yes.

it will not since it is not a multiple of 3. 99 is a multiple of three so the next multiple would be 102.

Amongst others they can be 2+3+15 = 20 which is a multiple of 10