# Can you guarantee that if you have a group of five numbers you can pick out three that will add up to a multiple of 3?

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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:
1. we have three sets that include every possible number that we can select
2. we can pick at most two numbers out of each of those sets
3. 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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# Can you guarantee that it is always possible to choose three numbers that will add up to a multiple of three from any set of five positive whole numbers?

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

# What are three numbers from 11 to 19 that add up to make a multiple of 10?

There are not any numbers from 11 through 19 that are multiples of 10.

# Three consecutive numbers that add up to a multiple of three?

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 numb

90 + 3 + 6

15

Three is.

# What three numbers adds up to 15?

1, 2 and 12 is one possible triplet.

# What three numbers add up to 87?

1,2 and 84 is one possible triplet. -100, 87 and 100 is another.

# What three numbers add up to 865?

Multitudes of combinations...in what context? Prime numbers or what?

# What are three numbers add up to 50?

Do they have to be all the same numbers? If not, you're a retard if you don't know that! (: just kiddin. I love you!

# How do five threes add up to 100?

They don't. Five threes add up to 15 and that is it! You can write an expression using 5 threes that makes 100, but that is not the same.

# What three numbers add up to 7749?

0 + 0 + 7749. Actually, there are infinitely many solutions to this question.

# How many numbers must be selected from the set 1 2 3 4 5 6 to guarantee that at least one pair of these numbers add up to 7?

1 and 6 . 2 and 5 . 3 and 4

12

# What three multiples add up to get 42?

Multiples don't add. Addends add, like 13 + 14 + 15 = 42 but I suspect you want factors,specifically prime factors. 2 x 3 x 7 = 42

# Which multiples of 3 add up to 36?

12 lots of 3 add up to 36
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# What odd three 3 number has digits that are all the same and add up to 9?

They are: 3+3+3 = 9