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Mathematics 9 Online
OpenStudy (anonymous):

A list contains exactly 6 different counting numbers. No number in the list is a multiple of any other in the list. What is the least possible total of these 6 numbers? help me!! @jim_thompson5910 @pooja195

OpenStudy (anonymous):

i got no number satisfied

OpenStudy (whpalmer4):

Do you remember the Sieve of Eratosthenes?

OpenStudy (anonymous):

i don't know what is that?

OpenStudy (whpalmer4):

Okay, well, then you don't remember it, no problem, we'll reinvent it :-) Make a list of numbers from 1 to 100 on a piece of paper. Cross out 1, because everything is a multiple of 1. Circle 2. Now cross out all of the multiples of 2. What's the next number that isn't crossed out? Well, it's 3. Circle 3. Cross out all the multiples of 3. 4 should already be crossed out, so that brings us to 5. Circle 5. Cross out all the multiples of 5 (half of them already will be). Keep on in this fashion until all numbers are crossed out, or you've got 6 numbers circled. Add up the circled numbers, and you've got your answer.

OpenStudy (whpalmer4):

You could actually go with just 1-50, no need to go to 100 for this.

OpenStudy (whpalmer4):

By our procedure, we'll have the 6 lowest numbers which are not multiples of anything (except 1), and that makes them so-called prime numbers.

OpenStudy (whpalmer4):

As it turns out, the answer is also a prime number!

OpenStudy (anonymous):

Yes i got the prime numbers to but no solution there

OpenStudy (whpalmer4):

What do you mean, no solution there?

OpenStudy (anonymous):

no prime number would satisfies

OpenStudy (whpalmer4):

No prime number would satisfy what? Your problem was to make the list of 6 counting numbers where no number on the list is a multiple of any other, and find the set of such numbers that would give you the smallest possible sum. If you take the first 6 numbers that are circled and sum them, that is the smallest possible sum.

OpenStudy (anonymous):

if i entry the number is 2,3,5,7,...,... what are 2 numbers else?

OpenStudy (whpalmer4):

Did you complete the sieve?

OpenStudy (anonymous):

so 2,3,5,7,11,13 it contains 8 numbers and the 1=3 times 3=2 times

OpenStudy (whpalmer4):

You only need the first 6. I don't know what your 1=3 times 3 = 2 times bit is about...

OpenStudy (whpalmer4):

2+3+5+7+11+13 = ?

OpenStudy (anonymous):

41

OpenStudy (anonymous):

then?

OpenStudy (whpalmer4):

Go read your problem statement again.

OpenStudy (anonymous):

ok i got it :)

OpenStudy (anonymous):

thank you very much

OpenStudy (whpalmer4):

Learning to recognize when you've found the answer (or that you haven't found it, or it isn't in the correct form) is an important skill...

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