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how do we find the first 40 positive integers divisible by 6?

SARAHann  Mar 25, 2017
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 #2
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When we want to find integers divisible by x (x=integer) we want to find integers of the form x*n when n is an integer as well.

 

Try it now, and let me know if you solved it

Ehrlich  Mar 25, 2017
 #3
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i'm sorry. this does'nt make sense

SARAHann  Mar 25, 2017
 #4
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*sigh*

 

 

6 is divisble by 6. 6=6*1

 

12 is divisble by 6. 12=6*2

 

18 is divisble by 6. 18=6*3

 

34545674756345453.34543575 is not divisble by 6, because (34545674756345453.34543575)/6 is not an integer

 

21 is not divisble by 6, because 21/6 is not an integer.

 

im sorry, i dont want to give you the answer, i want YOU to get to the final answer

Ehrlich  Mar 25, 2017
 #5
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ok i get the first part. but how do we find the last(40th) positive integer divisible by 6?

SARAHann  Mar 25, 2017
 #6
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If we know that when n is an integer 6*n is divisble by 6 starting with 1, and we want to find 40 integers that are divisble by 6 then for the last integer n will be........?

 

 

 

 

n=?

Ehrlich  Mar 25, 2017
 #7
avatar+310 
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my mistake- instead of "starting with 1" it should be "starting with n=1"

Ehrlich  Mar 25, 2017
 #8
avatar+302 
+1

oh heyyy i get it now. n=240? thats 6*40? thxxxxxxx XD

SARAHann  Mar 25, 2017

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