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What is the largest number that will always divide the sum of any \(19\) consecutive positive integers?

 Feb 10, 2022
 #1
avatar+36434 
0

I think that would be 38

 Feb 11, 2022
 #2
avatar+23200 
+2

If the first of 19 consecutive positive integers is  x  then the 19th integer is  x + 18.

 

Using the formula:  Sum  =  number · (first + last) / 2

                               Sum  =  19( x + (x + 18) ) / 2

                               Sum  =  19( 2x + 18 ) / 2

                               Sum  =  19( 2(x + 9) ) / 2

                               Sum  =  19( x + 9 )

 

So, I think that the answer is 19.

 Feb 11, 2022
 #3
avatar+36434 
0

Agree with geno....   I read the question as 19 consecutive EVEN integers.....   D'Oh!    surprise

ElectricPavlov  Feb 11, 2022

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