The sum of all integers from 43 to 107, inclusive, is divisible by whi

Question

The sum of all integers from 43 to 107, inclusive, is divisible by which of the following numbers?

A. 17
B. 18
C. 21
D. 24
E. 39

niks18 · Answer

Total numbers from 43 to 107  = 107-43+1=65

Sum of nos from 43 to 107  = \frac{65}{2}(43+107) = 65*75 = 3*5^3*13

from among the options only  39  divides it.

Option  E

pushpitkc · Answer

Sum of n positive integers is  \frac{n(n+1)}{2}

The sum of all integers between 43 and 107 can be found out as follows:

Sum of all positive integers till 107 is  107*54 
Similarly, the sum of all positive integers till 42 is  43*21 
Therefore, the sum of all integers from 43 to 107 is  107*54 - 43*21 = 5778 - 903 = 4875

4875  when prime factorized is  3*5^3*13  and the only number which divides this is 39(Option E)

HouseStark · Answer

I did not comprehend the highlighted part of the above.

it supposed to be 43*22
if you calculate using formula n(n+1)/2

GMATinsight · Answer

PROPERTY :

Sum of n positive integers,  1+2+3+.......+n = \frac{n(n+1)}{2}

Sum of any Arithmetic Progression  = \frac{n}{2}

Method-1:

Sum of all Integers from 43 to 107 =  (1+2+3+....+107)  -  (1+2+3+.....+42)

Sum of all Integers from 43 to 107 =   (\frac{1}{2})*107*108   -   (\frac{1}{2})*42*43

Sum of all Integers from 43 to 107 =   107*54   -   21*43

Sum of all Integers from 43 to 107   = 4875 = 25*195 = 5^2*5*39

Answer: Option E

Method-2:

Number of terms from 43 to 107 inclusive = 107-42 = 65

Sum of all integers from 43 to 107  = (\frac{65}{2})  = (\frac{65}{2})  = 13*5*75 = 13*3*5*5^2

Answer: Option E

abhishek13916

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The sum of all integers from 43 to 107, inclusive, is divisible by whi

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The sum of all integers from 43 to 107, inclusive, is divisible by whi

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