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Question

Find the sum of all 3 digit natural numbers, which are divisible by 8

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Solution

The three digit natural numbers divisible by 8 are 104,112,120,....,992.
Let Sn denote their sum. That is, Sn=104+112+120+128+,....+992.
Now, the sequence 104,112,120,...,992 forms an A.P.
Here, a=104,d=8 and l=992.
n=lad+1=9921048+1
=8888+1=112.
Thus, S112=n2[a+l]=1122[104+992]=56(1096)=61376.
Hence, the sum of all three digits numbers, which are divisible by 8 is equal to 61376.

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