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Number Theory Level 3

Find the sum of all three digit numbers which leave the remainder 2 when devided by 5


The answer is 98910.

Aman Sharma by Aman Sharma , 25, India

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2 solutions

Omkar Kulkarni
Jan 1, 2015

The terms involved form an A.P. with a = 102 a=102 , d = 5 d=5

a n = 997 a_{n}=997

a + ( n − 1 ) d = 997 a+(n-1)d=997

n = 180 n=180

Therefore S n S_{n} = n 2 ( 2 a + ( n − 1 ) d ) \frac {n}{2}(2a+(n-1)d)

S n = 98910 \boxed {S_{n}=98910}

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Hansraj Sharma
Aug 14, 2014

102,107,112........,997 997=102+(n-1)5 n=180 Sn=90[204+(179)5] => 98910

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Daniel Sbs - 6 years, 9 months ago

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