A number theory problem by Shriniketan

Number Theory Level 2

What is the sum of all three-digit numbers which leave a remainder 3 3 when divided by 5 5 ?

99900 99009 90909 99090

Shriniketan Ruppa by Shriniketan Ruppa , 15, India

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3 solutions

Razing Thunder
Jul 3, 2020 
1
2
3
4
5
 
x=[]
for i in range(100,1000):
    if i%5==3:
        x.append(i)
print(sum(x))

0 Helpful 1 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Jun 27, 2020

Let the sum be S S , then

S = 103 + 108 + 113 + ⋯ + 998 = ∑ n = 1 180 ( 5 n + 98 ) = 5 × 180 ( 180 + 1 ) 2 + 98 × 180 = 99090 \begin{aligned} S & = 103 + 108 + 113 + \cdots + 998 \\ & = \sum_{n=1}^{180} (5n+98) \\ & = 5 \times \frac {180(180+1)}2 + 98 \times 180 \\ & = \boxed{99090} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
X X
Jun 27, 2020

103 + 108 + . . . + 993 + 998 = 180 ( 103 + 998 ) 2 = 99090 103+108+...+993+998=\frac{180(103+998)}2=99090

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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