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Find the sum of all positive integers less than 1000 ending in 3 or 4 or 6 or 9.

 May 14, 2024
 #1
avatar+9673 
+1

\(\quad \text{Required sum}\\ = (3 + 13 + 23 + \cdots + 993) + (4 + 14 + 24 + \cdots + 994) + (6 + 16 + 26 + \cdots + 996) + (9 + 19 + 29 + \cdots + 999)\\ = 40(1 + 2 + \cdots + 99) + 100(3 + 4 + 6 + 9)\\ = 200200\)

.
 May 14, 2024
 #2
avatar+129883 
+1

Note that

3 + 9 =  2(6)

13 + 19 = 2(16)

23 + 29 = 2(26)

etc.

 

And

4 + 6 = 2(5)

14 + 16 = 2(15)

24 + 26 = 2(25)

etc.

 

So

 

2 ( 6 + 16 + 26 + ....+ 986 + 996)  =  2 * [ 6 + 996] (100/2)  =  100200

 

+

 

2 ( 5 + 15 + 25 + ... + 985 + 995 )  = 2 * (5 + 995) (100/2) = 100000 

 

100200 + 100000 = 

 

200200

 

cool cool cool

 May 14, 2024

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