Sum of Numbers

If all 6 are replaced by 9 then sum of all the numbers from 1 to 100 is?

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madhavimca

  • Oct 11th, 2008
 

Sum of 1....100=(100(101))/2 [formula n(n+1)/2]
The no '6' occurs at 10 times in 1st place from 1....100
so 10*3=30 [6 is incremented by 3]
At 10th place, no '6' occurs 10 times from 1....100
so 10*30=300
sum=5050+300+30 =5380

Ans:5380

vivek.rick

  • Nov 30th, 2009
 

n=100, sum of nos=n(n+1)/2
i.e.. 100(100+1)/2=5050.
Single 6 occurs 9 times from 1 to 100 and 9-6=3.
So 9*3=27.
Double 6 occurs 1 time, 99-66=33.
So the new sum is 5050+27+33=5110.

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Sorry, In my previous posting I missed one number. I will correct it now

Sum for 1 to 100 is 5050.

5050 - (6+16+26+36+46+56+60+61+62+63+64+65+66+67+68+69+76+86+96) + (9+19+29+39+49+59+90+91+92+93+94+95+96+97+98+99+79+89+99) = 5380

 

Sum for 1 to 100 is 5050.

5050 - (6+16+26+36+46+56+60+61+62+63+64+65+66+67+68+69+76+86+96) + (9+19+29+39+49+59+90+91+92+93+94+95+99+97+98+99+79+89+99) =5380

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krishna127

  • Apr 1st, 2010
 

Sum of 1 to 100 is 5050. Number 6 is repeted 10 times between 1 to 100
so (3*10=30)
Now add the 30 to actual answer
(5050+30=5080).
Therefore the answer is 5080

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chirutha48

  • May 29th, 2010
 

6 in tens digits are also replaced by 9's. so +30
6 in units place replaced by 9's so +3
therfore sum of 1 to 100 5050
5050+3*10+30*10=5380

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9-6=3

and every change in 9 with 3 will get a diffenrence of 3
=>   there will be 10 no's with 3 in units place whose place value is 1
         =>     there will be increase in sum by (3*1)*10=30
=>   and also there will be 10 no's with 3 in tens place whose place value is 10
          =>    there will be increase in sum by (3*10)*10=300
=>   total increase will be 300+30=330
       there fore the sum will be 1+2+3+.......+100+330=  5050+330=?

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SAIKIRAN

  • Sep 17th, 2011
 


1416

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