1057@aditya how can u say dat...
evn 1+2+3+4....+11 is 66.. :P
find the smallest number that can be represented by a sum of 9,10 and 11 consecutive integers.....?
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12 Answers
262i dont think there is any such number.
sum of 9 consecutive integers is always even
sum of 11 consecutive integers is always odd
contradiction.
11@aditya- sum of 11 consecutive integers is not always odd. ex-1,2,...,11
i think u meant sum of 10 consecutive integers is odd.
also sum of 9 consecutive integers isnot always even. ex- 1,2,...,9
21Getting 495,where 40,45,51 are the starting integers of the 11,10,9 no. consecutive series respectively.
1057maybe u'all are correct....i dun know de answer but your answer does match de options given....
21Solution:
let the starting integers of the 9,10,11 no. series be x,y,z.
Let the sum of each be S.
S=9x+(0+1+....+8)=9x+36=9(x+4)
S=10y+(0+1+....9)=10y+45=5(2y+9)
S=11z+(0+1+.....10)=11y+55=11(y+5)
Therefore S has to be a multiple of 5,9,11.
The least S=5*11*9=495(ans)
[Note that 0 would also satisfy but then for 2nd sum we would get 2y+9=0,which is not possible for integral y]
262ohh !! my bad.
for 9 , let N = a +(a+1) + ... +(a+8) = 9a+36 =9(a+4)
for 10, N = b+(b+1) + ..... + (b+9) = 10b +45 = 5(2b+9)
for 11 , N= c+(c+1) + ..... +(c+10) = 11c +55 = 11(c+5)
therefore N is surely of the form - (9*5*11 )k = 495 k (k→Natural number)
and 495 satisfies the condition
hence smallest number is 495
