Latest elitmus Aptitude Question SOLUTION: a series of natural number 1+2+3+4+5+6+7+8+9+10 is divisibel by 11 so we has to find how many such series is possible upto n where n is less than 1000. a.90 b.99 c.180 d 96

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(#M40156441) ELITMUS Numerical Ability question number series Keep an EYE

a series of natural number 1+2+3+4+5+6+7+8+9+10 is divisibel by 11 so we has to find how many such series is possible upto n where n is less than 1000.
a.90
b.99
c.180
d 96

Asked In elitmus rajat (8 years ago)

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vinay kumar  (7 years ago)

the total multiple of 11 less then 1000 are 90 .
and for every multiple there are two such series which are divisible by 11.
. .
. sum of series = n(n+1)/2
for multiple 11 we can take n=10 and 11
for 22 '' '' ''' n=21 and 22
and so on and for 990 multiple of 11 we can take n=989 and 990
so total series formed are 90*2= 180

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Mallikarjuna Reddy  (8 years ago)

99 is the answer
sum of n natural numbers is n(n+1)/2 for n(n+1)/2 is divisible by 11 then n(n+1) should be divisible by 22 because 2 and 11 co-primes now find out possible n
n can take 22,44,66,88,....,990 i.e., 22 multiples less than 1000 there are 45 such numbers less than 1000
n can also take 21,43,65....989 i.e., 1 less than 22 multiples again there are 45 such numbers less than 1000
we need to look other possibilities as n(n+1) should be divisible by 22 then n(n+1) should be divisible by 2 and 11 so find whether there are consecutive numbers whose product is divisible by 11 and 2 they are
10*11
11*12
121*122
231*232
452*452
561*562
671*672
781*782
891*892
those are obtained by multiples of 11 which end with 1 bcz next number is even which is divisible by 2 so 11,121,231,451,561,671,781,891 are multiples of 11 so hence there are 9(above mentioned) possibilities
Total=45+45+9=99

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Priyanka  (7 years ago)

ans is 99 they are asking how many such series is possible
if we go through 11 to 99 then 9 such series are possible 10(1+2+...9),20(1+2+3..) already given series is divisible by 11..from 100 to 999 there is 90 such series possible for ex 100(1+2+3+4+5..) for 110(1+2+3+4.....)....990(1+2+3...+9)
total 90+9 =99 such series are possible
if i am wrong plzz correct me

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KUMAR HARSH  (7 years ago)

please admin gives us correct ans

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Vikash kumar  (7 years ago)

180 is the correct answer.

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kamakshi gupta  (7 years ago)

Sum of the series=n(n+1)/2
for being it divisible by 11 , there should be n= multiple of 11 or (n+1)= multiple of 11

for 1st condition n can be:
11,22,33,..........,990 ( a+ (n-1)d= 990 ,a=11, d=11 => n=90

for 2nd condition (n+1) can be:
10,21,32,...........,989 ( a+ (n-1)d= 989 ,a=10, d=11 => n=90

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Shubham Kanwal  (8 years ago)

@MALLIKARJUNA REDDY .Why did u take 2 as a co-prime of 11? There are also others co-prime of 11 like 3,4....
And how did u calculated how many nos are there b/w 22 and 990?

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Mallikarjuna Reddy  (8 years ago)

sorry the ans I think is 180 bcz all multiples of 11 for n is satisfied 11

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