Latest Elitmus Aptitude Question SOLUTION: Q. 1+2+3+4 which is divisible by 10. Here n is 4. So how many series are possible, starting with 1 such that sum of series is divisible by 10. n should be less than 1000.

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(#M40031122) ELITMUS Numerical Ability question Keep an EYE

Q. 1+2+3+4 which is divisible by 10. Here n is 4. So how many series are possible, starting with 1 such that sum of series is divisible by 10. n should be less than 1000.

Asked In Elitmus (10 years ago)

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RAKESH  (10 years ago)

Tthis one is CORRECT
sum of first n natural no's=n(n+1)/2
given that, sum should be multiple of 10 and n should less than 1000
so n can be
4,24,44,...,984 gives (984-4)/20 +1=50 series
15,35,55,...,995 gives (995-15)/20 +1=50 series
19,39,59,...,999 gives (999-19)/20 +1=50 series
20,40,60,...,980 gives (980-20)/20 +1=49 series
total no. of series=50+50+50+49=199
ans 248

Is this solution Helpfull? Yes (33) | No (5)  

RAKESH  (10 years ago)

sum of first n natural no's=n(n+1)/2
given that, sum should be multiple of 10 and n should less than 1000
so n can be
4,24,44,...,984 gives (984-4)/20 +1=50 series
15,25,35,...,995 gives (995-15)/10 +1=99 series
19,39,59,...,999 gives (999-19)/20 +1=50 series
20,40,60,...,980 gives (980-20)/20 +1=49 series
total no. of series=50+99+50+49=248
ans 248

Is this solution Helpfull? Yes (14) | No (7)  

RAKESH  (10 years ago)

sum of n natural no's =n(n+1)/2
sum should be less than 1000
so n should be less than 45
now n(n+1)/2 should multiple of 10 which is satisfied for
n=4,15,19,20,24,25,35,39,40,44
so no. of possible series is =10

Is this solution Helpfull? Yes (2) | No (3)  

ANSHUL KUMAR  (10 years ago)

n(n+1)/20
n:- 4,14,24,34............994

994-4/10=99
99+1=100

100 series possible

Is this solution Helpfull? Yes (1) | No (7)  

Bharadwaj  (10 years ago)

****
n(n+1)
should be a multiple of 20****

case1: n or n+1 is a multiple of 20

possibilities: 19,39,59...999 , 20,40,60....980 (n

Is this solution Helpfull? Yes (1) | No (3)  

Ajay Sharma  (10 years ago)

Sn=n(n+1)/2
Therefore, n(n+1)/2 must be divisible by 10
Or, n(n+1) must be divisible by 20

CASE 1: take the following consecutive numbers
4,5,6 [4*5 divisible by 20]
14,15,16 [15*16 divisible by 20]
24,25,26 [24*25 divisible by 20]
34,35,36 [35*36 divisible by 20]
44,45,46 [44*45 divisible by 20]
.
.
.
984,985,986 [984*985 divisible by 20]
994,995,996 [995*996 divisible by 20]
From above steps it is clear that there is
an alternative number around the 5's multiple
which is producing the combination to
produce a number divisible by 20.
COUNT_1=100

CASE 2: take the following consecutive numbers
19,20,21 [19*20 and 20*21 both are divisible by 20]
39,40,41 [39*40 and 40*41 both are divisible by 20]
59,60,61 [59*60 and 60*61 both are divisible by 20]
79,80,81 [79*80 and 80*81 both are divisible by 20]
.
.
.
979,980,981 [979*980 and 980*981 both are divisible by 20]
From above steps it is clear that both the numbers
around the 20's multiple are producing the combination to produce a number divisible by 20.
COUNT_2= 49*2=98

CASE 3:
The series 1 to 999 is divisible by 20
COUNT_3=1

Total Count = Count_1 + Count_2 +Count_3 = 100+98+1 = 199

Is this solution Helpfull? Yes (1) | No (1)  

Bharadwaj  (10 years ago)

****
n(n+1)
should be a multiple of 20****

case1: n or n+1 is a multiple of 20
possibilities: 19,39,59...999 , 20,40,60....980 (n

Is this solution Helpfull? Yes | No (2)  

Nikhil Sharma  (10 years ago)

ans is 50+50+50+49 = 199

Is this solution Helpfull? Yes | No (2)  

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