Latest Elitmus Aptitude Question SOLUTION: How many possible combination of four consecutive numbers between 1to 1000 whose sum is divisible by 10.ex;-16+17+18+19=70

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How many possible combination of four consecutive numbers between 1to 1000 whose sum is divisible by 10.ex;-16+17+18+19=70

Asked In Elitmus pammi kumari (9 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

BHASKAR MANDAL  (9 years ago)

it is possible for all the numbers having unit digit as
(1,2,3,4) and (6,7,8,9)
so, b/n 1 to 10 difference there are two possibility (i,e, 1-10,11-20,21-30,etc)
so total possibility
= 1000*2/10
= 200
200 is the answer

Is this solution Helpfull? Yes (60) | No (21)  

Anjaneyulu  (9 years ago)

ans is 199
2 possibles for every 10
20 possibles for every 100
200 possibles for 1000
but we will not consider 1(1,2,3,4 this not possible)
so 200-1=199

Is this solution Helpfull? Yes (43) | No (10)  

srinu  (9 years ago)

11+12+13+14=50 which is divisible by 10 and
16+17+18+19=70 which is divisible by 10.
So for 10 to 20 there is two possibilities, hence for 20 to 30 there will be two possibilities .
so in order to count how many possibilities for 10 to 1000, let
10,20,30,.....1000, we want to find no. of terms
last term=a+(n-1)d
1000=10+(n-1)10
=>100=1+n-1
=> n=100
so there will be 100*2 chances
=200

Is this solution Helpfull? Yes (15) | No (2)  

victor munna  (9 years ago)

ans-198 bcz 1 & 1000 not include it is given in question b/w 1 to 1000.

Is this solution Helpfull? Yes (3) | No (10)  

Gouri shankar Kumawat  (8 years ago)

let numbers be x,x+1,x+2,x+3
sum=4x+6=2(2x+3)
now for x=1,6,11.........996 it gives the multiple of 10.
so total no. is 996=1+(n-1)5
n=200

Is this solution Helpfull? Yes (3) | No   

Anuj Awasthi  (8 years ago)

ans is 199
2 possibles for every 10
20 possibles for every 100
200 possibles for 1000
but we will not consider 1(1,2,3,4 this not possible)
so 200-1=199
becoz between we will not consider

Is this solution Helpfull? Yes (2) | No (1)  

suman kumari  (9 years ago)

ans is 33

Is this solution Helpfull? Yes (1) | No (11)  

Pranshu Preet Kaur  (9 years ago)

ans is 200

11+12+13+14=50 ( divisible by 10 )
16+17+18+19=70 ( divisible by 10)
from 10 to 20 there is two possibilities
for 20 to 30 there will be two possibilities .
so in order to count how many possibilities for 10 to 1000, let us assume that the nos. are in AP

10,20,30,.....1000
l=a+(n-1)d
1000=10+(n-1)10
=>100=1+n-1
=> n=100
so 100*2 combinations are possible.

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

swati priya  (9 years ago)

there are total of 200 combination of number between 1 to 1000
since in each 1 to 10 , 11 to 20 , 21 to 30 ...............991 to 1000 there are only two combination of four consecutive numbers whose sum is divisible by 10.

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Ehtesham  (9 years ago)

four consecutive numbers : a+b+c+d=10k;

1+2+3+4=10;
6+7+8+9=30;
11+12+13+14=50;

now give a look ,here among the possible combination there is pattern followed
1+5=6
2+5=7
3+5=8
4+5=9
similarly all combinations follow same pattern.
now find last combination consecutive digit :
will come from 1+(n-1)*5;
take n=200;
then 1+(200-1)*5=996;
so last consecutive combination :996+997+998+999=3990

so total number of such combinations are: (996-1/5)+1=200

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RAJNISH SINGH  (8 years ago)

1 to 10 = (1+2+3+4 , 6+7+8+9)=2 combinations
2 to 20=(11+12+13+14 , 16+17+18+19)=2 combinations
so total combinations between 1 and 100=10*2=20
like this total combinations 1 to 1000=20*2=200

Is this solution Helpfull? Yes | No (1)  

Nitin Maheshwari  (8 years ago)

(x+(x+1)+(x+2)+(x+3))/10=(4x+6)/10=(2x+3)/5
(2x+3)/5=399
x=996

Is this solution Helpfull? Yes | No (1)  

avinash kumar jha  (8 years ago)

10-2p,
1000-200
ans-200-1=199

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Gautam Anand  (8 years ago)

1+2+3+4=10
6+7+8+9=20

11+12+13+14=50
16+17+18+19=60
as show on
for ever 10 (like 1-10,11-20,21-30....)number we get 2 set of consecutive number
so,
possible combination is=(1000/10)*2=200

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