Latest Elitmus Aptitude Question SOLUTION: Asked on 25th may 2014 in elitmus Q. Find then sum of numbers between 1 to 200 which when divided by either 5 or 7 leaves remainder of 2?

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Asked on 25th may 2014 in elitmus

Q. Find then sum of numbers between 1 to 200 which when divided by either 5 or 7 leaves remainder of 2?

Asked In Elitmus Rahul kumar (10 years ago)

Solved Devendra Marghade Read Solution (8)

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Read Solution (8) :

Devendra Marghade  (10 years ago)

6341

Number which leaves remainder 2, when divided by 5 are 7,12,17,22,....,197
So 197=7+(n-1)5, n=39,Sum of these numbers=(39/2)[2*7+38*5]=3978
Similarly,
Number which leaves remainder 2, when divided by 7 are 9,16,23,30,.....,198
So 198=9+(n-1)7, n=28,Sum of these numbers=(28/2)[2*9+27*7]=2898

As above both the sum includes some common numbers 37,72,107,142,177
we have to subtract once,the sum of these common numbers(=535),to find the sum of all the numbers
So Sum of all the numbers=3978+2898-535= 6341

Is this solution Helpfull? Yes (68) | No (15)  

ashes  (10 years ago)

everything @devendra m has done is right but he forgot to include 2.so answer is 6343 not 6341

Is this solution Helpfull? Yes (46) | No (6)  

shaik khader  (10 years ago)

numbers when divided by 5 leaves a reminder 2 is 2,7,12.....197 (total 40nos) and the sum is (197+2/2)*40=3980;
numbers when divided by 7 leaves a reminder 2 is 2,9,16.....198 (total 29nos) and the sum is (198+2/2)*29=2900;
numbers when divided by 5&7 ie 35 leaves a reminder 2 is 2,37,.....177 (total 6nos) and the sum is (177+2/2)*6=537;
ie sum=3980+2900-537=6343

Is this solution Helpfull? Yes (45) | No (2)  

kumar rahul dev  (10 years ago)

TAKING LCM OF(5,7)=35 HENCE NUMBER WILL BE MULTIPLE OF 35 AND ADDING 2 AT THE END.
FIRST NUMBER=35+2=37
2ND- 72;
3rd-107 , 142, 177. adding all this we get ans = 535

Is this solution Helpfull? Yes (13) | No (4)  

Ishan  (10 years ago)

Numbers leaving remainder 2 when divided by 7:
2 9 16 23.....198

using a+(n-1)d=198
2+(n-1)7=198
n=29
Sn= na+d(n)(n-1)/2
29*2+7*29*28/2
2900



Numbers leaving remainder 2 when divided by 5:
7 12 17 22......197

using a+(n-1)d=197
7+(n-1)5=197
n=39

Sn=na+d*n*(n-1)/2
39*7 + 5*38*39/2
=3978

total =3978+2900=6878

Is this solution Helpfull? Yes (3) | No (6)  

Mohammad Zeeshan  (10 years ago)

MANIKANDAN I. common numbers means the numbers which give reminder =2 when divided by both 7 and 5...when u are calculating the total sum, these get added twice so u have to subtract their sum once

Is this solution Helpfull? Yes (1) | No   

Mani kandan  (10 years ago)

hai devendra, i can't understand "common numbers 37,72,107,142,177
we have to subtract once,the sum of these common numbers(=535"

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Dark Knight  (9 years ago)

hey devendra and ashes i guess if the answer is 6341 or 6343 then when dividing by 7 and 5 we will not get the remainder as 2

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