Latest AIEEE Aptitude Question SOLUTION: sum of all three digit numbers which leave the remainder 3 when divided by 5?

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(#M40000320) AIEEE question sum of all three digi Keep an EYE

sum of all three digit numbers which leave the remainder 3 when divided by 5?

Solution

smallest 3 digit number which leave the remainder 3 when divided by 5 =103
largest 3 digit number which leave the remainder 3 when divided by 5 =998
3 digit number after 103 which leave the remainder 3 when divided by 5 =108
an ap is formed: 103,108........998
first term=103
common difference=5
let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n
998=103+(n-1)5
180=n
we know that the formula for sum in AP = no of terms*(first term + last term)/2
sum of ap=180*(103+998)/2 = 90*1101
=99090
This answer is given by reya and is correct answer

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Asked In AIEEE (14 years ago)

Solved reya Read Solution (11)

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Read Solution (11) :

reya  (13 years ago)

smallest 3 digit number which leave the remainder 3 when divided by 5 =103
largest 3 digit number which leave the remainder 3 when divided by 5 =998
3 digit number after 103 which leave the remainder 3 when divided by 5 =108
an ap is formed: 103,108........998
first term=103
common difference=5
let the total no. of three digit numbers which leave the remainder 3 when divided by 5=n
998=103+(n-1)5
180=n
sum of ap=180/2 *(103+998)=90*1101
=99090

Is this solution Helpfull? Yes (5) | No (1)  

Dr.Dipin Singh  (14 years ago)

Numbers can be represented by 5x+3 103+108+......998 >>>>> 180 terms Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180 = 5*19710 +3*180 = 99090

Is this solution Helpfull? Yes (4) | No (1)  

C.T.Narayanan  (14 years ago)

109090

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

Dr.Dipin Singh  (14 years ago)

Numbers can be represented by 5x+3
103+108+......998 >>>>> 180 terms


Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180
= 5*18710 +3*180 = 94090

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Sukham  (14 years ago)

99090

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devender singh saini   (14 years ago)

two ways we can solve
first 103+108+113+........998=sum of this A.P series =99090

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Tushar Garg  (14 years ago)

103,108,113,118......993,998
Total numbers = 20X9=180
sum=100X20X(1+2+3+...+9)+10X9X2X(0+1+2+..+9)+90X(3+8)=99090 ANS

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Pargat Pal Singh  (14 years ago)

Just a correction to Dipin's sol

Numbers can be represented by 5x+3 .........103+108+......998 >>>>> 180 terms
...
Sum = 103+108+......998 = 100+3+105+3+.....995+3 = 5*(20+21+...199)+3*180 = 5*19710 +3*180 = 98540+ 540 =99090

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

Dimple  (14 years ago)

sum = 103+108+........+ 998 = 180*(103+998)/2 = 90*1101 = 99090

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

Rahul Kumar Saini  (13 years ago)

99090

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Johnson   (14 years ago)

6630 = (300*20)+3(20*21/2)

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