Latest TCS Aptitude Question SOLUTION: When a number is successively divided by 5, 3 and 2 leaves a remainder of 0, 2 and 1 respectively. What are the remainders when the same number is divided by 2, 3 and 5 successivel

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When a number is successively divided by 5, 3 and 2 leaves a remainder of 0, 2 and 1 respectively. What are the remainders when the same number is divided by 2, 3 and 5 successively?

Asked In TCS divyasree (8 years ago)

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Subhendu Pradhan  (8 years ago)

2*x+1
3(2*x+1)
5(3(2*x+1))=5(6x+5)=30x+25 it will be the number.

when divided by 2 remainder will be 1, quotient will be 15x+12;
when divided by 3 remainder will be 0, quotient will be 5x+4;
when divided by5 remainder will be 4.
so remainders will be 1,0,4

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goram uday sree  (8 years ago)

z=2*1+1=3, y=3z+2=11, x=5y=55
55 divide by 2 ;1 ,27 divide by 3;0 ,9 divide by 5 ;4

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NIKITHA BOLLINENI  (8 years ago)

when divided by 5 we get X=5L
When divided by 3 we get L=3k+2
when divided by 2 we get k=2M+1
NOW X=5*(3(2M+1)+2)=30M+25
ASSUME M=0
WHEN 25 IS DIVIDED BY 2 WE GET REMAINDER 1 AND QUOTIENT 12
WHEN 12 IS DIVIDED BY 3 WE GET 0 REMAINDER AND QUOTIENT 4
WHEN 4 IS DIVIDED BY 5 WE GET 4
SO ANS IS 1,0,4

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KROVVIDI ANANTH DUTT  (8 years ago)

The remainders will be 1,0,4 respectively...................

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JENSI KARUNYA.S  (8 years ago)

ans: 1,2,0 respectively.

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ditty k joy  (8 years ago)

5 0
3 2
2 1

(1*3+2)=5
5*5+0=25
the no is 25

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Neha singh  (8 years ago)

as the no is divided by 5 ,3 and 2 leaves remainder 0,2,1 respectively..
so the only no that is divided by 5,3,2 leaves remainder is 0,2,1 is 5 ..again 5 is dived by 2,3,5 leaves remainder 1,2,0 ans is 1,2,0

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abhishek  (8 years ago)

first take 5 as a divisor ,N as a dividend and Q1 as a quotient
now from the division meathod we know that
DIVIDEND =DIVISOR *QUOTIENT +REMINDER
so now N =5*Q1+0..........(1)
Similarly
Q1=3*Q2+2.......(2)
Q2=2*Q3+1........(3)
put Q3 as a zero then ...
eqn 1,2,3 will be
25,5,1......
now take 25 as a new DIVIDEND and sole with new numbers
which are 2,3,5
so reminder will be 1,0,4.... ans.
thank you

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Abdul jaffar  (8 years ago)

Sol: By given condition, we have
A+B+C = 13...(1)
C+D+E = 13...(2)
E+F+G = 13...(3)
G+H+I = 13...(4)
Add up (1)+(2)+(3)+(4), we get
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45

Hence,C+E+G = 52 - 45 = 7...(5)
By (5), we have E = 7-C-G

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Ayanendu Banerjee  (8 years ago)

let the final quotient be x
now it was obtained by dividing the dividend by 2 leaving 1 as remainder
so to find the first dividend the expression looks like
2*x+1.
repeating the same process for the 2nd dividend we get 3*(2*x+1)+2.
and finally the actual number is calculated as 5*(3*(2*x+1)+2).
This expression can be rearranged to be made to look like 30*x+25.

now successive division of the number by 2,3,5 respectively we get 1,0,4 as remainders which are the required answers

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