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Numerical Ability
LCM and HCF
A three digit number was divided successively in order by 4, 5 and 6 leaving out the remainders. The remainders were respectively 2, 3 and 4. How many such three digit
numbers are possible?
3
5
7
9
Read Solution (Total 16)
-
- three digit no in the form of 120x+94 ie possible no is 7
- 12 years agoHelpfull: Yes(19) No(1)
- When a three digit number is divided by
4,5,6 successively, the final quotient will be
a single digit (as 4*5*6 = 120 and the max
three digit number 999 is not more than
9*120).
Let the number be x and the quotient after
division by 4 be a and the quotient after
division by 5 be b and the quotient after
division by 6 be c.
Then, x = 4a+2,
a = 5b +3
b = 6c + 4.
Now start with c=1, b=10, a=53, x=214
when c=2, b=16, a=83, x=334,
c=3, b=22, a=113, x=454 and so on with
an increment of 120. Thus the 3 digit
numbers possible are 214, 334, 454, 574,
694, 814, 934.
The answer is 7 such numbers are
possible. - 10 years agoHelpfull: Yes(9) No(2)
- N=k LCM(a,b,c)-P where p is the diff between divisor n remainder
- 12 years agoHelpfull: Yes(6) No(1)
- n=60k+2; differnce z 2
- 12 years agoHelpfull: Yes(4) No(1)
- answer is pakka 7..... others answers are wrong
- 12 years agoHelpfull: Yes(3) No(6)
- 15 numbers r possible..cz required number is of the form-60k+2..where k ranges from 2 to 16..therefore 15 such 3 digit numbers r possib;le
- 12 years agoHelpfull: Yes(2) No(8)
- friends i am getting answer as 8...please check these numbers....118,238,358,478,598,718,838,958......correct me if i am wrong....
- 12 years agoHelpfull: Yes(2) No(2)
- answer is 14 because 3 digit number divided by 5 leaves remainder 3 is either 3 or 8 at his unit place i.e 103,108,113,118..........and so on but only even number on divided by 4 leaves remainder 2 so we have only 108,118,128.............. now by checking i found that 118 is first 3 digit number that leaves remainder 2,3,4 by divided by 4,5,6 and after that 178 is next option and at difference of 60 all options are satisfied so in between 118-999 14 options satisfied a condition......
- 11 years agoHelpfull: Yes(2) No(0)
- The no. is
4(5(6x + 4) + 3) + 2 = 120x + 94,
so 7 three digits no. are possible - 7 years agoHelpfull: Yes(2) No(0)
- i m not satisfied ur answer
plz explain clearly
how can solve it. - 12 years agoHelpfull: Yes(1) No(1)
- let the no. be x which goes on sucessive division by 4,5,6.
x=4y+2
y=5z+3
z=6c+4
substituting z in y and y in x
we get x=120c+94 - 8 years agoHelpfull: Yes(1) No(0)
- Ans=7
come to back multiplication
- 12 years agoHelpfull: Yes(0) No(3)
- pls see the word successively. it means a lot.
- 12 years agoHelpfull: Yes(0) No(0)
- someone pls explain it clearly
- 11 years agoHelpfull: Yes(0) No(0)
- yes ans is:7
- 10 years agoHelpfull: Yes(0) No(0)
- When a three digit number is divided by 4,5,6 successively, the final quotient will be a single digit (as 4*5*6 = 120 and the max three digit number 999 is not more than 9*120).
Let the number be x and the quotient after division by 4 be a and the quotient after division by 5 be b and the quotient after division by 6 be c. Then, x = 4a+2, a = 5b +3 b = 6c + 4.
Now start with c=1, b=10, a=53, x=214
when c=2, b=16, a=83, x=334, c=3, b=22, a=113, x=454 and so on with an increment of 120. Thus the 3 digit numbers possible are 214, 334, 454, 574, 694, 814, 934.
The answer is 7 such numbers are possible - 8 years agoHelpfull: Yes(0) No(0)
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