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when a number divided by D leaves remainder of 8and when by 3D leaves remainder of 21.what is the remainder left,when twice the number diveded by 3D.
a-32
b-4567
c-3
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- 3
Number is 21 and D is 13.
when 21 is divided by 13 leaves remainder of 8and when by 39 leaves remainder of 21.
3 is the remainder left
42 is divided by 39. - 13 years agoHelpfull: Yes(11) No(4)
- Let number is n
and k,l,m are positive integers.
Then:
n=kd+8 (1)
n=l(3d) + 21 (2)
from above two equations:
(3*l-k)d+13=0
which means either d|13 or (k-3*l)|13
but d can't be 1 and 13 is prime so d=13
so,
n=13k+8, n=39*l+21
k,l >= 0, n is also integer,
k=1, l=0 is one solution (For which answer of question is 3)
and others are: k-3*l=1 or k=3*l+1
- 13 years agoHelpfull: Yes(3) No(6)
- suppose the no. is a
a = Dx +8 ...1
a = 3Dy + 21 ...2
So from 1 & 2 we get
D(x-3y) = 13
13 is prime so x-3y has to be 1
let x = 4 and y = 1
We get a = 60
2a = 120
So 120 when divided by 3D = 39
we get remainder as 3
So answer is 3 - 12 years agoHelpfull: Yes(3) No(0)
- ans is 3
the number is 60 and D=13
as 60/13=4 (remainder 8)
and 60/3*13=60/39=1 (remainder 21)
now 2x/3D=2*60/3*13=120/39=3 (3 remainder) - 13 years agoHelpfull: Yes(1) No(0)
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