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Numerical Ability
LCM and HCF
A number when divided by D give remainder 23 and when by 12D give 104.What will be remainder when divide by 6D?
a. 23
b. 47
c. 52
d. 104
Read Solution (Total 7)
-
- N = ad + 23 when N divided by D quotient is a
N = 12bD + 104 when N divided by 12D quotient is b
let b = 1
N = 12D + 104
aD + 23 = 12D + 104
D(a – 12) = 81
D = 81/(a – 12)
least value of a = 15 so (a – 12) divides 81 exactly
so D = 27
N = 15 × 27 + 23 = 428
6D = 27 × 6 = 162
428/162 = Remainder is 104 - 11 years agoHelpfull: Yes(42) No(8)
- it's a trick, when n/12D gives 104. We can write n/6D as 2n/12D i.e (n+n)/12D. So it would give 104+104=208 as remainder, but acc. to remainder theorem we need to multiply d answer by the factor dat we hav divided with and that was 2. So the answer would b 104, this trick is working in each n every case.
- 11 years agoHelpfull: Yes(20) No(2)
- Let N be the no.,so according to wat's given, D>23 =>6D>138 =>12D>276.
Also N = 12D*m + 104 = 6D*2m + 104,here,when N is divided by 6D, the remainder is either 104 or a no. lesser which means 104 could be again divided by 6D.
But we know that 6D > 138 ie. 6D > 104.
Therefore 104 cannot be further divided by 6D.Therefore the remainder is 104. - 11 years agoHelpfull: Yes(6) No(2)
- Ans 104. D>23 i.e. 24,25.... when we divide 104 by 24*6 reminder should be 104
- 11 years agoHelpfull: Yes(1) No(2)
- when divide by 6D give 104
- 11 years agoHelpfull: Yes(1) No(1)
- i think answer will be 23
- 10 years agoHelpfull: Yes(1) No(1)
- N = ad + 23 when N divided by D quotient is a
N = 12bD + 104 when N divided by 12D quotient is b
let b = 1
N = 12D + 104
aD + 23 = 12D + 104
D(a – 12) = 81
D = 81/(a – 12)
least value of a = 15 so (a – 12) divides 81 exactly
so D = 27
N = 15 × 27 + 23 = 428
6D = 27 × 6 = 162
428/162 = Remainder is 104 - 5 years agoHelpfull: Yes(1) No(0)
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