A no. N when divided by divisor D leaves remainder 23. when same no. N is divided by 12D remainder is 104. what will be the remainder when the no is divided by 6D ?
Soln : If I start from second statement that N = a*12D + 104, so this is same N which when divided by D gives a remainder of 23. i.e. 104 = b*D + 23 or D is a divisor of 104-23 = 81 and also D is greater than 23. So D = 27 or 81.Thus in any case when same N is divided by 6D, remainder will be given by 104 as a*12D is certainly divisible by 6D. So the required remainder is = 104.
D has to be greater than 23 as remainder on division is 23.
So, 6D > 104
Hence, remainder will be 104
Else,
N = Dx + 23
N = 12Dy + 104
Equating them we will get.
D(x - 12y) = 81
So, D has to be a factor of 81 and greater than 104. So, possible values are 27 and 81
So, 6D is 162 or 324 (both greater than 104)
So, remainder will be 104
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