#Numbers-2



A number N, when divided by a divisor D, gives a remainder of 52. The number 5N, when divided by D, gives a remainder of 4.How many values of D are possible?

SOLUTION-

We know that,
Dividend = Divisor*Quotient + Remainder
So, a number N will be of a type,
N=D*k + 52
Now, from the above equation, we can say that D>52

Now, when 5N is divided by D,
i.e,
5N = 5(D*k+52) = Dk1 + 4
=>5Dk + 260 = Dk1 + 4
Now, 5Dk is divisible by D and when 260 is divided by D remainder is 4,
=>5Dk+256=Dk1=>5k + 256/D=k1
So, D must be 256 or its factors as k1 is always a naturak number,
Factors of 256 = 2^8
Now, 52<D<=256 (where D is a natural number)
Values of D satisfying are 64,128,256
Hence, 3 values of D are possible ANS

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