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When positive integer n is divided by 5, the remainder is 2
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14 Sep 2020, 21:49
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When positive integer n is divided by 5, the remainder is 2; when n is divided by 6, the remainder is 3. If 0<n<100, how many possible values of n are there?
Re: When positive integer n is divided by 5, the remainder is 2
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15 Sep 2020, 07:00
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lazyashell wrote:
When positive integer n is divided by 5, the remainder is 2; when n is divided by 6, the remainder is 3. If 0 < n < 100, how many possible values of n are there?
A. 1 B. 2 C. 3 D. 4 E. 5
When it comes to remainders, we have a nice property that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
When positive integer n is divided by 5, the remainder is 2 This means the possible values of n are: 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97
When n is divided by 6, the remainder is 3 This means the possible values of n are: 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99
How many possible values of n are there? We need to find possible values of n that are in BOTH lists. They are: 27, 57 and 87 So there are 3 possible values of n
When positive integer n is divided by 5, the remainder is 2
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10 Jul 2021, 07:11
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Theory: Dividend = Divisor*Quotient + Remainder
When positive integer n is divided by 5, the remainder is 2 => n = 5x + 2 [ x being the quotient] ...(1)
When positive integer n is divided by 6, the remainder is 3 => n = 6y + 3 [ y being the quotient] ...(2)
There are multiple ways to solve the problem. Two of them are listed below
Method 1: Compare the two equations for n (Faster Method) n = 5x + 2 = 6y + 3 => x = \(\frac{6y+1}{5}\) So, only those values of y which will give x as integer will give common values for n in (1) and (2) We need to get units digit of 6y as 4 or 9 to make 6y+1 a multiple of 5 So, y can be 4 giving x as \(\frac{6*4+1}{5}\) = 5 => n = 6*4 + 3 = 27 y can also be 9 giving 6y as 54 and x as \(\frac{6*9+1}{5}\) = 11. => n = 6*9 + 3 = 57 y can be 14 giving x as an integer. n = 6*14 + 3 = 87 y = 19 will become too big for n < 100 So, 3 values of n such that 0 < n < 100 are possible.
Method 2: Write all values of n in both the cases and compare them n = 5x + 2 Put x=0,1,2,3,4... and find the corresponding values of n x=0, n = 5*0 + 2 = 2 and so on.. So values n will be 2, 7, 12, 17, 22,27.... So numbers ending with 2 and 7
n = 6y + 3 Put y=0,1,2,3,4... and find the corresponding values of n x=0, n = 6*0 + 3 = 3 and so on.. So values n will be 3, 9, 15, 21, 27,.... If we write all the numbers then we can find the common numbers as 27,57,87
So, Answer will be C Hope it helps!
Watch the following video to MASTER Remainders
gmatclubot
When positive integer n is divided by 5, the remainder is 2 [#permalink]