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Re: How many two-digit numbers are there whose remainder when divided by 1
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09 Jul 2021, 08:55
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Carcass wrote:
How many two-digit numbers are there whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5?
A. 3 B. 4 C. 5 D. 6 E. 7
IMPORTANT: When it comes to remainders, we have a nice rule 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.
Let's start with the second piece of information... The remainder is 5 when the number is divided by 6 So, the possible values are: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95,... (we'll stop here, since the questions specifies that the number is a 2-digit number)
The remainder is 1 when the number is divided by 10 So, the possible values are: 1, 11, 21, 31,... Notice this this basically tells us that the number has a UNITS DIGIT of 1
So, go back to examine the first set of possible values: {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} 3 values have a UNITS DIGIT of 1
So, there are 3 values that satisfy both conditions. Answer: A
Re: How many two-digit numbers are there whose remainder when divided by 1
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04 Jan 2022, 09:47
We need to find the total number of two-digit numbers whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5
Let's solve the problem using two methods
Let n be the two digit number which gives 1 remainder when divided by 10 and 5 remainder when divided by 6
Dividend = Divisor * Quotient + Remainder n when divided by 10 gives 1 remainder Dividend = n Divisor = 10 Quotient = a (Assume) Remainder = 1 => n = 10a + 1 ...(1)
n when divided by 6 gives 5 remainder Dividend = n Divisor = 6 Quotient = b (Assume) Remainder = 5 => n = 6b + 5 ...(2)
Method 1: Writing Values
n = 10a + 1 [ From (1) ] Putting a = 1, 2,... we get values of n as n = 11, 21, 31, 41, ...., 91 (we need to consider only two digit values)
n = 6b + 5 [ From (2) ] Putting b = 1, 2,... we get values of n as n = 11, 17, 23, 29, ...., 95 (we need to consider only two digit values) For common values we need unit's digit to be 1 => units' digit of 6b + 5 = 1 => units' digit of 6b = 11-5 = 6 This will happen when b = 1, 6, 11, 16, => n = 6b + 5 = 11 (for b=1) n = 41 (for b=6) n = 71 (for b=11) n = 101 (for b= 16.. but we need only two digit values of n)
=> Common values of n in both the cases is 11, 41 and 71 => 3 values
Method 2: Algebra
n = 10a + 1 [ From (1) ] and n = 6b + 5 [ From (2) ] => 10a + 1 = 6b + 5 => 10a = 6b + 4 => a = \(\frac{6b+4}{10}\) Now, only those values of "b" which also make "a" an integer will give us common values of n in both the cases
=> 6b + 4 has to be a multiple of 10 => unit's digit of 6b = 10-4 = 6 => b = 1, 6, 11 (As shown above)
So, Answer will be A Hope it helps!
Watch the following video to learn the Basics of Remainders
Re: How many two-digit numbers are there whose remainder when divided by 1
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22 Feb 2023, 02:09
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