How many two-digit numbers are there whose remainder when divided by 1

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

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2 digit numbers whose remainder when divided by 10 is 1 are
11 , 21 , 31 , 41 , 51 , 61 , 71 , 81 , 91
Out of above , numbers whose remainder when divided by 6 is 5 are
11 , 41 and 71

Answer A
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Of all the 2-digit numbers that have a remainder of 1 when divided by 10, only 11, 11 + LCM(6, 10) = 11 + 30 = 41, and 41 + 30 = 71, when divided by 6, yield a remainder of 5.

Answer: A
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Isn’t GCF(6,10) 2 how did you get to 30?

I tried the n=24x + …. But I can’t find the value of …
Could someone please advise? Thanks
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We actually meant LCM(6, 10) instead of GCD(6, 10); you're absolutely right that GCD(6, 10) is 2. Thanks for pointing it out.

If the "n" in your equation represents a two digit number that produces a remainder of 1 when divided by 10 and a remainder of 5 when divided by 6; I don't think you'll be able to find the numbers using any equation of the sort n = 24x + ... You can proceed as follows in order to find the numbers that way:

Since n produces a remainder of 1 when divided by 10, n = 10k + 1 for some k.

Since n produces a remainder of 5 when divided by 6, n = 6s + 5 for some s.

Let's add 19 to each equation:

n + 19 = 10k + 20

n + 19 = 6s + 24

Notice that 10k + 20 is divisible by 10 and 6s + 24 is divisible by 6. Since n + 19 is both divisible by 10 and by 6, n + 19 must be divisible by the LCM of 10 and 6, which is 30. Thus, the smallest possible value of n + 19 is 30; which yields the smallest possible value of n is 30 - 19 = 11.

Once we find the smallest possible value of n, we can just add LCM of 6 and 10 to find other two digit numbers which satisfy the requirements of the question; the numbers are 11, 11 + 30 = 41 and 41 + 30 = 71.
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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!

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