How many two-digit numbers are there...

[swerve]

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

The OA is A.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

[Brent@GMATPrepNow]

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

Cheers,
Brent

[[email protected]]

Hi swerve,

We're asked 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. This question can be solved with a bit of 'brute-force' arithmetic and some note-taking. To start, let's focus on the smaller list: the number of 2-digit numbers that give a remainder of 1 when divided by 10... there are only 9 possible numbers:

11, 21, 31, 41, 51, 61, 71, 81, 91

From here, we just have to divide each number by 6 and keep track of the remainders (and you might spot a pattern):
11/6 = 1r5
21/6 = 3r3
31/6 = 5r1
41/6 = 6r5
51/6 = 8r3
61/6 = 10r1
71/6 = 11r5
81/6 = 13r3
91/6 = 15r1

Thus, the total number of integers that 'fit' all of the given information is 3.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

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

The OA is A.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

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

Answer: A