how to solve efficiently

[buoyant]

How many two-digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?

a) None
b) One
c) Two
d) Three
e) Four

[spoiler]OA: d)[/spoiler]

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Hi buoyant,

This type of question is rooted in pattern-matching. Once you find the patterns behind this question, it won't be hard to solve.

Instead of trying to do every step all at once, I suggest that you break the prompt into "pieces":

First, name the 2-digit numbers that are evenly divisible by 10:

10, 20, 30, ......90

Now, name the 2-digit numbers that have a remainder of 1 when divided by 10:

11, 21, 31,.....91

Now that we've established the numbers that fit the first 2 "restrictions" in the prompt, let's factor in numbers that are ALSO divisible by 6:

30, 60, 90

And ALSO have a remainder of 1 when divided by 6:

31, 61, 91

Final Answer: D

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

[GMATGuruNY]

How many two-digit whole numbers yield a remainder of 1 when divided by 10 and also yield a remainder of 1 when divided by 6?

a) None
b) One
c) Two
d) Three
e) Four

Let x = the two-digit integer.

x has a remainder of 1 when divided by 10.
In other words, x is 1 more than a multiple of 10:
x = 10a + 1, where a is an nonnegative integer.

x has a remainder of 1 when divided by 6.
In other words, x is 1 more than a multiple of 6:
x = 6b + 1, where b is a nonnegative integer.

Since x = 10a + 1 and x = 6b + 1, we get;
10a + 1 = 6b + 1
10a = 6b
5a = 3b
a = (3/5)b.

If b=5, then a=3, in which case x = 10a + 1 = 10*3 + 1 = 31.
If b=10, then a=6, in which case x = 10a + 1 = 10*6 + 1 = 61.
If b=15, then a=9, in which case x = 10a + 1 = 10*9 + 1 = 91.
A total of 3 options for x:
31, 61, and 91.

The correct answer is D.

[Mathsbuddy]

2-digit common multiples of 6 and 10 are:
30, 60 and 90
Therefore the answer is 3 (D).
(The actual value of the remainder is redundant information)