Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Want to solve 700+ level Algebra questions within 2 minutes? Attend this free webinar to learn how to master the most challenging Inequalities and Absolute Values questions in GMAT

What you'll gain: Strategies and techniques for approaching featured GMAT topics, an introduction to Manhattan Prep's world class GMAT instructors, and more.

Re: How many two-digit numbers are there whose remainder when divided by 1
[#permalink]

Show Tags

31 Jan 2016, 11:46

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
_________________

When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long

Re: How many two-digit numbers are there whose remainder when divided by 1
[#permalink]

Show Tags

28 Aug 2018, 12:42

2

Top Contributor

Bunuel 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

How many two-digit numbers are there whose remainder when divided by 1
[#permalink]

Show Tags

Updated on: 25 Oct 2019, 09:37

1

Bunuel 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

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.

Re: How many two-digit numbers are there whose remainder when divided by 1
[#permalink]

Show Tags

22 Oct 2019, 22:45

ScottTargetTestPrep wrote:

Bunuel 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

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

Answer: A

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

How many two-digit numbers are there whose remainder when divided by 1
[#permalink]

Show Tags

25 Oct 2019, 09:38

David nguyen wrote:

ScottTargetTestPrep wrote:

Bunuel 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

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

Answer: A

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

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.
_________________