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:

x is a positive integer less than 500. When x is divided by [#permalink]
05 Nov 2006, 12:47

x is a positive integer less than 500. When x is divided by 7, the
remainder is 1; when x is divided by 3, the remainder is 2. How many
such numbers are possible? _________________

Notation: A number n when divided by k has remainder r, we say that
n = r Mod k

E.g., 28 = 4 mod 6
4 = 1 mod 3
0 = n mod n

Thm: If n = r Mod k then n*t = r*t Mod k (t an integer obviously)

E.g., (for this problem)

7 = 1 mod 3 so 7*t = t Mod 3 so 7 = 1 Mod 3, 14 = 7*2 =2 Mod 3, 21 = 7*3 = 3 Mod 3 = 0 Mod 3, 28 = 7*4 = 4 Mod 3 = 1 mod 3, 35 = 2 Mod 3, ...

So the sequence is {1, 2, 0, 1, 2, 0, ...}

The question asks " How many + numbers n < 500 of the form 7*k + 1 are equal to 2 mod 3?" (so maybe it is by now clear that it's 1/3 of the numbers of this form...)

8 = (7*1 + 1) Mod 3 = 7*1 Mod 3 + 1 Mod 3 = 2 mod 3
15 = (7*2 + 1) Mod 3 = 2 Mod 3 + 1 Mod 3 = 0 mod 3
22 = (7*3 + 1) Mod 3 = 3 mod 3 + 1 Mod 3 = 1 mod 3
29 = 2 Mod 3
36 = 0 Mod 3
43 = 1 Mod 3
...
498 = (7*71 + 1) Mod 3 = 71 Mod 3 + 1 Mod 3 = 0 Mod 3

So there are 71 numbers less than 500 of the form 2 Mod 7 and 1 out of 3 is of the form 2 Mod 3 so it's just a matter of checking the endpoints and so the asnwer is 24 because we start with 2 Mod 3 and end with 0 mod 3.

That seems complicated and long-winded, but the principles can help you do these problems in no time (maybe there's a good review of modular arithmetic somewhere on the web).

Re: Number Properties [#permalink]
07 Nov 2006, 21:39

Paayal wrote:

x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

Vikram, 24 is answer for this question. What I was asking was what is 'x' in the question.

I agree..it is a series of numbers. But can you please post a few initial numbers in the series? _________________

The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.

Re: Number Properties [#permalink]
09 Nov 2006, 18:54

ak_idc wrote:

Paayal wrote:

x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?

Vikram, 24 is answer for this question. What I was asking was what is 'x' in the question.

I agree..it is a series of numbers. But can you please post a few initial numbers in the series?

x is the # of "numbers" between 2 and <500 satisfying the given conditions....

gmatclubot

Re: Number Properties
[#permalink]
09 Nov 2006, 18:54