How many positive integers less than 500 have a remainder of 1 when

\(\frac{500}{7} =71.\) There are 71 integers less than 500 that are divisible by 7.

Quick check for the last integer: 7(71)+1 = 497+1 = 498. 498 is less than 500.
So, there are 71 integers less than 500 that have a remainder of 1 when divided by 7.


Integers that have a remainder of 1 when divided by 7 = 7k+1.
Integers that have a remainder of 2 when divided by 3 = 3n+2


Possible values of
7k+1: 8 15 22 29 36 43 50 57 64 71.......498
3n+2: 8 16 23 29 38 44 50 56 65 71.......498

(It is easier and faster to check for the integers of the form 3n+2 that match with integers of the form 7k+1 than to make an entire list of all the integers of the form 3n+2.)



Every third term after the first term (8) is common between 7k+1 and 3n+2.

\(\frac{71}{3} = 23. \)
23+1 = 24. The number of common integers = 24.



Therefore, the answer is C : 24.

gmatophobia can you please explain the concept after finding the first common term?

hope this helps: the LCM was taken and added with the lowest value. Then the given condition N<500 was substituted with the new arrived equation

\(N=7a+1=3b+2\\
N+13=7()=3()\\
N=LCM(3,7)-13=21z-13.\\
z=1,2....24\)

for \(0<N<500\)

Bunuel, do you have more such questions for practice?

How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

A. 20

B. 21

C. 24

D. 26

E. 72

When 7 divides 500 it gives 71 integers as quotient 497 being the last divisible number.
Hence there are 71 integers that leave remainder 1 498(497+1) being the last one.

When 3 divides 500 it gives 166 integers as quotient 498 being the last divisible number.
Hence there are 166 integers that leave remainder 1 499(498+2) being the last one.
But this is not helpful since we are looking for a common integer that leaves 1 and 2 remainder when divided by 7 and 3 individually.

So, only 71 integers are there among which common ones would leave remainders 1 and 2. These common can be found by dividing 71 by 3 which gives us 23.xx
Since we are looking for integer decimal is not possible hence we round up it to 24. Answer C.

Estimation also helps but that depends on the choices available. Had 23 been there among the choices it would have been difficult. But that is unlikely in these tests.

Answer C.

­How many positive integers less than 500 have a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

When faced with a GMAT Quant question you're not sure how to answer, keep in mind that GMAT Quant questions are often designed to be answerable in multiple ways. So, it can often work to just try anything that might get you to the answer.

In this case, one approach that we can use is to look for a pattern starting with the smallest integer we can find such that it has a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3.

That integer is 8.

Then, we can continue with integers 1 greater than a multiple of 7 and look for a pattern to the ones that are 2 greater than a multiple of 3.

Doing so, we see the following:

8, 15, 22, 29, 36, 43, 50, 57, 64, 71

The integers in green fit our constraints. So, apparently, one in three integers that's 1 more than a multiple of 7 is two more than a multiple of 3.

There are 500 - 8 = 492 integers between 8 and 500.

492/7 = 70.X of those integers are one more than a multiple of 7.

70/3 = 23.X

23 + 1 (for 8) = 24 total integers that fit our constraints.

A. 20
B. 21
C. 24
D. 26
E. 72

Correct answer: C­

gmatophobia


Is this the equation in the form of an=a0+ (n-1) d.
If yes, I am curious as to how you plugged in 500 for an?

Isn't it supposed to be integer values less than < 500 as per question prompt?

 
­
21q + 8 < 500
21q < 492
q < 23.someting

Hence, q can take 24 values, from 0 to 23 inclusive. Consequently, 21q + 8 can also take the corresponding 24 values.

Hope it helps­

­
n = 7a + 1
n = 3b + 2

Since there is no common remainder, the first such value is found by hit and trial. 
If a = 1, n = 8. 8 is of the form 3*2 + 2 and hence 8 is the smallest such number. 
Next such number will be 21 (LCM of 7 and 3) more than 8. Next will be another 21 more. and so on

So all numbers of this form will be 21k  + 8 where k starts from 0 and takes all positive integer values. To get the greatest value less than 500, 
21k + 8 < 500
21k < 492
k < 23.something
Hence k can take all values from 0 to 23. These will be 24 total values.

Answer (C)

­The first commmon term is a number such that the remainder when it's divided by 7 is 1 and the remainer when it's divided by 3 is 2.

Now, notice that any number that's greater than the first common term by a muliple of 7 will be such that, when it's divided by 7, the remainder is 1.

For example:

8 + 7 = 15 when divided by 7 has a remainder of 1.

15 + 7 = 22 when divided by 7 has a remainder of 1.

etc.

Similarly, any number that's greater than the first common term by a muliple of 3 will be such that, when it's divided by 3, the remainder is 2.

For example:

8 + 3 = 11 when divided by 3 has a remainder of 2.

11 + 3 = 14 when divided by 3 has a remainder of 2.

etc.

Thus, any number that is BOTH greater by a multiple of 7 than the common term and greater by a multiple of 3 than the common term will fit the criterion of having a remainder of 1 when divided by 7 and a remainder of 2 when divided by 3?

Of course, since 21 is the LCM of 3 and 7, any number that is both greater by a multiple of 7 than the common term and greater by a multiple of 3 than the common term will be greater than the common term by a multiple of 21.

So, since 8 is first common term, we'll find that the numbers we're looking for fit the following pattern:

8 + 21 = 29

29 + 21 = 50

50 + 21 = 71

etc.

can you please explain the concept after finding the first common term?
Please share concept. calculation part i understood­

Posted from my mobile device


Why is the +1 added to 23 at the end of the solution?

­
Watch very simple two methods to this questions



Answer: Option C

Sample question for Method-2
https://gmatclub.com/forum/if-a-line-wh ... 40113.html

---
GMATinsight
(Private Tutoring) | Contact for a FREE Demo Session | http://www.GMATinsight.com
YouTube (1100+ GMAT Videos sorted in playlists): http://www.Youtube.com/GMATinsight

­There are 24 terms when you count 0 to 23, wherein 0 is the first term.
You might be missing the first term. Check Bunuel's explanation.

given x%7=1 and x%3=2 ,where x is any integer and % sign reads as n divided by 7 will give a remainder of 1.
first such no. is 8
second is 29
and the last one is 491
the combination above will certainly form an AP
so very easily can be no. of terms satisfying this AP can be found as 8+(n-1)21=491 i.e. 24

Why do we ad 1 to 23 in the end?­