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

Question

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

Show Spoiler

gmatophobia · Accepted Answer

Let's assume that the number is  n

...a remainder of 1 when divided by 7...

n = 7q_1 + 1

q_1  → quotient when n is divided by  7

...a remainder of 2 when divided by 3...

n = 3q_2 + 2

q_2  → quotient when n is divided by  3

Both the equations can be merged into a single equations

n = 	ext{LCM}(7,3)q + 	ext{first common term}

q  → quotient when n is divided by  21

To find the first common term, let's write a few of the terms of each of the sequences-

n = 7q_1 + 1  ⇒  1, 8, 15, 22, 29, 36, ....

n = 3q_2 + 2  ⇒  2, 5, 8, 11, ....

Hence, the first common term = 8

n = 21q + 8

This equation represents an arithmetic progression.

The first term of the equation is  8 , i.e. when  q = 0

Last term =

500 = 21q + 8

21q = 492

q = 23.XX

Number of terms =  23 - 0 + 1 = 24

Option C

Bunuel · Answer

6. Remainders 
     Theory   
 REMAINDERS 
 Remainders: Tips and hints 
 Divisibility and Remainders on the GMAT 
 Remainders on the GMAT 
 A Remainders Shortcut for the GMAT 
 All About Negative Remainders on the GMAT 
 A Tricky Question on Negative Remainders 
 Compilation of tips and tricks to deal with remainders 
 Cyclicity and the remainders on the GMAT 
 GMAT Question Patterns: Remainders 
     Questions   
 Units digits, exponents, remainders problems  
 DS Questions 
 PS Questions

For other subjects:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

Dumsy_1711 · Answer

\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.

Runjhun26 · Answer

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

SN9633 · Answer

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

AnkurGMAT20 · Answer

Bunuel , do you have more such questions for practice?

unraveled · Answer

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.

MartyMurray · Answer

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

Yes2GMAT · Answer

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?

Bunuel · Answer

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

KarishmaB · Answer

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) GMAT-Club-Forum-7d1gqs83.png

MartyMurray · Answer

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.

Tushwan · Answer

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

ANVEYSHIKA · Answer

Posted from my mobile device

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

GMATinsight · Answer

Watch very simple two methods to this questions https://www.youtube.com/watch?v=W-kIqlIR6to&feature=youtu.be Answer: Option C Sample question for Method-2 if-a-line-which-has-the-equation-7x-17y-1000-is-plotted-on-x-y-pla-240113.html GMAT-Club-Forum-n208a46u.png

unraveled · Answer

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.

pappal · Answer

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

valeriugirlea · Answer

Why do we ad 1 to 23 in the end?

AnkurGMAT20 · Answer

valeriugirlea  because we calculate the number of terms between two integers by

(Last - First) + 1

For example: If first term is 2 and last term is 11, then

(11-2) + 1 = 10

Hence 10 numbers.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards