How many positive integers are there from 0 to 1000 that leave a remai

Question

How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?

A. 19
B. 20
C. 24
D. 32
E. 36

Are You Up For the Challenge: 700 Level Questions

GMATinsight · Accepted Answer

Answer: Option E

Please the explanation in the video attached here.

https://www.youtube.com/watch?v=AOlXE3VRrlc

nick1816 · Answer

N= 3 mod 7

N= 3mod 28 or 10 mod 28 or 17 mod 28 or 24 mod 28 ......(1)

Also N = 2 mod 4.....(2)

From (1) and (2)

N= 10 mod 28

N= 10, 38........,990

Number of values N can take = 990-10/ 28 +1 = 36

Ksmew · Answer

nick1816

can you pls explain why you added 1 at the end step??

i solved this like - 
28K + 10 <= 1000
K <= 990/28
K <= 35

skhemani · Answer

All numbers between 0 and 1000 such that 
 n = 7k+3 = 4p + 2
=> 7k + 1 = 4p 
4p is even so 7k has to be odd which means we can quickly check all odd multiples of 7 to get a series:-
7x1, 7x5, 7X9, 7x13......and so on
The last digit of the series is less than 1000. Quick thinking here can be 7x140 = 980 so we can check odd numbers near 140. 7x141 = 987 and voila! (7X143 >1000)
so we have 1, 5, 9........141
nth term 141: 
1+4(n-1) = 141
and n = 36 
Ans E

Please, share if you have a faster and simpler method for such questions

nick1816 · Answer

Ksmew 
k<= 35

Hence k can be 0, 1, 2, 3......., 34 or 35. You gotta consider 0 because, at k=0, 28k+10 =10 is a positive integer.

so total values k can take is 35+1=36

OR

10, 38,........, 990 are in arithmetic progression(difference between subsequent terms is constant(28).

Hence,

L = F + (n-1) d

L= last term of the sequence 
F= First term
n = total number of the terms
d= difference between subsequent terms

990 = 10+(n-1)*28

n-1 = \frac{990-10}{28}

n = \frac{990-10}{28}+1

Both ways are one and the same thing tho.

Ksmew · Answer

thank you for explaining!!!!

i understood my mistake.

ScottTargetTestPrep · Answer

Solution:
 
Let x be such a number. Since x leaves a remainder of 3 on division by 7, x can be expressed as x = 7k + 3 where k is some positive integer. Similarly, since x leaves a remainder of 2 on division by 4, x can also be expressed as x = 4s + 2 where s is some positive integer.

Notice that x + 18 = 7k + 21 = 4s + 20 is divisible by both 7 and 4; therefore, x + 18 is divisible by LCM(7, 4) = 28. Since the smallest value of x + 18 is 28, the smallest value of x is 10.

Now that we know the smallest value of x, we can find all values of x simply by adding LCM(7, 4) to 10, so x can be any of the values 10, 38, 66, etc.

To find the greatest value of x, let’s divide 1000 - 10 = 990 by 28, obtaining 990/28 = 35 (ignore the remainder). Thus, the greatest value of x is 28 * 35 + 10 = 990.

Now, we simply need to find the number of terms in the arithmetic sequence 10, 38, 66, … , 990; which is given by (990 - 10)/28 + 1 = 35 + 1 = 36.

Answer: E

Kinshook · Answer

Asked: How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?

x = 7k + 3 = 4m + 2
4m - 7k = 1
If m=2; k = 1; 8-7 = 1; (k,m) = (1,2) is a solution
if m increases by 7 then k increases by 4 for each successive solution

(k,m) = {(1,2),(5,9),(9,16),.....}
x_{max} = 990
x_{min} = 10
Common difference = 4*7 = 28
Number of solutions = 980/28 + 1 = 35 + 1 = 36

IMO E

Kinshook · Answer

Asked: How many positive integers are there from 0 to 1000 that leave a remainder of 3 on division by 7 and a remainder of 2 on division by 4?

x = 3mod7 
x = {3,10,17,24}mod28

x = 2mod4
x = {2,6,10,14,18,22,26}mod28

x = 10mod28

x_{min} = 10
x_{max} = 990
Number of solutions = (990-10)/28 + 1 = 35 + 1 = 36

IMO E

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

How many positive integers are there from 0 to 1000 that leave a remai

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 are there from 0 to 1000 that leave a remai

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