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

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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

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