x is a positive integer less than 500. When x is divided by 7, the rem

+1 D
..
the terms common to both 7 and 3 will be of the form 8 29......491
491=8+(n-1)21
==>n=24

the nubmer which when divided by 7 leaves remainder 1 should be of the form 7k+1
this number when divided by 3 leaves remainder 2.
so,
(7k+1)-2 should be divisible by 3 or 7k-1 should be divisible by 3.
we now put the values of k starting from 0 to find first number divisible by 3
we find 1st number at k= 1
thus smallest number will be 7(1)+1 = 8
now,
next number will be = 8+ lcm of 3 & 7
i.e 29
now we will find number of all such values less than 500 by using the formula for last term of an a.p

8+(n-1)21=500

n=24.42
or n= 24

Answer:- D

y = 21x+8 represents all the numbers that fits the description.
Finding the range of x that yields y between 0 and 500 gives us the answer of 24.

So D it is.

Took me more than 2 mins to solve.

That's how I did it.

x is between 1-499 included.
Smallest number that can be written as 7n+1 (7*0+1) is 1
Largest number that can be written as 7n+1 is (7*71 +1) is 498

So there are total 72 numbers that can be written as 7n+1

Because x can also be written as 3m+2, we have to see how many numbers that can be written as 7n+1 also can be written as 3m+2

7*0+1 can not be written in the form 3m+2
7*1+1 can be written in the form 3m+2
7*2+1 can not be written in the form 3m+2
7*3+1 can not be written in the form 3m+2
7*4+1 can be written in the form 3m+2

Here we see a sequence that one out of every 4 numbers can be written as 3m+1

72/4= 24 numbers can be written as 3m+1

D is the answer

now x is less than 500 !!

and ,

let x= 7q+1
or, x= 3p+2

so, 7q+1 = 3p+2
or, q = (3p+1)/7

since p and q are integers !!

the following values will satisfy

p=2, q=1 --> x=8
p=9, q=4 --> x=29
p=12,9=7 --> x=50

we could see an Ap with common difference of 21 !!

the last number less than 500 to satisfy the criteria is 491 !!

we can now use the AP formula to calculate the no of terms

491 = 8+(n-1)21
491 = 21n-13
n= 504/21
n=24

Least possible number of x ( divided by 7, the remainder is 1 ) = 8

The sequence is -

{ 8 , 15 , 22 , 29, 36 , 43 , ..........}

Least possible number of x ( divided by 3, the remainder is 2 ) = 5

{ 5 , 8 , 11 , 14 , 17 , 20 , 23 , 26 , 29..........}

Combine these the series becomes -

{8, 29...........}

Thus difference is 21 , possible nummber of numbers is 499/21 => 23.76

Now use AP series formula

Common difference = 21
First Term = 8

Test with No of terms as (D) 24 and (C) 23

Answer will be (D) as the 24th term of this AP series is 491

Excellent problem from an excellent user... Nothing but sheer respect for great Bunuel

We are given that when x is divided by 7, the remainder is 1. Thus, x can be any integer that is 1 more than a multiple of 7:

1, 8, 15, 22, 29, …

We are also given that when x is divided by 3, the remainder is 2. Thus, x can be any integer that is 2 more than a multiple of 3:

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, ...

The first integer that both sets have in common is 8 and the next is 29. The difference is 21, and so that is the common difference we can use to determine all values that x can take on. The next few numbers in this evenly spaced set will be 29 + 21 = 50 and 50 + 21 = 71, etc.

We have an evenly spaced set of numbers. We see that the largest number under 500 that works for x is 491. Now, let’s determine the number of values of x:

(largest integer in the set - smallest number in the set)/21 + 1

(491 - 8)/21 + 1 = 483/21 + 1 = 23 + 1 = 24

Answer: D

let n=number of x
least value of x=8
8+7*3*(n-1)<500
21n<513
n<24.4
n=24
D

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