When positive integer n is divided by 3, the remainder is 2. When n is

n=3Q1+2
n=7Q2+5

3Q1+2=7Q2+5
3Q1=7Q2+3
Q1=7Q2/3+1

Approach 1: So Q2 has to be a multiple of 3. Q2 can be 0,3,6,9,12,15....
But when we replace Q2 in 7Q2+5, only 0,3,6,9 and 12 are less than 100

Therefore, n can be 5,26,47,68 or 89--->5

Approach 2: 1st possible value is 1 and then add 1 to multiples of 7. Therefore, possible values of Q1=1,8,15,22,29,36...

But only when Q1=1,8,15,22 and 29 will n<100

n, therefore, can be 5,26,47,68 and 89--->5

Answer : E

Find out n/7=R=5= 5,12,19,26,33,40,47,54,61,68,75,82,89,96
out of all these 5,26,47,68 & 89 gives R=2 when n/3; therefore there are 5 digits in total

Hence Answer is E

Thanks,

Answer = E = 5

Possible values of n which gives remainder 5 when divided by 7:

12, 19, 26, 33, 38, 46, 53, 60, 67, 74, 81, 88, 95

In red are the 5 values which gives remainder 2 when divided by 3

38 divided by 7 doesn't give a remainer of 5.

Although I prefer to tackle questions like these conceptually or by using SMART numbers, the algebraic approach outlined above looks like the best way to solve the problem.

As per the question stem:

3x + 2 = 7y + 5

3x - 3 = 7y

3(x-1) = 7y

Therefore (x-1) is a multiple of 7.

Possible values of x for x-1 being a multiple of 7 include; 1, 8, 15, 22, 29, 36, ...etc

The answer choices that fit the less than 100 parameter are: 1,8,15,22,29.

Choice E. 5.

Hi Guys,

The first integer when divided by 3 leaves the remainder 2 is 5.
Also when 5 is divided by 7 the remainder is 5 itself.
To find out the next number which satisfies both the condition, take LCM of 3 and 7 which is 21.
So the next number which will satisfy both the conditions is 26(5+21).
The next number is 47(26+21) , 68(47+21) and 89(68+1).
So the number less than 100 which satisfy the conditions 5,26,47,68 and 89.
Answer is 5. Choice is E.

VERITAS PREP OFFICIAL SOLUTION:

So n is a number 2 greater than a multiple of 3 (or we can say, it is 1 less than the next multiple of 3). It is also 5 greater than a multiple of 7 (or we can say it is 2 less than the next multiple of 7)

n = 3a + 2 = 3x – 1
n = 7b + 5 = 7y – 2

No common remainder! When we have a common remainder,the smallest value of n would be the common remainder. Say, if n were of the forms: (3a + 1) and (7b + 1), the smallest number of both these forms is 1. When 1 is divided by 3, the quotient is 0 and the remainder is 1. When 1 is divided by 7, the quotient is 0 and the remainder is 1. But that is not the case here. So then, what do we do now? Let’s try and work with some trial and error now. n belongs to both the lists given below:

Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50…
Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75…

Which numbers are common to both the lists? 5, 26, 47 and there should be more. Do you see some link between these numbers? Let me show you some connections:
– 26 is 21 more than 5.
– 47 is 21 more than 26.
– 21 is the LCM of 3 and 7.

How do we explain these? Say, we identified that the smallest positive number which gives a remainder of 2 when divided by 3 and a remainder of 5 when divided by 7 is 5 (note here that when we divide 5 by 7, the quotient is 0 and the remainder is 5). What will be the next such number? Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… away from 5 i.e. it will be a multiple of 3 and a multiple of 7 away from 5. The smallest such multiple is obviously the LCM (lowest common multiple) of 3 and 7 i.e. 21. Hence the next such number will be 21 away from 5. We get 26. Use the same logic to get the next such number. It will be another 21 away from 26 so we get 47. By the same logic, the next few such numbers will be 68, 89, 110 etc. How many such numbers will be less than 100? 5, 26, 47, 68, 89 i.e. 5 such numbers.

Answer: E.

Can anyone pls. explain how N became 21k+5?

Think of it this way. You have been given that a number leaves a reminder of 2 when divided by 3 and the same number when divided by 7, leaves a remainder of 5. Now as 3 and 7 are prime numbers, there is only one way of denoting such a number so that we take the above mentioned divisibilities by both 3 and 7 into account. LCM of 3,7 = 21 and thus Bunuel mentions that the number can be written as 21r+5.

You can also solve this by looking at a few values that this number can take: 5,26,47, 68, 89 etc. These numbers satisfy the 2 conditions of divisibilities with 3 and 7 mentioned above. This way, you don't have to worry about how all of it led to 21r+5.

[/quote]

Think of it this way. You have been given that a number leaves a reminder of 2 when divided by 3 and the same number when divided by 7, leaves a remainder of 5. Now as 3 and 7 are prime numbers, there is only one way of denoting such a number so that we take the above mentioned divisibilities by both 3 and 7 into account. LCM of 3,7 = 21 and thus Bunuel mentions that the number can be written as 21r+5.

You can also solve this by looking at a few values that this number can take: 5,26,47, 68, 89 etc. These numbers satisfy the 2 conditions of divisibilities with 3 and 7 mentioned above. This way, you don't have to worry about how all of it led to 21r+5.[/quote]

Thanks a lot...

Hi robinpallickal,
to your Q that how N=21k+5..
first we find the first number that is ccommon to two eqs..
in case of 3.. it is 2,5,8 and so on
in case of 7.. it is 5,12,19..
so if 5 is the first number next number will be 5+ LCM of 3 & 7=5+21.. next 5+21*2.. next 5+21*3.. and so on...
therefore you get the eq 21k + 5
hope it is clear

When positive integer n is divided by 3, the remainder is 2 i.e. n = 3x + 2

When positive integer n is divided by 7, the remainder is 5 i.e. n = 7y + 5

For some Values of x and y the values of 3x + 2 must be equal to 7y + 5 as both are the values of n

So to find the common Solution

n = 3x + 2 = 7y + 5

PROPERTY: In such liners equations
1) The Value of x always changes by he value of coefficient of y i.e. 7 in this case and
2) The Value of y always changes by he value of coefficient of x i.e. 3 in this case and
3) the value of n always changes by the LCM of coefficients of x and y i.e. 21 in this case

Let's find the first solution of 3x + 2 = 7y + 5
i.e. x = (7y + 3) / 3
i.e. For some Integer value of y, x must be an Integer too

First Value of y = 3 and First value of x = 8 and n = 26
Second Value of y = 6 and First value of x = 15 and n = 47
Third Value of y = 9 and First value of x = 22 and n = 68
Forth Value of y = 12 and First value of x = 29 and n = 89

Answer: Option D

Here is what i would do in such a question
the first number that satisfies such a case is 5
hence the expression for such an integer becomes => 5+21P
hence between 0 and 100 => 5 values => E

Love this approach, so quick and easy. I did the same thing but got the answer wrong because of a careless calculation error. I hate it when i do that.!!

n/3 remainder 2 = (5, 8 , 11, 14, 17, 20, 23, 26....)
n/7 remainder 5 = ( 5, 12, 19, 26.....)


Possible numbers which are divisible by 3 ( having remainder 2 ) > numbers which are divisible by 7 ( having remainder 5) further the first possible value of n is 5 then 26- with a gap of 21 numbers

Now The best thing to do find all the numbers less than 100, that leaves remainder 2 when divided by 3 and remainder 5 when divided by 7 -

5 , (5 + 21) 26 , 47 (26+ 21) , 68 ( + 21) , 89 ( 68+21)

So, there are only 5 numbers and the only answer is (E)

\(3a+2=7b+5\)

\(3a=7b+3\)

Now we have to find integer values for b that will yield integer values for a, such that \(3a + 2 < 100\)

\(b=0, a =1\)
\(b=3, a =8\)
\(b=6, a =15\)
\(b=9, a =22\)
\(b=12, a = 29\)

We stop, since the next b 15 would make n larger than 100

n=3p+2
Put p=0,1,2,..
n=2,5,8,..

n=7q+5
Put q=0,1,2,..
n=5,...

LCM of 3&7 is 21 & first common element is 5
n=21r+5
Put r=0,1,2,3,4,...

n=21*4+5=89<100
n=21*5+5>100

Hence max value is 4 and we have 0,1,2,3,4=5 values.

E.

lowest possible value of n=5
interval between successive values of n=product of divisors 3*7=21
let x=additional possible values of n
5+21x<100
x<4.5
1+4=5 possible values of n<100
E