How many 3 digit numbers are there which leaves remainder 2 on division by 7?

Note: For solving questions like these, the basic concept of Arithmetic progression and its formula should be known. If you know the formulas then they are the direct application of the value and solving it. Since we had already obtained common difference d = 3 and n = 300, we could also have used formula for sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].

1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.Video transcript1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.First 3 Digit Number Exactly  Divisible by 7Last 3 Digit Number Exactly  Divisible by 7Sum of All 3 Digit Numbers Divisible by 7How many 3 digit number are there which leave a remainder 3 on division by 7?How many 3 digit numbers are divisible by 7 find their sum?How many 3 digit numbers are there which leaves remainder 2 on division by 7?What number has a remainder of 3 divided by 7?How many 3 digit numbers which leaves a remainder 3 on division by 7 find the sum of all such terms?What number has a remainder of 3 divided by 7?How many 3 digit numbers are there which leaves remainder 2 on division by 7?How many 3 digit numbers are there those are divisible by 7?Solution1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.Video transcript1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.First 3 Digit Number Exactly  Divisible by 7Last 3 Digit Number Exactly  Divisible by 7Sum of All 3 Digit Numbers Divisible by 7How many three digit number which leaves remainder 3 on division by 7?What is the smallest 3 digit number divisible by 7?How many numbers of 3 digit who divided of 7?How many 3 digit numbers are there which leaves remainder 2 on division by 7?4 = 2x27 and 13 are prime numbers.LCM ( 4, 7, 13) = 364We know that, the largest 4 digit number is 9999Step 1: Divide 9999 by 364, we get9999/364 = 171Step 2: Subtract 171 from 99999999 - 171 = 9828Step 3: Add 3 to 98289828 + 3 = 9831Therefore 9831 is the number.1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.Video transcript1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.First 3 Digit Number Exactly  Divisible by 7Last 3 Digit Number Exactly  Divisible by 7Sum of All 3 Digit Numbers Divisible by 7How many 3 digit number are there which leave a remainder 3 on division by 7?How many 3 digit numbers are divisible by 7 find their sum?How many 3 digit numbers are there which leaves remainder 2 on division by 7?What number has a remainder of 3 divided by 7?How many 3 digit numbers which leaves a remainder 3 on division by 7 find the sum of all such terms?What number has a remainder of 3 divided by 7?How many 3 digit numbers are there which leaves remainder 2 on division by 7?How many 3 digit numbers are there those are divisible by 7?Solution1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.

Dividend = (divisor× quotient) + remainder

Here, divisor = 7

Remainder = 3

First 3 digit no. is 100

For, dividend = 100, quotient = 13.6…

Hence, first 3 digit no. divisible by 7 with remainder 3 comes with quotient>13.6… = 14

When quotient = 14, dividend = 101

Last 3 digit no. divisible by 7 with remainder 3 = 997

an = 997

a = 101

d = 7

an = a + (n-1)× d

∴ n-1 = 128

∴ n = 129

Hence,129 numbers are present.

Video transcript

Let's take the number 7 and divide it by 3. And I'm going to conceptualize dividing by 3 as let me see how many groups of 3 I can make out of the 7. So let me draw 7 things-- 1, 2, 3, 4, 5, 6, 7. So let me try to create groups of 3. So I can definitely create one group of 3 right over here. I can definitely create another group of 3. So I'm able to create two groups of 3. And then I can't create any more full groups of 3. I have essentially this thing right over here left over. So this right over here, I have this thing remaining. This right over here is my remainder after creating as many groups of 3 as I can. And so when you see something like this, people will often say 7 divided by 3. Well, I can create two groups of 3. But it doesn't divide evenly, or 3 doesn't divide evenly into 7. I end up with something left over. I have a leftover. I have a remainder of 1. So this is literally saying 7 divided by 3 is 2 remainder 1. And that makes sense. 2 times 3 is 6. So it doesn't get you all the way to 7. But then if you have your extra remainder, 6 plus that 1 remainder gets you all the way to 7. Let's do another one. Let's imagine 15 divided by 4. Let me draw 15 objects-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Now, let me try to divide it into groups of 4. So let's see, that's one group of 4. That's another group of 4. And then that's another group of 4. So I'm able to create three groups of 4. But then I can't create a fourth full group of 4. I am then left with this remainder right over here. I have a remainder right over here of 3. I have 3 left over. So we could say that 15 divided by 4 is 3 remainder 3. 4 goes into 15 three times. But that only gets us to 12. 4 times 3 is 12. To get all the way to 15, we need to use our remainder. We have to get 3 more. So 15 divided by 4, I have 3 left over. Now, let's try to think about this doing a little bit of our long division techniques. So let's say that I have 4. Let's say I want to divide 75 by 4. Well, traditional long division techniques. 4 goes into 7 one time. And If you're looking at place value, we're really saying the 4 is going into 70 ten times, because we're putting this in the tens place. And then we say, 1 times 4 is 4. But really, once again, since it's in the tens place, this is really representing a 40. But either way, we subtract 4 from the 7. We get a 3. And then we bring down this 5. And we say 4 goes into 35. Well, let's see. 4 times 8 is 32. 4 times 9 is 36. That's too big. So it goes 8 times. 8 times 4 is 32. You subtract 35 minus 32 is 3. And 4 doesn't go into 3 anymore. So here I have this 3 left over. I have a remainder of 3. So you could say that 75 divided by 4 is equal to 18 remainder 3.

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• Video transcript
• 1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.
• First 3 Digit Number Exactly  Divisible by 7
• Last 3 Digit Number Exactly  Divisible by 7
• Sum of All 3 Digit Numbers Divisible by 7
• How many 3 digit number are there which leave a remainder 3 on division by 7?
• How many 3 digit numbers are divisible by 7 find their sum?
• How many 3 digit numbers are there which leaves remainder 2 on division by 7?
• What number has a remainder of 3 divided by 7?

Solution

1st three-digit number which leaves the remainder 3 when divided by 4 = 1032nd three-digit number which leaves the remainder 3 when divided by 4 = 1073rd three-digit number which leaves the remainder 3 when divided by 4 = 111…nth three-digit number which leaves the remainder 3 when divided by 4 = 999So, the arithmetic sequence is 103, 107, 111, … , 999Common difference = 107 − 103 = 4To get the nth three-digit number from the 1st three-digit number, we must add the common difference (n − 1) times to the 1st three-digit number.∴ 999 = 103 + (n − 1) × 4⇒ 999 − 103 = 4n − 4 ⇒ 4n − 4 = 896 ⇒ 4n = 900 ⇒ n = 225Thus, there are 225 three-digit numbers which leaves the remainder 3 when divided by 4.

To get the sum of 3 digit numbers divisible by 7, first we have to find the first and last 3 digit numbers divisible by 7. 

First 3 Digit Number Exactly  Divisible by 7

The first and also the smallest 3 digit number is 100.

To find the first 3 digit number divisible by 7, we divide the very first 3 digit number 100 by 7 

100/7  =  14.29

We have decimal in the result of 100/7.

Clearly the first 3 digit number 100 is not exactly divisible by 7.

Let us divide the second 3 digit number 101 by 7.

101/7  =  14.43

We have decimal in the result of 101/7 also.

So, the second 3 digit number 101 is also not exactly divisible by 7

Here, students may have some questions on the above process.

They are,

1. Do we have to divide the 3 digit numbers by 7 starting from 100 until we get a 3 digit number which is exactly divisible by 7?

2. Will it not take a long process?

3. Is there any shortcut instead of dividing the 3 digit numbers 100, 101, 102.... one by one?

There is only one answer for all the above three questions.

That is, there is a shortcut to find the first three digit number which is exactly divisible by 7.

SHORTCUT

What has been done in the above shortcut ?

The process which has been done in the above shortcut has been explained clearly in the following steps.

Step 1 :

To get the first 3 digit number divisible by 7, we have to take the very first 3 digit number 100 and divide it by 7. 

Step 2 :

When we divide 100 by 7 using long division as given above, we get the remainder 2. 

Step 3 :

Now, the remainder 2 has to be subtracted from the divisor 7.

When we subtract the remainder 2 from the divisor 7, we get the result 5 (That is 7 - 2 = 5).

Step 4 :

Now, the result 5 in step 3 to be added to the dividend 100. 

When we add 5 to 100, we get 105.

Now, the process is over. 

So, 105 is the first 3 digit number exactly divisible by 7.

This is how we have to find the first 3 digit number exactly divisible by 7.

Important Note :

This method is not only applicable to find the first 3 digit number exactly divisible by 7. It can be applied to find the first 3 digit number exactly divisible by any number, say k.

Last 3 Digit Number Exactly  Divisible by 7

The last and also the largest 3 digit number is 999.

To find the last 3 digit number divisible by 7, we divide the very last  3 digit number 999 by 7. 

999/7  =  142.71

We have decimal in the result of 999/7

Clearly the last 3 digit number 999 is not exactly divisible by 7.

Let us divide the preceding 3 digit number 998 by 7.

998/7  =  142.57

We have decimal in the result of 998/7 also.

So, the preceding 3 digit number 998 also is not exactly divisible by 7

Here, students may have some questions on the above process.

They are,

1. Do we have to divide the 3 digit numbers .......997, 998, 999 by 7  until we get a 3 digit number which is exactly divisible by 7 ?

2. Will it not take a long process ?

3. Is there any shortcut instead of dividing the 3 digit numbers ...........997, 998, 999 one by one ?

There is only one answer for all the above three questions.

That is, there is a shortcut to find the last three digit number which is exactly divisible by 7.

SHORTCUT

What has been done in the above shortcut ?

The process which has been done in the above shortcut has been explained clearly in the following steps.

Step 1 :

To get the last 3 digit number divisible by 7, we have to take the very last 3 digit number 999 and divide it by 7. 

Step 2 :

When we divide 999 by 7 using long division as given above, we get the remainder 5. 

Step 3 :

Now, the remainder 5 has to be subtracted from the dividend 999.

When we subtract the remainder 5 from the dividend 999, we get the result 994 (That is 999 - 5 = 994).

Now, the process is over. 

So, 994 is the last 3 digit number exactly divisible by 7.

This is how we have to find the last 3 digit number exactly divisible by 7.

Important Note :

The process of finding the first 3 digit number exactly divisible by 7 and the process of finding the last 3 digit number exactly divisible by 7 are completely different.

Be careful! Both are not same.

The methods explained above are not only applicable to find the first 3 digit number and last 3 digit number exactly divisible by 7. They can be applied to find the first 3 digit number and last 3 digit number exactly divisible by any number, say k.

Sum of All 3 Digit Numbers Divisible by 7

Let us see how to find the sum of all 3 digit numbers divisible by 7 in the following steps.

Step 1 :

The first 3 digit number divisible by 7 is 105.

After 105, to find the next 3 digit number divisible by 7, we have to add 7 to 105. So the second 3 digit number divisible by 7 is 112.

In this way, to get the succeeding 3 digit numbers divisible by 7, we just have to add 7 as given below. 

105, 112, 119, 126,...................994

Clearly, the above sequence of 3 digit numbers divisible by 7 forms an arithmetic sequence. 

And our aim is to find the sum of the terms in the above arithmetic sequence. 

Step 2 :

In the arithmetic sequence

105, 112, 119, 126,...................994, 

we have

first term  =  105

common difference  =  7

last term  =  994

That is,

a  =  105

d  =  7

l  =  994

Step 3 :

The formula to find the numbers of terms in an arithmetic sequence is given by

n = [(l - a) / d] + 1

Substitute a = 105, l = 994 and d = 7.

n  =  [(994 - 105) / 7] + 1

n  =  [889/7] + 1

n  =  127 + 1

n  =  128

So, number of 3 digit numbers divisible by 7 is 128.

Step 4 :

The formula to find the sum of n terms in an arithmetic sequence is given by

=  (n/2)(a + l)

Substitute a = 105, d = 7, l = 994 and n = 128.

=  (128/2)(105 + 994)

=  64 x 1099

=  70336

So, the sum of all 3 digit numbers divisible by 7 is 70336.

Note :

The method explained above is not only applicable to find the sum of all 3 digit numbers divisible by 7. This same method can be applied to find sum of all 3 digit numbers divisible by any number, say k.

How many three digit numbers are there which leave remainder 3 dividing with 7?

How many 3 digit numbers are there divided by 7?

How many 3 digit numbers are there which leaves remainder 2 on division by 3?

How many 3 digit numbers are there which leaves remainder 5 on division by 7?