Question

How many 3 digit positive integers exist that leave remainder 5 when divided by 7?

Answer 1

Hello!

• Min. three digit number = 100
• Max. three digit number = 999

ATQ
AP : 103, 110, 117 ... 999

Then,

$\sf a_{n} = \sf a + (n - 1)d$

⇒ 999 = 103 + (n - 1) 7

⇒ 999 - 103 = (n - 1) 7

⇒ n - 1 = $\dfrac{\sf 896}{\sf 7}$

⇒ n = 128 + 1

⇒$\boxed{\sf n = 129}$

Cheers!

Answer 2

Answer

There exists are 129 positive 3 digit integers that, when divided by 7, leave remainder 5.

Concept

An arithmetic progression or arithmetic sequence is a set of numbers in which the difference between the terms is always the same.

For example, 2,4,6,8,10,12.... is an arithmetic progression with a common difference of 2.

The formula to calculate the $n^{th}$ term of a sequence ($a_n$) is given as;

${\displaystyle \ a_{n}=a_{1}+(n-1)d}$

where; $a_1$ is the initial term, d is the difference in terms.

Given:

Difference, d = 7

Find:

1) $n^{th}$ term of a sequence

2) Initial term, $a_1$

3) No. of 3 digit positive integers that leaves remainder 5 when divided by 7.

Solution:

We find the smallest 3 digit number that leaves remainder 5 when divided by 7 by trial and error method.

Smallest 3 digit no =  100

By long division method, 100 divded by 7 gives 2 as remainder.

Moving to the next number, i.e. 101, when divded by 7 gives 3 as the remainder.

Similarly, 103 when divded by 7 gives 3 as the remainder.

Therefore, the smallest 3 digit number that leaves remainder 5 when divided by 7 is 103.

Adding 7 to 103 gives the next 3 digit no that leaves remainder 5 when divided by 7. Thus the terms of the sequence are obtained by adding 7 to the preceeding term. i.e.,

103 + 7 = 110

110 + 7 = 117....

We now find the largest 3 digit number that leaves remainder 5 when divided by 7 by trial and error method.

Largest 3 digit no = 999

By long division method, 999 divded by 7 gives 5 as remainder.

Therefore, the largest 3 digit number that leaves remainder 5 when divided by 7 is 999.

Therefore the sequence is as follows;

103, 110, 117.......... 999

Calculating Further using Arithmethic Progressions

$a_1 = 103\\d = 7\\a_n = 999\\\\a_n = a_1 + (n-1)d\\999 = 103 + (n-1)7\\999 = 103 + 7n - 7\\999-103 + 7 = 7n \\903 = 7n \\n = \frac{903}{7} \\n = 129$

There are 129 positive 3 digit integers that leave remainder 5 when divided by 7