Question

How many numbers between 650 and 780 leave the remainders 4 and 6 when divided by 7 and 11, respectively?

Answer 1

18 numbers between 650 and 780 have remainder 4 when

divided by 7

12 numbers between 650 and 780 have remainder 6 when

divided by 11

Explanation:

* Lets explain how to solve the problem

- The first number divisible by 7 between 650 and 780 is 651

∴ The first number between 650 and 780 has reminder 4 when

divided by 7 is 651 + 4 = 655

- The last number divisible by 7 between 650 and 780 is 777

- But we add 777 by 4 the answer will be 781 out the range of the

numbers, so will take the number just before 777 and divisible by 7,

the number is 770

∴ The last number between 650 and 780 has reminder 4 when

divided by 7 is 770 + 4 = 774

- Now lets use the rule of the arithmetic series

∵
`a_(n)=a_(1)+(n-1)d`, where
`a_(1)` is the 1st term, d is

the constant difference between each two consecutive terms and

n is the position of the term in the series

∵
`a_(1)` = 655

∵
`a_(n)` = 774

∵ d = 7

∴ 774 = 655 + (n - 1)(7)

- Subtract 774 from both sides

∴ 119 = (n - 1)(7)

- Divide both sides by 7

∴ 17 = n - 1

- Add both sides by 1

∴ n = 18

18 numbers between 650 and 780 have remainder 4 when

divided by 7

- By checking the number 655 when divided by 11 give remainder 6

∴ The first number between 650 and 780 has remainder 6 when

divided by 11 is 655

- By checking the number 744 when divided by 11 give remainder 4,

we need remainder 6 so add 2 to 774 to get 776

∴ The last number between 650 and 780 has reminder 6 when

divided by 11 is 776

- By using the same rule rule of the arithmetic series

∵
`a_(1)` = 655

∵
`a_(n)` = 776

∵ d = 11

∴ 776 = 655 + (n - 1)(11)

- Subtract 774 from both sides

∴ 121 = (n - 1)(7)

- Divide both sides by 11

∴ 11 = n - 1

- Add both sides by 1

∴ n = 12

12 numbers between 650 and 780 have remainder 6 when

divided by 11