A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder Class 8

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder Class 8

Question 1.

A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7? Class 8

Solution:

Let first number be 7x + 3 and second number be 7y + 5.

Their sum = 7x + 3 + 7y + 5

= (7x + y) + 8

= (7x + y) + 7 × 1 + 1

So, remainder in this case will be 1.

Their difference = 7x + 3 - 7y - 5

= 7(x - y) - 2

= 7 (x - y) - 7 - 2 + 7

= 7 (x - y) - 7 + 5

So, in this case, remainder will be 5.

Their product = (7 + 3) (7y + 5)

= 49xy + 35x + 21y + 15

= 7(7xy + 5x + 3y) + 7 × 2 + 1.

So, in this case, remainder will be 1.

Question 2.

Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity. Class 8

Solution:

Let the numbers be 4, 5 and 6.

Now, 5² - 4 × 6 = 25 - 24 = 1

For 7, 8 and 9, 8² - 7 × 9 = 64 - 63 = 1.

We always obtain 1.

Algebraic equation : Let numbers be m, m + 1, m + 2.

Now, (m + 1)² - m (m + 2)

= m² + 2m + 1 - m² - 2m = 1.

Yes, it is true.