The product of any whole numbers each of which leaves a remainder of 1 on dividing by 8, also leaves a remainder of 1 on dividing by 8. Why?

First I picked the any two numbers that leave a remainder of 1 on dividing by 8;
9 and 17

9 x 17 = 153 also leaves a remainder of 1 on dividing by 8.

Then I did; 9 = 8 + 1; also 17 = 16 + 1

and 9 x 17 = (8 + 1) x (16 + 1) = (8 x 19) + 1

I am stuck on how to explain how this is so, not sure if there is another way of presenting this.

Thank you for the help.

The product of any whole numbers each of which leaves a remainder of 1 on dividing by 8, also leaves a remainder of 1 on dividing by 8. Why?

First I picked the any two numbers that leave a remainder of 1 on dividing by 8;
9 and 17

9 x 17 = 153 also leaves a remainder of 1 on dividing by 8.

Then I did; 9 = 8 + 1; also 17 = 16 + 1

and 9 x 17 = (8 + 1) x (16 + 1) = (8 x 19) + 1

I am stuck on how to explain how this is so, not sure if there is another way of presenting this.

Thank you for the help.
If a number leaves a remainder of 1 when divided by 8, then that number is equal to some other number times 8 plus 1.

Take two number that have a remainder of 1 when divided by 8. If we let these two numbers be x and y, where a, b, x, y are all integers, we can let:
x = 8a + 1
y = 8b + 1

Then the product of these numbers is:
xy = (8a + 1)(8b + 1) = 64ab + 8a + 8b + 1 = 8(8ab + a + b) + 1

Where (8ab + a + b) is another integer, we have that whatever x and y equal, if x and y have a remainder of 1 when divided by 8, then the product of x and y will have a remainder of 1 when divided by 8.

3. I appreciate your help. Thnk you.