Find the last digit of a number

Whether natural numbers x, y, z, u satisfy the relation:

7x + 5y - 2z -2u = 0, then find the last digit of the number A when

A = (11x + 9y) *(z + u - x).

Solution:

The problem gives us an equality between the numbers x, y, z and u.

7x+5y – 2z – 2u = 0 (the hypothesis)

If you look at this equality you can see that the relation could be processed moving two terms on the right side of the equality.

-2z and -2u moved to the right side with the changed sign become:    2z and 2u.

7x + 5y = 2z + 2u. We can remove the common factor 2.

7x + 5y = 2 (z + u).

From this form of the last relationship we note the following: 

to the right we have a bracket multiplied by 2, which means that this product is an even number.

If on the right side of the equality it is an even number, then we also have an even number on the left of equal sign.

So, 7x + 5y is an even number.

Here the sum is even.

How do we get an even number in the sum? if we add two even numbers or two odd numbers!

How 7 and 5 are odd, it follows that x and y are either both even or both odd (in other words: they have the same parity).

We conclude that the amount x + y is even (if we add two even numbers the result is an even number, or if we add two odd numbers the result is an even number also).

The problem asks us the last digit of the number A

A = (11x + 9y) *(z + u - x).

.

Number A is a product of two brackets. We see that z is assembled with u and this amount (z + u) is in the hypothesis multiplied by 2:    7x + 5y = 2 (z + u).

.

To be able to use the hypothesis equality we multiply by 2 the number A and we replace 2(z + u) by 7x+5y. And we get:

2 A = (11x + 9 y) * 2 * (z + u - x) = (11x + 9 y) (2 (x + u) - 2 x) = (11x + 9 y) (7 x + 5 y - 2 x) = (11x + 9 y) (5x + 5y) = (11x + 9 y) * 5 (x + y)

We have demonstrated that x and y have the same parity, and x + y is an even number, x + y is divisible by 2.

For parenthesis (11x + 9y) we note:

If x and y are both even 11x + 9y is an even number, so it is divisible by 2.

If x and y are both odd 11x is odd, 9y is also odd and the sum of two odd numbers is always an even number, so 11x + 9y is divisible by 2.

It follows that number 2A is divisible by the number 2 from the bracket 11x + 9y, once it is divisible by 2 by the even number x + y and it is also divisible by 5.

Number A will be divisible by 2 and 5.

Therefore, number A is divisible by 2 and 5, so it is divisible by 10.

From the criterion of divisibility by 10 (any number that has the last digit “0” is divisible by 10) the last digit of the number A is „0”.