**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”.