# Task № 5.1

## Condition:

In a correct equation, identical digits were replaced with the same letters, and different digits with different letters. The result is

$$
P+\mathrm{P}+\mathrm{A}+3+Д+\mathrm{H}+\mathrm{U}+\mathrm{K}=\mathrm{U} \mathrm{U}
$$

What can U be equal to?
# Answer:

$\circ 1$

$\circ 2$

$\checkmark 3$

$\circ 4$

० 5

० 6

○ 7

○ 8

$\circ 9$

$\circ 0$

Exact match of the answer - 1 point

## Solution.

Notice that the expression involves 9 different letters, meaning all digits except one are used. On the left, we have the sum of eight different single-digit numbers. The sum of all single-digit numbers is $0+1+2+3+4+5+6+7+8+9=45$. Let's see which two-digit numbers with identical digits can be obtained by summing eight numbers.
$44=45-1, 33=45-12, 22=45-23, 11=45-34$. The last two are not possible because the maximum sum of two single-digit numbers is $17=8+9$. Also, 44 is not suitable because to get it, we would need to use all numbers except 0 and 1, but then U cannot be 4 because 4 is already used on the left. The variant 33 is quite suitable. $12=3+9$, and we get the equality:

$$
0+1+2+4+5+6+7+8=33
$$

## Condition:

In a correct equality, identical digits were replaced by identical letters, and different digits by different letters. The result is

$$
П+\mathrm{O}+\mathrm{Б}+\mathrm{E}+Д+\mathrm{И}+Ш+\mathrm{b}=\mathrm{AA} .
$$

What can $A$ be equal to?

## Answer:

$\circ 1$

$\circ 2$

$\checkmark 3$

$\circ 4$

$\circ 5$

○ 6

○ 7

○ 8

० 9

$\circ 0$

Exact match of the answer - 1 point

Solution by analogy with problem № 5.1.

## Condition:

In a correct equality, identical digits were replaced by identical letters, and different digits by different letters. The result is

$$
\mathrm{C}+\mathrm{y}+П+\mathrm{E}+\mathrm{P}+\mathrm{K}+Л+\mathrm{A}+Д=\mathrm{AA}
$$

What can $A$ be equal to?

## Answer:

$\circ 1$

$\circ 2$

$\circ 3$

$\checkmark 4$

$\circ 5$

० 6

○ 7

○ 8

० 9

$\circ 0$

Exact match of the answer - 1 point

Solution by analogy with problem № 5.1.

## Condition:

In a correct equality, identical digits were replaced by identical letters, and different digits by different letters. The result is

$$
Г+\mathrm{P}+\mathrm{A}+Д+\mathrm{y}+\mathrm{C}+\mathrm{H}+\mathrm{И}+\mathrm{K}=\text { ИИ. }
$$

What can И be equal to?

## Answer:

$\circ 1$

$\circ 2$

$\circ 3$

$\checkmark 4$

$\circ 5$

० 6

○ 7

○ 8

० 9

○ 0

Exact match of the answer - 1 point

Solution by analogy with problem № 5.1.

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