Only "7","0","+"allowed

$222222 = 2 \times 3 \times 7 \times 11 \times 13 \times 37$ $= 7 \times 31746$ $= 7 \times ( 3 \times 10^{4} + 1 \times 10^{3} + 7 \times 10^{2} + 4 \times 10^{1} + 6 \times 10^{0} )$ One way to type this on the calculator looks like this $77777 + 70777 + 70777 + 777 + 707 + 707 + 700$ In the ten thousands place, the number 7 is added 3 times, in the thousands place, the number 7 is added once, and so on and so forth. The number of 7's in this particular method is 21, but in any method of addition, the number of 7's used is equal to $3 + 1 + 7 + 4 + 6 = 21$ .

Thus, the answer is $\boxed{21}$

Firstly, we need to understand that it doesn't matter how many additions you make, only how many times you use the number $7$ . Then we want to minimize the number of times we use $7$ .

So first we see how many times $70,000$ goes into $222,222$ , which is $3$ . --------------------- $3$

$70,000 \times 3 = 210,000$

$222,222 - 210,000 = 12,222$

Now we want to now how many times $7,000$ goes into $12,222$ , which is $1$ .------------------ $1$

$12,222 - 7,000 = 5,222$

$700$ goes into $5,222$ $7$ times.------------------------------------------------------------------------------------------------ $7$

$700 \times 7 = 4,900$ ; $5,222 - 4,900 = 322$

$70$ goes into $322$ $4$ times.-------------------------------------------------------------------------------------------------------- $4$

$70 \times 4 = 280$ ; $322 - 280 = 42$

$42 / 7 = 6$ ----------------------------------------------------------------------------------------------------------------------------------------------- $6$

Adding these: $3+1+7+4+6 = 21$

This can all be simplified to $222,222 / 7 = 31746$ .

Again $3+1+7+4+6 = 21$ , adding the sum of the digits. This will work in general cases of this problem, but might not be obvious from the beginning.

So the answer is $\boxed{21}$

222222/7 = 31746 ----> 3 + 1 + 7 + 4 + 6 = 21

Let's assume that we use $a+b+c+d+e$ number of $"7"$ keys where we type the $70000$ $a$ times, $7000$ $b$ times, $700$ $c$ times, $70$ $d$ times, and $7$ $e$ times.

$a \times 70000 + b \times 7000 + c \times 700 + d \times 70 + e \times 7 = 222222$
$7(10000a + 1000b + 100c + 10d + e) = 222222$
$10000a + 1000b + 100c + 10d + e = 31746$

$"10000a + 1000b + 100c + 10d + e"$ can be written as the number $"\overline{abcde}"$

$\overline{abcde}=31746$

Therefore, $a+b+c+d+e = 3+1+7+4+6 = \boxed{21}$

I'll work backwards, trying to reach 0 from 222,222 by subtraction.

Note that subtracting 777 for example is the same as subtracting 700, 70, and 7, with the same amount of 7's. If we subtract numbers made up just of 7's, we may end up adding ten 7's from the same place value, which is the same as one from a higher place value. It is thus optimal to subtract in this way:

$222,222 - 70,000 - 70,000 - 70,000 - 7,000 - 700 - 700 - 700 ... = 0$

Subtracting the highest $7*10^n$ that we can until we reach 0. This way we will never repeat the same place value ten times, making this the optimal method. Doing this we can count that the amount of 7's is $\boxed{21}$

Because we must use the least number of times of 7 so first we use 70000*3=210000 which use 7 ,

_ 3 times _

Now it is left 222222-210000=12222 next we use 7000 which use 7 , _ 1 times _

Now it is left 12222-7000=5222 next we use 700*7=4900 which use 7, _ 7 times _

Now it is left 5222-4900= 322 next we use 70*4=280 which use 7, _ 4 times _

Now it is left 322-280=42 use 7 another. 6 times

So all we use is 3+1+7+4+6= 21 times