4. A two-digit number $N$ was multiplied by 2, the digits of the result were swapped, and then the number was divided by 2. The result was the same number $N$. How many such numbers $N$ exist?

Answers:

A) none (-) B) exactly 4 (-) C) at least 10 (+) D) at least 14 (+) E) at least 15 (-)
Solution. Since the result turned out to be the same number, two identical digits were swapped. This means that $2 \mathrm{~N}$ should have two such digits. Let's consider several cases:

1) When multiplying by 2, there was no carry-over to the next place value. Obviously, as $N$, the numbers 11, 22, 33, 44 fit and only they do.
2) When multiplying by 2, there was a carry-over to the tens place. Then the first digit of the number $2 \mathrm{~N}$ is 1. The last digit is always even. Note that if there was a carry-over in the units place during multiplication, then in $2 \mathrm{~N}$ only the last digit (the units digit) will be even. This means that in this case $2 \mathrm{~N}=11 *$, so $N$ can be 55, 56, 57, 58, or 59.
3) Now let's assume there was only a carry-over to the tens place. Then the number $2 \mathrm{~N}$ has two identical even last digits. That is, $2 \mathrm{~N}$ is equal to 100, 122, 144, 166, or 188. Then $N$ is equal to 50, 61, 72, 83, or 94.

Thus, there are a total of 14 different numbers.