2022/017) find positive integer n such that $n^2-10n -22$ is same as product of digits of n

Let us fix the lower bound of n

now $n^2 -10n - 22 = (n^2 - 10n + 25) - (22+25)$ adding 25 for completing the square

$= (n-5)^2 - 47 > = 0$ as it has to be positive

so $(n-5)^2 >= 57$

or $(n-5) >= \sqrt{47}$ 

or $ n > 5 + \sqrt{47}$

as n is integer so n > 5 + 6$ as $6 < \sqrt{47} < 7$ so 5 + 6 is the largest value which does nt satisfy the condition

so $n >11$

now  n = 12 this gives product of digits = 2 and $n^2 -10n -22 = 12^2 - 10 * 12 - 22 = 2$ so they are same

so 12 is a sloution

for $n  > 12$ we have $22 < 2n$ and hence $n^2 - 10n - 22 > n^2 - 10n - 2n$

or $n^2 - 10n - 22 > n(n-12)$ and as $n-12$ is positive we have

$n^2 - 10n -22 >n$ for $ n > 13\cdots((1)$

We shall prove a little lemma:

For for any n the product of the digits of  n is less then n

proof:

let n be a k digit number starting with a.

the value of $n >= a* 10^{k-1}\cdots(2)$

the number that starts with a shall give the largest product when all rest (k-1) digits will be ragest that is 9

so product of digits  $ < a * 9^{k-1}\cdots(3)$

using (2) and (3) we get product of digts < the value n < value of exptession.

end of lemma

using the lemma and (1) we have for $n >12$ it is not possible for the expression . same as product of digits.