Question

determine the sum of all possible positive integer n, the product of whose digit equals to n^2-15n-27

Answer 1

Here. Hope it helps!

Answer 2

Answer:

The value of n is 17.

Step-by-step explanation:

To determine : The sum of all possible positive integer n, the product of whose digit equals to $n^2-15n-27$?

Solution :

We know,

$n^2-15n-27$

$n(n-15)-27$

If n is a more than 2-digit number, say 3-digit number, then product has to be ≤ 9 × 9 × 9 = 729

But (n(n-15)-27) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n).

So, n can be either one-digit or 2-digit.

If n is 1-digit then $n^2-15n-27 = n$

n = not an integer

So, n is a two-digit number

As product is positive.

So, $n(n-15)-27>0$

$n\geq 17$

As 2-digit product is less than equal to 81.

$n(n-15)-27\leq81$

$n(n-15)\leq108$

$n\leq20$

So, n can be 17,18,19,20.

If n=17, then $n(n-15)-27=17(17-15)=7$

If n=18, then $n(n-15)-27=18(18-15)\neq 1\times 8$

If n=19, then $n(n-15)-27=19(19-15)\neq 1\times 9$

If n=20, then $n(n-15)-27=20(20-15)\neq 0$

Only possible when n is 17.