Question

What is the sum of all natural numbers n such that the product of the digits of
n (in base 10) is equal to n² - 10 n - 36 ?

Answer 1

Answer:

please refer to the attachment

Answer 2

SOLUTION:-

Let product of digits of n be p(n)

Note: p(n)≤n

let n=am10^m + am-1 10^m-1+....+a0

≥am10^m

≥am9^m

≥amam-1..............a0

=) p(n)

Now n² -10n-36≤n

n²-11n-36≤0

-2.64≤n≤13.64

=) -3<n<14.............(1)

Also p(n)≥0

n²-10n-36≥0

(n-5)²≥61

n≥12.81.................(2)

from (1) & (2)

n∈ [12.81,14]

Since, n should be an integer Thus, the only integer lying in the interval [12.81,14] is 13.

n= 13

Hope it helps ☺️