Try this beautiful Integer based Problem from Algebra, from PRMO 2018.
Determine the sum of all possible positive integers n, the product of whose digits equals \(n^2 -15n - 27 \)
Algebra
Integer
multiplication
Answer:$17$
PRMO-2018, Problem 17
Pre College Mathematics
Product of digits = \(n^2 – 15n – 27 = n(n – 15) – 27\)
so at first we observe when n=one digit ,two digit and 3 digit numbers.....
If n is a more than 2-digit number, say 3-digit number, then product has to be\(\leq 9 × 9 × 9 = 729\) but \((n(n – 15) – 27)\) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit.
If n is 1-digit then \(n^2 – 15n – 27 = n\) \(\Rightarrow n\)= not an integer , so n is a two digit number
now we will observe for 2-digit numbers.....
Can you now finish the problem ..........
For Two-digit numbers:
As product is positive so n(n-15)-27>0\(\Rightarrow n\geq 17\)
Now two digit product is less than equal to 81
so \(n(n-15)-27\leq 1\)\(\Rightarrow n(n-15)\leq 108\) \(\Rightarrow n\leq 20\)
Therefore n can be \(17\),\(18\),\(19\) or \(20\)
Can you finish the problem........
For \(n\)= \(17\),\(18\),\(19\) or \(20\)
when n=17,then \(n(n-15)-27=7=1 \times 7\)
when n=18,then \(n(n-15)-27=27\neq 1\times 8\)
when n=19,then \(n(n-15)-27=49=1 \neq 9\)
when n=20,then \(n(n-15)-27=73=1 \neq 0\)
Therefore \(n\)=17
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