How Cheenta works to ensure student success?
Explore the Back-Story

Integer based Problem | PRMO-2018 | Question 20

Join Trial or Access Free Resources

Try this beautiful Integer based Problem from Algebra, from PRMO 2018.

Integer based Problem - PRMO 2018, Question 20

Determine the sum of all possible positive integers n, the product of whose digits equals \(n^2 -15n - 27 \)

  • $9$
  • $17$
  • $34$

Key Concepts




Check the Answer


PRMO-2018, Problem 17

Pre College Mathematics

Try with Hints

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

Subscribe to Cheenta at Youtube

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
Math Olympiad Program