Math Olympiad Number Theory PRMO USA Math Olympiad

Integer based Problem | PRMO-2018 | Question 20

Try this beautiful Integer-based Problem from Algebra from PRMO 2018, Question 20. You may use sequential hints to solve the problem.

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




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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

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