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# Integer based Problem | PRMO-2018 | Question 20

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

Algebra

Integer

multiplication

Answer:$17$

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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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

Algebra

Integer

multiplication

Answer:$17$

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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