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## Sum of two digit numbers | PRMO-2016 | Problem 7

Try this beautiful problem from Algebra based on Sum of two digit numbers from PRMO 2016.

## Sum of two digit numbers | PRMO | Problem 7

Let s(n) and p(n) denote the sum of all digits of n and the products of all the digits of n(when written in decimal form),respectively.Find the sum of all two digits natural numbers n such that $n=s(n)+p(n)$

• $560$
• $531$
• $654$

### Key Concepts

Algebra

number system

Answer:$531$

PRMO-2016, Problem 7

Pre College Mathematics

## Try with Hints

Let $n$ is a number of two digits ,ten’s place $x$ and unit place is $y$.so $n=10x +y$.given that $s(n)$= sum of all digits $\Rightarrow s(n)=x+y$ and $p(n)$=product of all digits=$xy$

now the given condition is $n=s(n)+p(n)$

Can you now finish the problem ……….

From $n=s(n)+p(n)$ condition we have,

$n=s(n)+p(n)$ $\Rightarrow 10x+y=x+y+xy \Rightarrow 9x=xy \Rightarrow y=9$ and the value of$x$ be any digit….

Can you finish the problem……..

Therefore all two digits numbers are $19,29,39,49,59,69,79,89,99$ and sum=$19+29+39+49+59+69+79+89+99=531$

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## Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on roots of equations.

## Roots of Equations | PRMO | Problem 8

Suppose that $a$ and $b$ are real numbers such that $ab \neq 1$ and the equations $120 a^2 -120a+1=0$ and $b^2-120b+120=0$ hold. Find the value of $\frac{1+b+ab}{a}$

• $200$
• $240$
• $300$

### Key Concepts

Algebra

Roots

Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

## Try with Hints

The given equations are $120 a^2 -120a+1=0$ and $b^2-120b+120=0$.we have to find out the values of $a$ and $b$….

Let $x,y$ be the roots of the equation $120 a^2 -120a+1=0$then $\frac{1}{x},\frac{1}{y}$ be the roots of the equations of $b^2-120b+120=0$.can you find out the value of $a$ & $b$

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…$a=x$ and $b=\frac{1}{y}$ (as $ab \neq 1)$

Can you finish the problem……..

$\frac{1+b+ab}{a}$=$\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}$=$\frac{(x+y)+1}{xy}=240$