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If a certain number is divided by the sum of its two digits, the quotient is $7$ and the remainder is $3$. If the digits are reversed and the resulting number is divided by the sum of the digits, the quotient is $3$ and remainder is $7$. Find the number.

My Attempt:

Let the number be $10x+y$.

According to question: $$\dfrac {10x+y}{x+y}=??$$.

I could not get how to make the equation using quotient and remainder. Please help.

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    $\begingroup$ Hint: the first constraint reads $10x+y = 7(x+y) + 3$. Write the second constraint and solve the system. $\endgroup$ – mlc Mar 31 '17 at 15:32
  • $\begingroup$ Read the problem more carefully and write down the full equation for each of the two conditions. That gives you two equations in two unknowns. $\endgroup$ – Ross Millikan Mar 31 '17 at 15:40
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HINT:$$10x+y=7(x+y)+3=7x+7y+3$$ and $$10y+x=3(x+y)+7=3x+3y+7$$ Can you proceed now?

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  • $\begingroup$ How did you make the equations? $\endgroup$ – pi-π Mar 31 '17 at 15:34
  • $\begingroup$ It's the division algorithm $dividend=quotient\cdot divisor+remainder$ $\endgroup$ – user35508 Mar 31 '17 at 15:35
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To generalize a bit, if the quotient/remainder pairs are $a,b$ and $b,a$ and $a+b=10$, then the two digits of the numbers in the solution are $a$ and $b$.

So in the case of a problem where the quotient is 6 and remainder 4 and vice versa, the answer is $64$ and $46$

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The way it is given, it can be translated in the equations as follows:

I am assuming that the orignal number is $10x+y$ and so the reversed number will be $10y+x$.

ATQ:

$$10x+y=7(x+y)+3$$ and $$10y+x=3(x+y)+7$$

You have two equations and two unknowns ($x$ and $y$), solving these two will give you $x=7$ and $y=3$. So the orignal number is $10\times 7+3=73$.

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