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Number $S$ is obtained by squaring the sum of digits of a two digit number $D$. If the difference between $S$ and $D$ is $27$, then the two digit number $D$ is?

My thoughts:

Let the two digit number $D$ be $AB$.

And so $S=(A+B)^2$

If $\,S-D=27,\,$ then $\,(A+B)^2 -AB=27$

$$A^2 + 2AB + B^2 - AB=27$$

Now how to obtain the value of $D$ further?

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    *To get the best possible answers, you should explain what your thoughts on the problem are so far*. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself.2012-12-25
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    sure. let me edit my thoughts on the same.2012-12-25
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    I think you've confused yourself with your notation here, `2AB-AB` can't be simplified since `AB` is D and `2AB` is twice the product of D's digits2012-12-25
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    @jozefg, Sorry for the confusion. I just tried to get the value somehow. thats it.2012-12-25
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    @jozefg you are totally right, i was also confused till i read your comment... lol2012-12-25
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    Everyone seems to have missed a second solution. I don't know your level of math skill, so I tried to keep things as simple as possible in case you had no knowledge of modular arithmetic.2012-12-25

3 Answers 3

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To try and approach systematically:

We know $|S-D|=27$.

Let $D$ is comprised of the digits $A$ and $B$, with each of $A, B$ each a single digit integer. Let $A$ be the left-most digit (the "ten's digit of $D$), and let $B$ be the right-most digit (the "one's" digit of $D$).

So $S = (A + B)^2$ and $D = 10 \cdot A + B.\;\;\;\;$(*)

Then $$|S - D| = 27 \iff |(A + B)^2 - (10A + B)| = 27, \quad (0 < A < 10, \;0 \le B < 10)$$

$$\quad\quad\quad\quad\quad\quad\quad\quad\iff (A+B)^2 - (10A + B) = 27,\quad\text{OR}\quad (10 A + B) - (A + B)^2 = 27$$ $$\text{with}\quad(0 < A < 10, \;0 \le B < 10)$$


(*) For example, if $D = 54$, then $D = 5 \cdot 10 + 4$, $S = (5 + 4)^2 = 81,\; S - D = 27$;
$\quad\;$ and if $D = 73 = 7\cdot 10 + 3\;\;, S = (7 + 3)^2 = 10^2 = 100; \;S - D = 27.$

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    great. This helps. Thank you.2012-12-25
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I believe I see 2 solutions for this problem. First let's redefine $D$ algebraicly as $D=10A+B$. So our equation is

$$(A+B)^2-10A-B=27$$

I don't know if you know any modular arithmetic. You may at least be aware of the divisibility test for $9$. The sum of the digits of a number is closely related to its remainder when divided by $9$. So it is a reasonable guess that we try to determine the remainder of both sides when divided by $9$. This simplifies the equation to

$$(A+B)^2-A-B\equiv0\pmod9$$ $$(A+B)^2-(A+B)\equiv0\pmod9$$ $$(A+B)(A+B-1)\equiv0\pmod9$$

So we have the product of 2 consecutive integers has a remainder of $0$ when divided by $9$, or in other words this product is divisible by $9$. Since 2 consecutive numbers share no common factors besides $1$, either $A+B$ is divisible by $9$, or $A+B+1$ is divisible by $9$. $D=99$ is too large and $D=10$ is too small, leaving 2 possibilities.

Our first possibility is $A+B=9$. This gives us

$$S=9^2=81$$ $$D=S-27=81-27=54$$

Our second possibility is $A+B-1=9$ or $A+B=10$. This yields

$$S=10^2=100$$ $$D=S-27=100-27=73$$

Both values check out. $D$ is either $54$ or $73$.

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    wo, wo, great, dude :D2012-12-25
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    awesome. thanks a lot.2012-12-25
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it is: $81 = (5 + 4)^2$ and $81 - 54 = 27$. so the $S=81$ and $D=54$