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The product of two consecutive even numbers is $3248$. Actually, I am interested in finding the larger number out of the two.

Say one of the even numbers is $a$ and the consecutive even number is $a+2$, then $a(a+2) = 3248$, so we can get the larger number by finding the roots of the above quadratic equation.

I was also thinking of another method of applying the product and the sum of the roots!. from which I conclude that one of the roots is negative and the other root is positive.

Any other method of obtaining the larger number without calculating the roots of the quadratic equation?

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7 Answers

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You can also take two consecutive numbers as $a-1$ and $a+1$ the we have:

$a^2-1=3248$

$a^2=3249$

etc...

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A great idea! by Lord Shark the Unknown is to just add $1$ to the equation, so that we will get a complete squared term, that is $a(a+2) + 1 = 3248+1 = 3249$ and that means $(a+1)^2 = 3249$, that is $a = 56$ as $57^2 = 3249$ we can also achieve this by just thinking that $60^2 = 3600$ and the number must be less than 60 and then clever hinting!. So here we have our larger number is $56$!

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We can do it a bit mentally as well knowing that $50 . 60 = 3000$ and $60 . 60 = 3600$. So, the two consecutive even numbers must lie between $50$ and $60$.

Let us concentrate only on those consecutive even numbers whose product would result in $8$ as the last digit. These two consecutive even numbers are "$52$ and $54$" and "$56$ and $58$".

Now, $3248 = 56 . 58$

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The other methods are clearly more awesome. But this is its much lamer cousin.

We have that the product of two consecutive even numbers to be 3248. We can see that the maximum power of $2$ that divides $3248$ is $16$. Since two consecutive even numbers can't be divisible by 4, one has to be divisible by $8$ and the other by $2$. Now, $3248/16=203=7 * 29$. So, $29$ needs to be paired with $2$ and $7$ with the larger $8$, yielding $29*2$ and $28*2$. Clearly, the first one is larger and hence, our answer.

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When a (positive) number can be written as $ab$ with $a,b$ close together, try Fermat's factoring technique. Find the squares larger than the number, say we call the number $n = 3248.$ The first few squares larger than this are $57^2 = 3249, \; $ $58^2 = 3364, \;$ $59^2 = 3481, \; $ $ 60^2 = 3600 $

Fermat would take the differences and see when squares occurred. As you have seen, $3249 - 3248 = 1,$ so $57^2 - 3248 = 1^2,$ so $57^2 - 1 = 3248.$

Extra credit: factor $8051.$

Detail: This can work for numbers divisible by $4$ and odd numbers. It cannot work for any number $n \equiv 2 \pmod 4.$ In such a case, try the same technique for $n/2$ which is odd.

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$\sqrt{3248} \approx 56+$ This suggests $56 \times 58$

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let $x$ & $x+2$ be $2$ consecutive even numbers$$ \begin{align} (x)(x+2)&=3248 \\ x^2 + 2x -3248 &= 0 \\ x^2 + 58x - 56x - 3248 &= 0 \\ x(x + 58) - 56(x + 58) &= 0 \\ (x - 56)(x + 58) &= 0 \\ \end{align} $$$x = 56$ since the value could not be in negative.

$x + 2 = 58$ which is the largest number.

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