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## Question

What kind of poem am I?

(12 + 144 + 20 + (3 x (4^.5)))/7 + 5 x 11 = (9 ^2) + 0

a limerick

A dozen a gross and a score,

Plus three times the square root of four,

Divided by seven,

Plus five times eleven,

Is nine squared and not a bit more

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I like it! Here's another one I found...

4 + (6! - 0.5(12^2 + (403 + 1))) = 2(15^2)

Four plus the difference between

The factorial of six and the mean

Of twelve squared and four

Hundred three (plus one more)

Equals double the square of fifteen.

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And I haven't personally checked this but I'll take it on faith...

The integral of z-squared dz,

From 1 to the cube root of 3,

Times the cosine,

Of 3 pi by 9,

Is the log of the cube root of e.

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I wrote the second limerick in 2005. I had read the first one (the "a dozen a gross, and a score" one) which was written by the late John Saxon and was inspired to write my own. I was in the middle of a 4th grade teacher inservice meeting, but, in between round table discussions, I worked on finding a way to get the numbers just right so that the limerick would contain a true equation. It took about an hour. I knew I wanted mean average, a square, and a factorial in the limerick, and I knew that the first, second, and fifth lines would contain the words "between", "mean", and something-"teen", so I worked around that. Now, I have something really cool to show my students each year.

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And I haven't personally checked this but I'll take it on faith...

<div style="margin:20px; margin-top:5px">

<div class="smallfont" style="margin-bottom:2px">Spoiler for answer: <input type="button" value="Show" style="width:45px;font-size:10px;margin:0px;padding:0px;" onClick="if (this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display != '') { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = ''; this.innerText = ''; this.value = 'Hide'; } else { this.parentNode.parentNode.getElementsByTagName('div')[1].getElementsByTagName('div')[0].style.display = 'none'; this.innerText = ''; this.value = 'Show'; }">

</div><div class="alt2" style="margin: 0px; padding: 6px; border: 1px inset;"><div style="display: none;">The integral of z-squared dz,

From 1 to the cube root of 3,

Times the cosine,

Of 3 pi by 9,

Is the log of the cube root of e. </div></div></div>

the integral of z^2 dz = z^3 / 3, from 1 to cube root of 3 is:

(cubrt3)^3 / 3 - 1^3 / 3

3/3 - 1/3

2/3

2/3 times the cosine of 3pi/9 = 2/3 * cos(pi/3) = 2/3 * 1/2 = 1/3

so the left side evaluates to 1/3

the right side is ln(e^1/3), so pull out the power: 1/3 * ln(e) = 1/3 * 1 = 1/3

so both sides are 1/3

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