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The Super Duper Amazing Number Trick - January 27, 2011
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Spiked Math Comic - The Super Duper Amazing Number Trick



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51 Comments

is it supposed to be 4 for every case?
pick n. square [n^2] add [n^2+n] divide [n+1] add around 23 [n+1+~23] subtract [(n-n)+1+~23] divide [(1+~23)/6] ... so... ~4? [where ~ is read "around" or "approximately"]


also, that integral seems like I'm going to need more than mental operations... I'll leave it for later

The motivation of today's comic comes from the following email sent to the entire Math Department which ended up with lots of follow up replies (to all):

The Year 2011

This year we will experience 4 unusual dates 1/1/11, 1/11/11, 11/1/11 and 11/11/11.

NOW take the LAST 2 digits of the year you were born and add the age you'll be this year. It SHOULD be 111............ how strange is that!

I got this e-mail as well, trying to explain that it wasn't some sort of magical prophecy got me nothing but blank stares.

This is just the lamest "puzzle" I've read.

You could define age by present year - birth year. So if you assume people are born between 1900 & 1999 : adding your age to the last two digits will give you the present year - 1900 = 111 for 2011.

That's not even a trick.

only for the ones born from 1900 to 1999 AD :) [2011 - abxx (age) + xx = 2000 - ab00 + 11]

your equation doesn't work for babies born 2010.

This puzzle is rated T. (well, soon to be 12 year olds can still be tricked)

I first saw this year trick in a version purportedly dating from the second world war. It was used to show a mysterious connection between Roosevelt, Churchill, Stalin and Hitler.

I agree you get four, right up until the last step ;-) The integral is doable (a log and an arctan), but I don't get raising it to the power of Pi as anything analytic. I can't remember how to evaluate the infinite sum of n^2/2^n, either...

Ok, WolframAlpha tells me that the sum goes to three. Are you sure there's not a typo in the integral part of things?

AMAZING! UNBELIEVABLE!

I do it three times, and works!

yeah, i get that k = 4, the sum is 3 and the integral is about 0.85. so .85^pi is about .6 * 3 = 1.8 ish

Curses. I keep getting 2.54968 and I'm too tired to work out where my error is.

@Scott. Assuming you're good up until the interval, you missed the ceiling function around the integral. ceiling(0.85) = 1, and 1^pi = 1, so the problem reduces to the sum (3).

I got k=4 in 7 alright. Then I proved 8. But how do I do the tricky bit with the integral in 9?

The integral to the power pi gives approximately 0.6 and not 1.
http://www.wolframalpha.com/input/?i=%28∫%28x%2F%28x%5E2%2Bx%2B2%29%29dx+from+0+to+4%29%5Eπ

The summation gives exactly 3. I think its a typo.

You use the CEILING function before raising it to pi!!!!!!
Example:
Ceiling (0.85...) = 1
Ceiling (2.13) = 3

Round it up!!!!! (To nearest whole number)

hahahah

Yes, the integral is about 0.85, then you take the ceiling of that (look at those brackets outside the integral), which is the smallest integer greater than .85; that's 1. 1^pi = 1.

Ah, that's a ceiling function, and not just square brackets. Feels like cheating!

Amazing!

When I saw that it was the greatest integer function raised to pi, I tried to solve it assuming that was equal to one. :)) So that we'll get a nice number (3)

It's not the greatest integer function (also called floor), it's the least integer function (also called ceiling).

Oops. That's right.

I gave up at step 8. I believe it can be done--but I can't.

That's my point--the computers can prove it--but I can't do it with elementary proof methods. I believe that, given sufficient time, I could, in theory, accomplish the result--but don't want to spend the time. So I gave up.

You and I put $20 each into a box. Then I sell you the box for $30. We both have made a $10 profit! Repeat for infinite money!

-20-30+40=-10

how is that good for me

uum, about that year + age trick. im born on 11th september 1989. At this moment i am 21 years old, soo 21+89 != 111, its 110 !!!

Add the age you will be this year, which gives 22+89 = 111.

september 11th? damn... that must have been a rough 12th birthday party.

Meh, 12 isn't so bad. A good friend of mine turned 18 on 9/11/01. Talk about a bad day to be legally required to submit your SSS card.

Well, i had my party and everything, and we found out afterwards when the parents came to pick all the kids up :)

16-36=25-45 -> Both giving -20 as a result

16-36+(81/4)=25-45+(81/4) -> Adding (81/4) to both sides to get two squares

4^2-2*4*(9/2)+(9/2)^2=5^2-2*5*(9/2)+(9/2)^2

-> a^2-2*a*b*+b^2=(a-b)^2

(4-(9/2))^2=(5-(9/2))^2 -> Non-expanded form of the squares

4-()=5-() -> Square root and subtraction from both sides

4=5

2+2=5

(1)^2 = (-1)^2
1 =(-1) (square root both sides . . . )

Choose any number between 1 and 10^100. Square it, then add the digits up (in base ten) and square that result. Add 23. Call that number "k." Now compute: (k+3)-(2k)/2. The new number you get is 3!!!!!!!

ceiling(x) =.... is the smallest integer not less than x

how would one compute this sum...

without the help of wolframalpha or what not

1. Find the sum of the terms x^n/2^n with n ranging from 0 to infinity.

2. Differentiate both sides of that equation w.r.t to x.

3. Multiply both sides by x.

4. Differentiate again.

5. Substitute x=1.

For full credit justify that termwise differentiation and the substitution in step 5 are all ok. :-)

shouldn't (8) be at least k instead of at most k? or am I thinking about this incorrectly.

It's worded somewhat strangely (because of course if you can color a graph in n colors and the graph has >=n+1 vertices you can color the graph in n+1 colors), but "at most" is correct. I'd say something like "For every planar graph, at most "k" colors are necessary for a coloring such that no two adjacent vertices receive the same color"

00+11=11 :(

Obviously k=4. Plugging into the sum and integral, I got approxiamtely 1.8, instead of 3. The details are here http://twitpic.com/3v8noe

how did you transform that sum into a limit??

i found a formula for the sum from 0 to an arbitrary integer N, and then set N to infinity

Did you try the path I outlined a few posts above this one?

IOW. Start with the function f(x)=2/(2-x). Let D be the operator d/dx.
Compute D(x*D(f(x))). Substitute x=1. From the Maclaurin series of the said function you can tell that this is also the sum of n^2/2^n with n ranging from 1 to infinity...

Fun fact: I actually chose 4 in the beginning.

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