Pascal's Bomb

Revision as of 18:59, 26 January 2021 by Awesome3.14 (talk | contribs)

Pascal's Bomb(Introduction)

Pascal's Bomb is a (not)widely known theorem, applying it will solve almost every problem. However, the concept is quite hard to grasp, yet it is very important.

Pascal's Bomb is 1% Adihaya Jayasharmaramankumarguptareddybavarajugopal's lemma, and 99% pure big brain.


In the year 69, Munkey man first developed the idea. It was then sent to Gmaas for review, approved by Gmaas, and became published.

Forgotten by the year 696, it was later re-discovered. In the year 4269, bestzack66 got on FTW and said "Pascal's" and MathHayden said "Bomb". Thus, it became a real theorem.

Later on, bestzack66 and MathHayden contributed to mankind by re-publishing the theorem, this time onto the AoPS wiki. It is now here for all AoPS users to learn from.


Pascal's Bomb:

69-> 420 -> ... -> 8947

~by purplepenguin2


Pascal's Bomb begins with 69. It becomes infinitely large, although many people believe that it ends with 8947. Pascal's Bomb is a series of Munkeys. To apply this, you can use Complete the Circle or the Buadratic Bormula. After you have substituted for one of the variables, you can proceed to solve, using Inches or Watts. This is applicable on all Maff problems.


Suppose that MathHayden has $69$ apples. He needs to distribute them among $2$ of his very distinguished friends(not including himself), who happen to be Gmaas worshipers. Each friend must get at least $14$ apples. How many $Possible$ distributions are there?

Solution 1

By applying the Pascal's Bomb, we Munkey it and get an answer of $\boxed{42}$.

IMPORTANT: Each time you use Pascal's bomb, you inadvertently create an alternate universe that rips apart the fabric of reality and power-washes the universe... Be careful to use Pascal's bomb ONLY WHEN SOLVING PROBLEMS GMAAS RELATED, since GMAAS's power is so great that only it can withstand it.

Solution 2

By Adihaya Jayasharmaramankumarguptareddybavarajugopal's lemma, the answer is $\boxed{42}$ again.

Solution 3(slower)

We first proceed to give each friend $14$ apples. We then have $41$ apples left to distribute among the two friends. The first one can have $0 \leq x \leq 41$ apples and the second will have $41-x$ apples. There are $\boxed{42}$ $Possible$ values of $x$, from $0$ to $41$, and that is the answer.

OMG this thing contains pretty much all the numbers that I fear

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