Tuesday, January 6, 2015

I came across a fun problem in an open math forum, and the work being done is similar to other projects I've been working on. No one had attempted it, perhaps because the poser of the question let the thread die, but I figured I would offer a solution for anyone still following the thread.

The Problem

Given the positive numbers 1, 2, 3, 4, 5, 6, and 7, how many combinations of these numbers (using every number one time exactly) add up to 100? You are allowed to combine numbers, so long as each number is only used once.
For example: 12 + 3 + 4 + 6 + 75 = 100.

An Answer

Consider the meaning of the decimal expansion of a number. For instance, the problem can be made simpler by realizing 12 = 10 + 2.
You can then try different combinations of your set where different elements have been multiplied by 10.
Then you can look at general cases such as: 10, 2, 3, 4, 5, 6, 70 = 100.

This case is congruent to 12, 3, 4, 5, 6, 70 and 13, 2, 4, 5, 6, 70 and 14, 2, 3, 5, 6, 70 and 15, 2, 3, 4, 6, 70 and 16, 2, 3, 4, 5, 70.

Repeat for the one's-place in 70.

Since you can't use zeros, you have 5 possibilities for the one's-place corresponding to the 10, and 4 more possibilities for the one's place corresponding to the 70. The resulting sum will still be 100.
For the case where 1 and 7 become tens, you have 20 possibilities.
For the case where 2 and 6 become tens, you have 20 possibilities.
For the case where 3 and 5 become tens, you have 20 possibilities.
For the case where 1, 2, and 5 become tens, you have 24 possibilities.
For the case where 1, 3, and 4 become tens, you have 24 possibilities.
No other combination of different elements being multiplied by 10 will result in a sum of 100, primarily because the ones digits don't sum to a multiple of ten.
The total combinations comes to 108 possible combinations.

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Monday, November 4, 2013

Proof by Poem

I was reading Reddit last night and came across this poem. Quite elegant!
To prove irrational square root two,
Suppose there are integers p and q
with p over q entirely reduced
and when their ratio is taken, root two is produced.

squaring both sides of this faulty equation
yields p squared over q squared is two, a relation
so p squared is even, a fact that is true
if and only if it is divisible by two.

But since two is prime, evenness of p squared
says that p is even, (a fact that Euclid declared).
We can now deduce a little bit more;
the number two q squared is divisible by 4.

But then q squared is divisible by two
And we know that this fact, is simply not true
Because then q is even, that just cannot be
because p,q are both even, contradiction, QED.
Via itsatumbleweed on Reddit.

Thursday, October 24, 2013

Math Club: 10/24/13 - LaTeX Workshop!

We're having another LaTeX workshop!

More courses and majors on campus are requiring students create more documents via TeX than ever before, so now is the time to learn; whether it's for one course or for your dissertation, TeX is one of the premier methods of professional document creation!

Join us to get started or improve your skills. Today, 10/24 at 3:10 in A&S 228. We're in a computer lab, but if you want to set these TeX tools up for your personal use, bring your laptop or ipad! We hope to see you there!

Be sure to check out our LaTeX Resources!