## Solution of the Minute

October 25th, 2009

“Lewis and Carol travel together on a road from A to B, then return on the same road, with the entire trip taking 3 hours. Sometimes that road goes uphill, sometimes downhill, and sometimes it is level. When the road goes uphill, their rate is 40 mph; downhill their rate is 60 mph; on level road, their rate is x mph.”

Even if you were given a numerical value for x, the distance from A to B would (in most cases) not be uniquely determined. But there is one value for x that would determine that distance uniquely. Compute this value of x [Note: uphill going is downhill returning!]

Credits – Southern California ARML, Oct. 2009, Team round

Original post here.

Solution follows.
[spoiler]
Let $l$ be the distance spent on level road, and let $s$ be the distance spent on sloped road (which way, uphill or downhill, doesn’t matter because it’ll be reversed when they turn back).

$s \cdot \dfrac{1}{40} + s \cdot \dfrac{1}{60} + 2 l \cdot \dfrac{1}{x} = 3$

This equation can be simplified to:

$\dfrac{s}{24} + \dfrac{2l}{x} = 3$

Or, multiplying both sides by 24:

$s + \dfrac{48l}{x} = 72$

The only value of $x$ for which $s+l$ can be uniquely defined is $x=\boxed{48}$.

[/spoiler]

## Representations of Number as Sum of Consecutive Integers

October 2nd, 2009

How many ways are there of writing XXXX as a sum of positive consecutive integers?

I found this discussion very interesting (after coming upon it after a search due to encountering one of these problems on FTW and failing horribly).

Basically, the result is that the number of ways of writing a number as a sum of consecutive integers is the number of odd divisors of that number, minus one. This is an awesome conclusion.

How many ways are there of writing 123123 as a sum of positive consecutive integers?

$123123 = 3 \cdot 7 \cdot 11 \cdot 13 \cdot 41$

Therefore, the number of factors (all of which must be odd, because there is no factor of $2$ in $123123$) are:

$2^5 = 32 \text{ factors}$

Since there are 32 odd factors of 123123, the number 123123 can be written as a sum of positive consecutive integers in 31 ways.

Each of these ways is one of the odd divisors of 123123 (we subtract 1 because we can’t count the divisor 1 as that would yield only 1 integer, 123123 itself, and we need plural integers anyways).

If we count the even divisors, that will result in some of the divisors being negative. (See thread.)

Ok, lame post, sorry. Guys, do some of the earlier problems, like that staircase one ^_^

## Limits

September 26th, 2009

Question from the tryout:

Find:

$\dfrac{\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x} \cdot \lim_{x \to 3^{-}} \lfloor x \rfloor}{\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} + \lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}}$

This problem just looks scary, doesn’t it? Well, let’s split up the four limits.

First limit, the top left.

$\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x}$

l’Hôpital it! Bwahaha.

$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{1}{5}\cos \dfrac{x}{5}}{2 \cos 2x} = \dfrac{\dfrac{1}{5}}{2} = \boxed{\dfrac{1}{10}}$

Now for the top right:

$\displaystyle\lim_{x \to 3^{-}} \lfloor x \rfloor$

The limit of the floor function approaching from the negative side is the lower number, so this equals $\boxed{2}$.

Bottom left:

$\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$

Indeterminate form (zero over zero), so once again we can use l’Hôpital’s rule on it.

$\displaystyle\lim_{x \to 3} \dfrac{2x}{1} = \boxed{6}$

Last limit is the bottom right one.

$\displaystyle\lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}$

This one is $\frac{\infty}{\infty}$, so once again we must use l’Hôpital’s rule. This time we apply it twice in succession (because I’m not sure whether or not you can just cancel out the $x$s or not since it’s technically dividing by infinity; ask your local calculus teacher).

$\displaystyle\lim_{x \to \infty} \dfrac{-8x}{4x} = - \dfrac{8}{4} = \boxed{-2}$

Now, we put these solved limits back into the original problem:

$\dfrac{\dfrac{1}{10} \cdot 2}{6 + (-2)} = \dfrac{\dfrac{1}{5}}{4} = \boxed{\dfrac{1}{20}}$

Andy? We both fail. XD;

## Recurrences

September 23rd, 2009
In the sequence $2001, 2002, 2003, \ldots$, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $2001 + 2002 - 2003 = 2000$. What is the $2004$th term in this sequence?

I know that nobody cares, but here’s my solution anyways.

This problem defines a linear recurrence $x_n$ such that:

$x_n = x_{n-3} + x_{n-2} - x_{n-1}$

Replacing $x_n$ with a constant in order to find the characteristic polynomial, $\lambda^n = x_n$:

$\lambda^n = \lambda^{n-3} +\lambda^{n-2} - \lambda^{n-1}$

Thus the characteristic polynomial of this linear recurrence is:

$\lambda^n - \lambda^{n-3} -\lambda^{n-2} + \lambda^{n-1} = 0$

Rearrange, and divide by $\lambda^{n-3}$.

$\lambda^3 + \lambda^{2} -\lambda - 1 = 0$

This is our characteristic polynomial. Factoring it:

$(\lambda^3 - 1) + \lambda(\lambda - 1) = 0$

Difference of cubes.

$(\lambda - 1)(\lambda^2 + \lambda + 1) + \lambda(\lambda - 1) = 0$

Combine like terms and factor $\lambda^2 + 2\lambda + 1$.

$(\lambda - 1)(\lambda + 1)^2 = 0$

The roots of the characteristic polynomial are $\{1, -1, -1\}$. The root $-1$ has a multiplicity of two, so the general solution to the recurrence $x_n$ is given by:

$x_n = c_1 \cdot 1^n + \left( d_0 + d_1 n \right) (-1)^n$

We can plug in the initial values given in the problem, $x_1$, $x_2$, $x_3$, and $x_4$ to find the constants.

$2001 = c_1 \cdot 1^1 + \left( d_0 + d_1 \cdot 1 \right) (-1)^1$ $2001 = c_1 - d_0 - d_1$ $2002 = c_1 + d_0 + 2d_1$ $2003 = c_1 - d_0 - 3d_1$

Solving this system of three linear equations gives the following values for $c_1$, $d_0$, and $d_1$:

$c_1 = 2002$ $d_0 = 2$ $d_1 = -1$

Therefore, the formula for the $n$th term of this sequence is given by:

$\boxed{ x_n = 2002 + ( 2 - n ) (-1)^n }$

And, finally, $x_{2004}$ is:

$x_{2004} = 2002 + (-2002)(1)$
$x_{2004} = 2002 - 2002$
$x_{2004} = \boxed{0}$

Hey, Andy, you must have solved it intuitively (instead of actually solving the recurrence and finding the general formula), could ya tell me/us how? =D

## Variation of a previous number theory problem

September 19th, 2009

Lol, just saw this on the math team forum, was posted by like last last year, but anyways…

In the prime factorization of the expression $100! \cdot 99! \cdot 98! \cdot \ldots \cdot 3! \cdot 2! \cdot 1!$ there is a $7^x$ term. Find x.

## K’s Semi-solution to “Hard Problem”

September 19th, 2009

I can’t think well after lunch, so I don’t have the solution. But I know you have to have the travelling times of each individual equal each other. And they have to share the bike. x=distance girl rides on bike, y=distance of boy on bike, z=distance of dog on bike.

$x+y+z = 38$ $16(38-x) + 6x = 15(38-y) + 5y = 10 (38-x) +4z$

I think I’m on the right track ,but if not, well, there’s a reason I’m not trying out for math team. Ok, people are coming over to work on the English project now.

Edit/Ben: You’re thinking on the right track with systems of linear equations, with three variables. However, I don’t see why you set $x+y+z = 38$, because $x+y+z$ is the sum of the time the girl, boy, and dog ride the bike, and that shouldn’t have anything to do with the distance they travel. As for the second equation, why are the three equal?

Continue >.< or leave it to Hanchan when you guys finish with your projects.

Edit/2: On second thought I shouldn’t really barge into your post, I should write a comment ^_^

## Number Theory – Solutions

September 18th, 2009

1) Written out in base 10, the number $345!$ ends in how many zeroes?

Each zero in a number represents a factor of 10, or a factor each of five and two.

Therefore, to find the number of zeros after a number, we need to find the number of factors of five and the number of factors of two in $345!$.

How many factors of five are in $345!$?

Simply divide 345 by five.

$\dfrac{345}{5} = 69$

However, we need to count the factors of five squared (25) twice, so we divide again by 25.

$\dfrac{345}{25} = 13$

And finally, let’s not forget the factors of five cubed which need to be counted three times.

$\dfrac{345}{125} = 2$

Summing these together:

$69 + 13 + 2 = \boxed{84}$

Thus there are 84 zeroes after the number $345!$.

2) The product of the 12-digit number $493,\!224,\!762,\!945$ and the 12-digit number $471,\!910,\!894,\!925$ is the 24-digit number $232,\!758,\!139,\!280,\!5A5,\!928,\!554,\!125$. Find the digit $A$.

Did you try to find the digits that affected the value of A and try multiplying things out?

A smarter solution, do this calculation mod 9. It’s called “casting out nines”. Cross out all the nines and all the numbers that add up to nine, etc, and add up the remaining numbers, subtracting nine when needed.

$493,\!224,\!762,\!945 \equiv 3 \mod 9$ $471,\!910,\!894,\!925 \equiv 5 \mod 9$

Therefore,

$493,\!224,\!762,\!945 \cdot 471,\!910,\!894,\!925 \equiv 15 \mod 9$

And

$15 \equiv 6 \mod 9$

The product mod 9 is:

$232,\!758,\!139,\!280,\!5A5,\!928,\!554,\!125 \equiv 2 + A \mod 9$

And since the products are supposed to be equal, they will also be equal mod 9.

$2 + A \equiv 6 \mod 9$ $\boxed{A \equiv 4 \mod 9}$

It’s as simple as that!

## ‘nother solution

September 14th, 2009

[Original Problem]

The sequence $\tfrac 12,\tfrac 53,\tfrac{11}8,\tfrac{27}{19},\dots$ is formed as follows: each denominator is the sum of the numerator and denominator of the previous term; each numerator is the sum of its own denominator and the previous denominator. The successive terms are approaching the real number $x$ as a limit. Compute $x$.

Solution follows.

## Do it again!

September 14th, 2009

Hanchan’s awesome skillz yesterday were epic — two wrongs truly do make a right! I want to like, repeat the experiment and see if you can get it right the wrong way again x33

Okay, so another ARML problem from the day before yesterday (Saturday).

The sequence $\tfrac 12,\tfrac 53,\tfrac{11}8,\tfrac{27}{19},\dots$ is formed as follows: each denominator is the sum of the numerator and denominator of the previous term; each numerator is the sum of its own denominator and the previous denominator. The successive terms are approaching the real number $x$ as a limit. Compute $x$.

I dunno, I’m sick and feel really horrible, but this is interesting ^_^

Currently playing: ♪ Seikan Hikou – Nakajima Megumi

## Two wrongs make a right!

September 13th, 2009

If $\displaystyle(\sin 1^\circ)(\sin 3^\circ)(\sin 5^\circ) \cdots (\sin 87^\circ)(\sin 89^\circ) = \frac{1}{2^n}$, compute the rational number $n$.

Hanning says (11:35 PM):
so then i used the classic method when dealing with these problems
Me says (11:35 PM):
lol, what's that?
Hanning says (11:35 PM):
like group 1 with 99 or something like that
Me says (11:35 PM):
smart =D
Hanning says (11:35 PM):
so i grouped 1 with 89 to make 90 which is a good number to work with when dealing with these things
and so on
Me says (11:35 PM):
lol, nice
Hanning says (11:35 PM):
so i checked what sin45 * sin 45 was which was 1/2
Hanning says (11:36 PM):
so then i hypothesized that sin1 *sin 89 would also be 1/2
and there u go
90/2 which is 45
Me says (11:36 PM):
um, sin1*sin89 doesn't equal 1/2
Hanning says (11:36 PM):
Me says (11:36 PM):
lol
and you missed something
it's sin1 times sin3 times sin5
Hanning says (11:36 PM):
i dont care xD
Me says (11:36 PM):
only odd numbers
XD so you got it right cuz you misread the problem? awesome XD
Hanning says (11:37 PM):
uhh
w8 im still right?
hahaha
Me says (11:37 PM):
hahahaha~~~
lawl
lollll