## random math problem time

November 10th, 2009

What is the remainder when 356 is divided by 7?

## 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]

## Problem of the Minute

October 20th, 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

## Easy Problem

October 20th, 2009

A tangent line from the point $(2 \cdot 1994, 2 \cdot 1994)$ to the circle $x^2 + y^2 = 1994^2$ touches the circle at the point $(a,b)$. Compute $a+b$.

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

## Problem of the Day #2

October 20th, 2009

Semicircles are drawn on two sides of square ABCD as shown. Point E  is the center of the square, and QAP is a line segment with QA = 7 and AP = 23. Compute AE.

(Click image for larger size.)

Credits – Southern California ARML, Oct. 2009, Individual round; image credit goes to David (I think)

## Problem of the Day

October 18th, 2009

Rectangle PQRS is inscribed in rectangle ABCD, as shown. If DR = 3, RP = 13, and PA = 8, compute the area of rectangle ABCD.

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

## If You’re Bored… Math! Math! Math!

October 17th, 2009

It’s time for more math! Especially if you’re bored like me….

Express:
$8952!$ in 3 sig figs and scientific notation

No I haven’t solved it yet.
And yes I chose a number sure to max out a graphing calculator (well, I hope)

## ‘nother easy problem

October 17th, 2009

People actually REPLY when I post easy problems! GASP!

(In this problem, $\lfloor \cdot \rfloor$ represents the Greatest Integer function.)

If $x=19941994$, find the value of $\left\lfloor \sqrt{x^2 - 10x + 29} \right\rfloor$.

## Math desu yo, Math desu yo!

October 17th, 2009

I felt like it.

Here’s the design I’m sponsoring for the 2009-10 Arcadia Math Team t-shirt.

$i^2$

keepin’ it real

Anyways, let’s go with a really easy problem this time, okay? Great!

Um, I actually have more trouble coming up with “easy” problems, compared to hard ones. Okay, I’ll go with this one; it’s not too hard.

A number x is 56 less than a perfect square, and 124 less than another perfect square. What is x?

from October 2009 ARML Practice