Posts Tagged ‘Math Problems’

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

arml-oct09-i4

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

arml-oct09-i2Rectangle 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

Please comment your answers. (Pfft. As if anybody cares about this post.)

USAMTS 2009-10 R1: Question 2

October 13th, 2009

The ordered pair of four-digit numbers (2025, 3136) has the property that each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number. Find, with proof, all ordered pairs of five-digit numbers and ordered pairs of six-digit numbers with the same property: each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number.

This is the second question from this year’s USAMTS, round one. Contest is over now, so post and discuss all you want.

It’s been like a month since I’ve last posted a math problem, so take some time to ponder upon this problem. It’s actually not that hard, and there’s a simple, systematic way of finding the solutions.