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Part Of The
Month: Problem
Past Problems Of The Month
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April: 2010
16 = 7 and 17 = 9 [The number of letters in the spelling of 16 (sixteen) is 7 and that of 17 (seventeen) is 9]
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April: 2010
A fishing boat is docked in the harbor. There is a rope ladder hanging over the side with its end touching the water. The rungs of the ladder are 1 meter apart and the tide is rising at 50 centimetres an hour.
At the end of 6 hours, how many of the rungs will be covered?
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None of the rungs of the ladder will be covered by water... the boat will rise with the tide. If it doesn't you have problems!
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March: 2010
Letters above the line are all formed with straight lines. Below the line, the letters contain curves.
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February: 2010
John had a bag of oranges to share with two friends. To the first of his friends, he gave half of the oranges plus one more. To his second friend he gave half of the remaining oranges plus one more. By this time, John had one orange left. How many did he start with?
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There are two ways to find the solution. Use the logic of empirically substituting increasingly higher numbers until you get the result of having 1 orange left or use algebra and nail the solution the first time.
let x = amount of oranges that John started with so then:
x-.5x-1 = .5x-1 = oranges left after giving oranges to first friend
.5(.5x-1)- 1 = final amount of oranges left after giving remaining oranges to second friend,
the problem states that the final amount of oranges left is 1 orange so:
.5(.5x-1)- 1 = 1
.25x -.5 - 1 = 1
.25x - 1.5 = 1
.25x = 2.5
x = 10
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January: 2010
Problem:A rectangle with sides of the ratio
phi is called a golden rectangle. If
you make such a rectangle smaller
by removing a square formed by
the smallest side, the sides of the
remaining smaller rectangle have
the same
ratio of
phi and
so on.
What is
the value
of phi?
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December: 2009
Using four unique integer values from 0 thru 9 and any combination of the operators x / + - , create two different equations which = 44
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There are several possible answers. Here are three:
(7 x 6) + 4 - 2 = 44
(8 x 6) - (4 x 1) = 44
(9 + 2) x (8 - 4) = 44
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November: 2009
A perfect cube that measured one foot on a side has been expanded such that the cube now has twice the surface area that it originally had. Find the volume of this larger cube.
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The new volume is approx. 2.83 ft³ (or 2 * √2). You must increase the lengths of the cube's edges by a factor of √2. Volume then = (√2)³
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October: 2009
If ABCD * 4 = DCBA, then A=2, B=1, C=7, and D=8. Since A is multiplied by 4 it must be an even number. This is the best place to start.
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September 2009
A car travels at 60 miles per hour over a certain distance. The car makes the return trip over the exact same distance at 30 miles per hour. What is the average speed of the car?
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Suupose the car traveled 120 miles (60 our, 60 back). Given the parameters it will take 3 hours to complete the trip, for an average of 40mph.
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August: 2009
If a chicken and a half can lay an egg an a half in a week and a half, how many eggs would this chicken lay in a month and a half (6 weeks)?
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We can say that 1 chicken would lay one ggin a week and a half. Therefore, 1 chicken could lay FOUR EGGS in a month an a half.
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July: 2009
How many R's?
Complete the sentences written below.
Do not use numbers like 1, 2, 3, and so
on ...
In the sentence below are .............. R's.
In the upper sentence there are ............. .
R's.
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In the sentence below are "precisely five" R's.
In the upper sentence there are "three"
R's.
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May: 2009
The trick is that zero factorial (0!) equals 1, so...
(0!+0!+0!+0!+0!)! = 5! = 120
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April: 2009
Problem: "How much will one cost?" "25 cents"
"How much will fifteen cost?" "50 cents"
"OK then, I'll take one hundred and sixteen"
"Thank you, that will be 75 cents please"
Explain.
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He is buying digits or house numbers, each separate digit costs 25c, so 116 is three digits so 3 x 25c = 75c.
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March: 2009
Problem: Using the document that
you can download by clicking Here.
Arrange 7 equilateral pentagons such that each pentagon is touching 3 or more of its neighbors without overlapping any surfaces. (Contact by point or line only)
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February: 2009
Problem: Four identical balls on four different tracks are released simultaneously. Which of the tracks (bent, straight, circular, or cyloidal) will deliverthe ball to the end of the slope fastest?
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Strategy: The shortest path, or the straight line, is not the quickest. Instead, the ball that rolls on the cycloidal path will be the fastest. The cycloidal path is the longest, but it has the quickest decent. This gives the ball the highest speed and does not slow down soon enough for the rest of the paths to catch up. |
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January: 2009
Problem: Two semicircles are inscribed in a square with side 8 m, as shown. Approximate the area of the shaded region to the nearest tenth of a sq cm. Use the approximation 3.14 for p.
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Strategy: Split the region into more familiar shapes.
Draw two lines as shown to divide the square into four congruent smaller squares. Find the area of each shaded region separately.
Square A, fully shaded, has an area of 16 sq m. In squares B and C, the shaded
region is found by removing the quarter circle from the square. Each of those areas
is then 16 – ( 1
4 × 3.14 × 42), which then simplifies to 3.44 sq m. Adding 16, 3.44, and
3.44 and rounding off, the area of the shaded region is then 22.9 sq m. |
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December: 2008
Problem: How many 3 digit numbers have different digits in descending order such as 73 or 960?
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Solution:
Let's start small and work our way up.
The smallest hundred's digit possible is 2 (i.e. the number 210). The ten's digit could only be 1 and the only possibility for the unit's digit is 0. = 1 number.
Next the hundred's digit could be 3. Then the ten's digit could be 2, leaving 1 or 0 for the units = 2 numbers. Or the ten's digit could be 1, leaving only 0 for the unit's = 1. So the number of numbers beginning with 3 is 2+1 = 3.
For a hundred's digit of 4: The ten's digit could be 3, leaving the 3 possible unit's digits 2,1,and 0 = 3. Or the ten's digit could be 2, leaving the 2 possible unit's digits 1 and 0 = 2. Or, finally, the ten's digit could be 1, leaving only 0 for the unit's digit = 1. The number of numbers beginning with 4 is 3 + 2 + 1 = 6.
Consider the table below:
hundred's digit |
number of 3-digit numbers |
cumulative total |
2 |
1= 1 |
1 |
3 |
2+1=3 |
4 |
4 |
3+2+1=6 |
10 |
5 |
4+3+2+1=10 |
20 |
6 |
5+4+3+2+1=15 |
35 |
7 |
6+5+4+3+2+1=21 |
56 |
8 |
7+6+5+4+3+2+1=28 |
84 |
9 |
8+7+6+5+4+3+2+1=36 |
120 |
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