Triangular Numbers & Word Problems: Chapter Level

Triangular Numbers :  From Math is Fun. 1, 3, 6, 10, 15, 21, 28, 36, 45...

Interesting Triangular Number Patterns: From  Nrich

Another pattern: The sum of two consecutive triangular numbers is a square number.

What are triangular numbers? Let's exam the first 4 triangular numbers:

The 1st number is  "1".
The 2nd number is "3" (1 + 2)
The 3rd number is "6" (1 + 2 + 3)
The 4th number is "10" (1 + 2 + 3 + 4)
.
.
The nth number is \(\frac{n(n+1)}{2}\)

It's the same as finding out the sum of the first "n" natural numbers.

Let's look at this question based on the song "On the Twelve Day of Christmas"
(You can listen to this on Youtube,)

On the Twelve Day of Christmas
On the first day of Christmas
my true love gave to me
a Partridge in a Pear Tree

On the second day of Christmas,
My true love gave to me,
Two Turtle Doves,
And a Partridge in a Pear Tree.

On the third day of Christmas,
My true love gave to me,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree

On the fourth day of Christmas,
My true love gave to me,
Four Calling Birds,
Three French Hens,
Two Turtle Doves,
And a Partridge in a Pear Tree.


The question is "How many gifts were given out on the Day of Christmas?"

Solution I"
1st Day:       1
2nd Day:      1 + 2
3rd Day       1 + 2 + 3
4th Day        1 + 2 + 3 + 4
.
.
12th Day      1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12

Altogether, you'll have 12 * 1 + 11 * 2 + 10 * 3 + 9 * 4 + 8 * 5 + 7 * 6 + 6 * 7 + 5 * 8 + 4 * 9 + 3 * 10 + 2 * 11 + 1 * 12 = 12 + 22 + 30 + 36 + 40 + 42 + 42 + 40 + 36 + 30 + 22 + 12 = 364

Solution II: The sum of the "n" triangular number is a tetrahedral number.
To get the sum, you use \(\frac{n(n+1)(n+2)}{6}\)

n = 12 and \(\frac{12 (13)(14)}{6}= 364 \)

Here is a proof without words.

Applicable questions: (Answers and solutions below)

#1 Some numbers are both triangular as well as square numbers. What is the sum of the first three positive numbers that are both triangular numbers and square numbers?

#2 What is the 10th triangular number? What is the sum of the first 10 triangular numbers?

#3 What is the 20th triangular number?

#4 One chord can divide a circle into at most 2 regions, Two chords can divide a circle at most into 4 regions. Three chords can divide a circle into at most seven regions. What is the maximum number of regions that a circle can be divided into by 50 chords? 

#5: Following the pattern, how many triangles are there in the 15th image? 

Answers: 
#1 1262  The first 3 positive square triangular numbers are: 1, 36 (n = 8) and 1225 (n = 49).

#2 55; 220    a. \(\frac{10*11}{2}=55\)   b. \(\frac{10*11*12}{6}=220\)

#3 210   \(\frac{20*21}{2}=210\)  

#4 1276 
1 chord :    2 regions or \(\boxed{1}\) + 1
2 chords:   4 regions or  \(\boxed{1}\)  + 1 + 2
3 chords:   7 regions or \(\boxed{1}\)  + 1 + 2 + 3
.
.
50 chords:\(\boxed{1}\)  + 1 + 2 + 3 + ...+ 50 = 1 +  \(\frac{50*51}{2}\) =1276

#5: 120
The first image has just one triangle, The second three triangles. The third 6 triangles total. 
It follows the triangular number pattern. The 15th triangular is  \(\frac{15*16}{2}\) =120 

2013 Mathcounts School and Chapter Harder Problems

You can now download and discuss with your friends this year's school and chapter problems.
Here is the link to the official Mathcounts website.

Some more challenging problems from this year's Mathcounts school/or chapter problems.

2013 school team #10 : Three concepts are testing here :
Hint: 
a. If you get rid of the remainder, the numbers will be evenly divided into 192, so you are looking at
those factors of 192 - 12 = 180

b. To leave a remainder of 12, those factors of 180 that are included in the Set must be smaller than 12, otherwise, you can further divide it.

c. To find the median, make sure to line up the numbers from the smallest to the largest and find the middle numbers. If there are even numbers of factors larger than 12, average the middle two. Otherwise, the middle number is the answer.

\(180=2^{2}\times 3^{2}\times 5 \) so there are (2 + 1) (2 + 1) (1 + 1) = 18 factors

The list on the left side gives you the first 9 and if you times those numbers with "5", you get 9 other factors,which are 5, 10, 20, 15, 18, 60, 45, 90 and 180.

Discard the factors that are smaller or equal to 12 and list all the other factors in order and find the median.

The answer is "36".



2013 Chapter Sprint:
#21: Dimensional change problem : The height of the top pyramid is \(\dfrac {2} {3}\) of the larger
pyramid so its volume is \(\left( \dfrac {2} {3}\right) ^{3}\) of the larger pyramid.

\(\left( \dfrac {2} {3}\right) ^{3}\times \dfrac {1} {3}\times \left( \dfrac {36} {4}\right) ^{2}\times 12
= \) \(96  cm^{3}\)

# 24:  According to the given:   \(xyz=720\)   and   \(2( xy+yz+zx)= 484 \) so
\(( xy+yz+zx )= 242\)

Since x, y and z are all integers, you factor 720 and see if it will come up with the same x, y and z values
for the second condition.

Problem writer(s) are very smart using this number because the numbers "6", "10", "12" would give you
a surface area of 252. (not right)

The three corrrect numbers are "8", "9", and "10" so the answer is \(\sqrt {8^{2}+9^{2}+10^{2}}=\) \(7\sqrt {5}\)

#25: Geometric probability: Explanations to similar questions and more practices below. 

Probability with geometry representations form Aops.

Geometric probability from "Cut the Knots".

#26: This one is similar to 2002 AMC-10B #21, so try that question to get more practices. 
2002 AMC-10B #21 link 

#27:
\(\dfrac {1} {A}+\dfrac {1} {B}=\dfrac {1} {2}\)
\(\dfrac {1} {B}+\dfrac {1} {C}=\dfrac {1} {3}\)
\(\dfrac {1} {C}+\dfrac {1} {A}=\dfrac {1} {4}\)
Add them up and you have  \(2 * (\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {12}\)

\((\dfrac {1} {A}+\dfrac {1} {B}+ \dfrac {1} {C})=\dfrac {13} {24}\)

\(\dfrac {1} {\dfrac {1} {A}+\dfrac {1} {B}+\dfrac {1} {C}} = \)\(\dfrac {{24}} {13}\) hours

#28: Hint : the nth triangular number is the sum of the first "n" natural numbers and \(\dfrac {n\left( n+1\right) } {2}\) is how you use to find the sum.
From there, you should be able to find how many numbers will be evenly divided by "7".

#29 : Circle questions are very tricky so make sure to find more problems to practice accuracy.

#30 :  
Solution I: 
Read the solution that is provided by Mathcounts.org here.
Solution II:
Case 1 : \(x-1 > 0\rightarrow x > 1\) Times ( x - 1) on both sides and you have
\(x^{2}-1>8\) so x > 3 or x < -3 (discard)

Case 2: \(x-1 < 0\) so \(x < 1\) \(\rightarrow x^{2}-1 < 8\) [You need to change the sign since it's negative.]-3 < x < 3. Combined with x < 1 you have the range as -3 < x < 1
The answer is 60%.

                                               2013 Mathcounts Target #7 and 8: 

Target question #8 is very similar to 2011 chapter team #10. 
It just asks differently.   
Read the explanations provided on the Mathcounts official website.
They are explained very well.
Let me know if there are other easier ways to tackle those problems.

Hope this is helpful !! Thanks a lot !! Good luck on Mathcounts state.

2013 Mathcounts Basic Concept Review: Some Common Sums/numbers

These are some common sums that appear on Mathcounts often.

\(1+2+3+\ldots +n=\dfrac {n\left( n+1\right) } {2}\)

 \(2+4+6+\ldots .2n=n\left( n+1\right)\)

\(1+3+5+\ldots .\left( 2n-1\right) =n^{2}\)

The above are all arithmetic sequences.

The sum of any arithmetic sequence is average times terms (how many numbers).
Besides, the mean and median are the same in any arithmetic sequence.
Combining these knowledge, along with distributive rules some times
(case in point, sum of multiples of n, etc...) will expedite the calculation.

The "nth" triangular number is \(\dfrac {n\left( n+1\right) } {2}\)

The sum of the first n triangular numbers is \(\dfrac {n\left( n+1\right) \left( n+2\right) } {6}\).

 \(1^{2}+2^{2}+3^{2}+\ldots +n^{2}\) = \(\dfrac {n\left( n+1\right) \left( 2n+1\right) } {6}\)

\(1^{3}+2^{3}+3^{3}+\ldots +n^{3}=\)\(\left[ \dfrac {n\left( n+1\right) } {2}\right] ^{2}\)

This Week's Work : Week 11 -- for Inquisitive Young Mathletes

Area of an equilateral triangle

You can also use angle ratio 30-60-90 to prove that; besides, if you times that by 6, you'll get the area of a regular hexagon.

More links on Pascal's Triangle from Art of Problem Solving:

Introducing Pascal's Triangle

Patterns in Pascal's Triangle

Other review links:

From Math is Fun

Pascal's Triangle and Its Patterns

Triangular numbers

Science link :

From NOVA : How Smart Can We Get ?