## Wednesday, July 6, 2016

### Mathcounts Prep : Algebra Manipulation

Note that:

$$\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}$$
$$\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}$$
$$\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}$$
$$\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}$$
$$\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}$$

Applicable questions:

Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?

Solution:
$$\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}$$= $$\dfrac {a} {b}$$

Question 2: If $$x^{2}+y^{2}=153$$ and x + y = 15, what is xy?

Solution:
$$\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}$$
$$15^{2}-2xy=153$$$$\rightarrow xy=36$$

Question 3: If $$\left( x+y\right) ^{2}=1024$$ , $$x^{2}+y^{2}$$ = 530 and x > y , what is x - y?

Solution:
$$\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}$$
1024 - 2xy = 530$$\rightarrow 2xy=494$$
$$\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}$$
$$\left( x-y\right) ^{2}=36$$
x - y = 6

Question 4: x + y = 3 and  $$x^{2}+y^{2}=89$$, what is $$x^{3}+y^{3}$$?

Solution:
$$\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}$$
9 - 2xy = 89 $$\rightarrow -2xy=80$$ so xy = -40
$$\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}$$
$$x ^{3}+y^{3}$$= 387

Question #5: If $$x+\dfrac {1} {x}=5$$, what is $$x^{3}+\dfrac {1} {x ^{3}}$$?

Solution:
$$\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}$$
$$5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}$$
125 - 3*5 = $$x^{3}+\dfrac {1} {x ^{3}}$$

Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and $$x^{2}+y^{2}+z^{2}=19$$, what is the arithmetic mean of the three product xy + yz + xz?

Solution:
$$\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}$$
$$7^{2}-2\left( xy+yz+xz\right) =19$$
xy + yz + xz = 15  so their mean is $$\dfrac {15} {3}=5$$

More practice problems (answer key below):

#1:If x + y = 5 and xy = 3, find the value of $$\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}$$.

#2: If x + y = 3 and $$x^{2}+y^{2}=6$$, what is the value of $$x^{3}+y^{3}$$?

#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.

#4:If $$x-\dfrac {6} {x}$$ = 11, find the value of $$x^{3}-\dfrac {216} {x^{3}}$$?

#5: If $$x+\dfrac {3} {x} = 9$$, find the value of $$x^{3}+\dfrac {27} {x^{3}}$$?

#6:If $$x+\dfrac {1} {x} = 8$$, what is $$x^{3}+\dfrac {1} {x ^{3}}$$?

#1 :$$\dfrac {19} {9}$$
#2: 13.5
#3: $$\dfrac {2} {5}$$
#4: 1529 [ $$11^{3}$$+ 3 x 6 x 11 =1529]
#5: 648   [$$9^{3}$$-3 x 3 x 9 = 648]
#6: 488   [ $$8^{3}$$– 3 x 8 = 488]

## Sunday, July 3, 2016

### Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below)

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

Solutions:
#1:
Solution I:
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II:
The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers.
2.5 (1000 + 100 + 10 + 1) x 24 = 66660

#2:
Solution I:
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II:

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is $$\dfrac{1 + 3+ 5 + 9} {4} = 4.5$$
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

#2: $$\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8$$
4.8 * 11111 =$$\color{red}{53332.8}$$