NCTM November calendar problems.

[guilage]

My teacher gave us the monthly calendar to solve. It has 30 problems of which I solved 20. But 10 of them are confusing and I can't remember what to do or how to solve them. I will post them here ad if you could help me, I would really appreciate it. I know 10 problems is a lot but I have no clue what to do... here they are:

1 - We are given three consecutive integers. The difference between the cubes of the two larger of the three consecutive integers is 66 more than the difference between the cubes of two smaller integers. What is the median integer?

2 - For each positive integer k, let Ak be the sum of the first k positive integers. If exactly x if the Ak consist(s) of one digit, exactly y of the Ak consist(s) of two digits, and exactly z of the Ak consist(s) of three digits, what is the product of x.y.z?

3 - A rectangular solid has a top face with surface area of 28tf^2,a front face with surface area of 20ft^2, and a side face with surface area of 70ft^2. What is the volume of this solid?

4 - The lines: x-2y=2, -3x+y=4, and 2x+y=4 intersect in pairs to determine the vertices of a triangle. Find the area of the triangle.

5 - Let f(x)= ax+b. Find all the real values of a and b such that: f(f(f(1)))=29 and f(f(f(0)))=2.

6 - Find all ordered pairs of real numbers (x,y) that satisfy the equations: 3^x . 9^y=81 and 2^x/8^y=1/128.

7 - Find the largest integer p such that 5^7 can be expressed as the sum of p consecutive positive integers.

8 - Find bases of a and b such that 386a=272b and 146a=102b

9 - If x^4+x^2+1=0 what is the value of (x^2+1/x^2)^3?

10 - Let f(x) be a function such that f(x)+2f(3-x)=4x+5 for every real number x. Find f(1).

any help is welcome, thank you!!

 

[soroban]

Hello, guilage!

Here's some help . . .

1) We are given three consecutive integers.
The difference between the cubes of the two larger of the three consecutive integers
is 66 more than the difference between the cubes of two smaller integers.
What is the middle integer?

Let the middle integer be \(\displaystyle x.\)
Then the three integers are: .\(\displaystyle x-1,\:x,\:x+1\)

\(\displaystyle \underbrace{\text{Diff. of cubes of two larger}}_{(x+1)^3-\:x^3}\;\underbrace{\text{ is }}_{=}\; \underbrace{\text{diff. of cubes of two smaller}}_{x^3-\x-1)^3}\;\underbrace{\text{ plus 66 }}_{+\: 66}\)

\(\displaystyle \text{We have: }\;(x+1)^3 - x^3 \;=\;x^2 - (x-1)^3}\)

\(\displaystyle \text{Expand: }\;x^3 + 3x^2 + 3x + 1 - x^3 \;=\;x^3 - (x^3 - 3x^2 + 3x - 1)\)

\(\displaystyle \text{Therefore: }\;6x \:=\:66 \quade\Rightarrow\quad \boxed{x \:=\:11}\)

3) A rectangular solid has a top face with surface area of 28 sq.ft., a front face with surface area of 20 ft^2,
and a side face with surface area of 70 ft^2. . What is the volume of this solid?

\(\displaystyle \text{Let the dimensions be }\;L,\:W,\:H.\)

The box looks like this:

                L
          *-----------*
         /           /|
      W /    28     / | H
       /           /  |
      *-----------* 70*
      |           |  /
    H |    20     | / W
      |           |/
      *-----------*
            L

\(\displaystyle \text{We have: }\;\begin{array}{ccccc}LW &=& 28 & [1] \\ LH &=& 20 & [2] \\ WH &=& 70 & [3] \end{array}\)

\(\displaystyle \text{Divide [1] by [2]: }\;\frac{LW}{LH} \:=\:\frac{28}{20} \quad\Rightarrow\quad \frac{W}{H}\:=\:\frac{7}{5} \quad\Rightarrow\quad W \:=\:\tfrac{7}{5}H\)

\(\displaystyle \text{Substitute into [3]: }\;\left(\tfrac{7}{5}H\right)(H) \:=\:70 \quad\Rightarrow\quad H^2 \:=\:50 \quad\Rightarrow\quad\boxed{H \:=\:5\sqrt{2}}\)

\(\displaystyle \text{Substitute into [2]: }\;L(5\sqrt{2}) \:=\:20 \quad\Rightarrow\quad\boxed{L \:=\:2\sqrt{2}}\)

\(\displaystyle \text{Substitute into [1]: }\;(2\sqrt{2})W \:=\:28 \quad\Rightarrow\quad\boxed{W \:=\:7\sqrt{2}}\)


\(\displaystyle \text{Therefore: }\;V \;=\;LWH \;=\;(2\sqrt{2})(7\sqrt{2})(5\sqrt{2}) \;=\;140\sqrt{2}\text{ ft}^3\)

 

[soroban]

Hello again, guilage!

Another one . . .

\(\displaystyle \text{9) If }x^4+x^2+1\:=\:0\text{, what is the value of }\left(x^2+\frac{1}{x^2}\right)^3 \:?\)

\(\displaystyle \text{Note that: }\:x \neq 0\)


\(\displaystyle \text{We have: }\;x^4 + x^2 + 1 \;=\;0\)

\(\displaystyle \text{Divide by }x^2\!:\;\;x^2 + 1 + \frac{1}{x^2} \:=\:0 \quad\Rightarrow\quad x^2 + \frac{1}{x^2} \;=\;-1\)

\(\displaystyle \text{Cube both sides: }\;\left(x^2+\frac{1}{x^2}\right)^3 \;=\;(-1)^3\)

\(\displaystyle \text{Therefore: }\;\left(x^2+\frac{1}{x^2}\right)^3 \;=\;-1\)

 

[guilage]

Thank you so much soroban!!

 

[Mrspi]

My teacher gave us the monthly calendar to solve. It has 30 problems of which I solved 20. But 10 of them are confusing and I can't remember what to do or how to solve them. I will post them here ad if you could help me, I would really appreciate it. I know 10 problems is a lot but I have no clue what to do... here they are:





6 - Find all ordered pairs of real numbers (x,y) that satisfy the equations: 3^x . 9^y=81 and 2^x/8^y=1/128.



any help is welcome, thank you!!

3[sup:1s3wnru9]x[/sup:1s3wnru9] * 9[sup:1s3wnru9]y[/sup:1s3wnru9] = 81

You can write 9[sup:1s3wnru9]y[/sup:1s3wnru9] as (3[sup:1s3wnru9]2[/sup:1s3wnru9])[sup:1s3wnru9]y[/sup:1s3wnru9]. When you raise a power to a power, you multiply the exponents. So, (3[sup:1s3wnru9]2[/sup:1s3wnru9])[sup:1s3wnru9]y[/sup:1s3wnru9] is 3[sup:1s3wnru9]2y[/sup:1s3wnru9]

81 is 3[sup:1s3wnru9]4[/sup:1s3wnru9]

Substitute into the original equation, and you have

3[sup:1s3wnru9]x[/sup:1s3wnru9] * 3[sup:1s3wnru9]2y[/sup:1s3wnru9] = 3[sup:1s3wnru9]4[/sup:1s3wnru9]

When you multiply powers of the same base, you ADD the exponents.

3[sup:1s3wnru9]x + 2y[/sup:1s3wnru9] = 3[sup:1s3wnru9]4[/sup:1s3wnru9]

Note that the bases are the same, and the expressions on the left and right side are equal. So, the exponents must be equal and we have

x + 2y = 4

Let's try the same approach on the second equation, 2[sup:1s3wnru9]x[/sup:1s3wnru9] / 8[sup:1s3wnru9]y[/sup:1s3wnru9] = 1/128

8 is 2[sup:1s3wnru9]3[/sup:1s3wnru9], so 8[sup:1s3wnru9]y[/sup:1s3wnru9] is (2[sup:1s3wnru9]3[/sup:1s3wnru9])[sup:1s3wnru9]y[/sup:1s3wnru9], or 2[sup:1s3wnru9]3y[/sup:1s3wnru9]

And 128 is 2[sup:1s3wnru9]7[/sup:1s3wnru9], so 1/128 can be written as 1/2[sup:1s3wnru9]7[/sup:1s3wnru9], or 2[sup:1s3wnru9]-7[/sup:1s3wnru9]

Substitute into the original equation:

2[sup:1s3wnru9]x[/sup:1s3wnru9] / 2[sup:1s3wnru9]3y[/sup:1s3wnru9] = 2[sup:1s3wnru9]-7[/sup:1s3wnru9]

When you divide powers of the same base, you SUBTRACT the exponents. So, 2[sup:1s3wnru9]x[/sup:1s3wnru9] / 2[sup:1s3wnru9]3y[/sup:1s3wnru9] becomes 2[sup:1s3wnru9]x - 3y[/sup:1s3wnru9], and

2[sup:1s3wnru9]x - 3y[/sup:1s3wnru9] = 2[sup:1s3wnru9]-7[/sup:1s3wnru9]

Again, the bases are the same and the expressions on the left and right are equal, so the exponents must be equal. This gives us

x - 3y = -7

Now...you have a system of two equations in two variables. Solve for x and y, and you should be able to answer the question.

 

[Deleted member 4993]

3 - A rectangular solid has a top face with surface area of 28tf^2,a front face with surface area of 20ft^2, and a side face with surface area of 70ft^2. What is the volume of this solid?

Let the sides be L, W & H

Then

(L*W)*(W*H)*(H*L) = 28 * 20 * 70

(L*W*H)[sup:4e4m6dzj]2[/sup:4e4m6dzj] = 7[sup:4e4m6dzj]2[/sup:4e4m6dzj] * 2[sup:4e4m6dzj]3[/sup:4e4m6dzj] * 10[sup:4e4m6dzj]2[/sup:4e4m6dzj]

V = L*W*H = 140 * (2)[sup:4e4m6dzj]1/2[/sup:4e4m6dzj] ft[sup:4e4m6dzj]3[/sup:4e4m6dzj]

 

[soroban]

Very funny, Subhotosh!

That was briliant!

 

[soroban]

Hello, guilage!

Here's another . . .

\(\displaystyle \text{5) Let }f(x)\:=\: ax+b\)

\(\displaystyle \text{Find all the real values of }a\text{ and }b\text{ such that: }\:f(f(f(1)))\,=\,29\:\text{ and }\:f(f(f(0)))\,=\,2\)

\(\displaystyle \text{We have: }\;f(x) \:=\:ax+b\)

\(\displaystyle \text{Then: }\:f(f(x)) \:=\:a(ax+b) + b \:=\:a^2x + ab + b\)

\(\displaystyle \text{And: }\:f(f(f(x))) \:=\:a(a^2x + ab + b) + b \:=\:a^3x + b(a^2+a+1)\)


\(\displaystyle \begin{array}{cccccc}f(f(f(1))) = 29\!: & a^3 + b(a^2+a+1) &=& 29 & [1] \\ f(f(f(0))) = 2\!: & 0 + b(a^2+a+1) &=& 2 & [2] \end{array}\)


\(\displaystyle \text{Subtract [1] - [2]: }\;\;a^2 \:=\:27 \quad\Rightarrow\quad\boxed{ a \:=\:3}\)

\(\displaystyle \text{Substitute into [2]: }\;\;b(3^2+3+1) \:=\:2 \quad\Rightarrow\quad\boxed{ b \:=\:\frac{2}{13}}\)

 

[soroban]

And another . . .

\(\displaystyle \text{8) Find bases }a\text{ and }b\text{ such that: }\:386_a\,=\,272_b\:\text{ and }\:146_a\,=\,102_b\)

\(\displaystyle \text{Note that }a\text{ and }b\text{ are positive integers.}\)


\(\displaystyle \text{We have: }\;\begin{array}{cccc}3a^2 + 8a + 6 &=& 2b^2 + 7b + 2 & [1] \\ a^2 + 4a + 6 &=& b^2 + 2 & [2] \end{array}\)

\(\displaystyle \text{From [2], we have: }\;b^2 \:=\:a^2 + 4a + 4 \:=\a+2)^2 \quad\Rightarrow\quad b \:=\:a + 2\;\;\;[3]\)

\(\displaystyle \text{Substitute into [1]: }\;3a^2 + 8a + 6 \:=\:2(a+2)^2 + 7(a+2) + 2\)

\(\displaystyle \text{This simplifies to: }\;a^2-17a - 18 \:=\:0 \quad\Rightarrow\quad (a-9)(a+2) \:=\:0\)

\(\displaystyle \text{Hence: }\;\boxed{a \;=\;9}\;\rlap{////}-2\)

\(\displaystyle \text{Substitute into [3]: }\;\boxed{b \:=\:11}\)


\(\displaystyle \text{Check: }\;386_9 \;=\;272_{11} \;=\;321 \qquad 146_9 \;=\;102_{11} \;=\;123\)

 

[Deleted member 4993]

Very funny, Subhotosh!

That was briliant!

Thank you - however it was just practice - having 3 children going through SAT - those look very familiar.....

 

[soroban]

Hello, guilage!

\(\displaystyle \text{4) The lines: }\:\begin{Bmatrix}x-2y&=&2 & [1] \\\text{-}3x+y&=&4 & [2] \\ 2x+y&=&4 & [3]\end{Bmatrix} \;\text{ intersect in pairs to determine the vertices of a triangle.}\)

\(\displaystyle \text{ Find the area of the triangle.}\)

This takes a bit of preliminary work, but the punchline is quite easy.


To find the vertices, solve the three systems of equations . . .

\(\displaystyle [1] \cap [2]:\quad \begin{array}{ccc}x - 2y \:=\:2 \\ \text{-}3x + y \:=\:4\end{array}\quad\Rightarrow\quad A(\text{-}2,\text{-}2)\)

\(\displaystyle [1] \cap [3]: \quad \begin{array}{ccc}x - 2y \:=\:2 \\ 2x+ y \:=\:4\end{array} \quad\Rightarrow\quad B(2,0)\)

\(\displaystyle [2] \cap [3]: \quad\begin{array}{ccc}\text{-}3x+y \:=\:4 \\ 2x + y \:=\:4\end{array}\quad\Rightarrow\quad C(0,4)\)


\(\displaystyle \text{The graph looks like this:}\)

                |
           (0,4)o C
                |
                |   B
      - - - - - + - o - - - -
                | (2,0)
          A o   |
         (-2,-2)|
                |

\(\displaystyle \text{Examine the slopes of the sides of the triangle.}\)

\(\displaystyle m_{AB} \:=\:\frac{0-(\text{-}2)}{2-(\text{-}2)} \:=\:\frac{1}{2}\)

\(\displaystyle m_{BC} \:=\:\frac{0-4}{2-0} \:=\:-2\)

\(\displaystyle AB \perp BC \quad\Rightarrow\quad \Delta ABC \text{ is a right triangle}\,!\)


\(\displaystyle \overline{AB} \:=\:\sqrt{(2-(\text{-}2)^2 + (0-(\text{-}2)^2} \:=\;\sqrt{20}\)

\(\displaystyle \overline{BC} \:=\:\sqrt{(4-0)^2 + (0-2)^2} \:=\:\sqrt{20}\)


\(\displaystyle \text{Area} \;=\;\tfrac{1}{2}(\overline{AB})(\overline{BC}) \:=\:\tfrac{1}{2}(\sqrt{20})({\sqrt{20}) \:=\:10\)

 

[soroban]

Hello, guilage!

Okay, one more . . .

\(\displaystyle \text{10) Let }f(x)\text{ be a function such that: }\;f(x)\;+\;2\,\cdot f\!(3-x)\;=\;4x\;+\;5\,\text{ for every real number }x.\)

\(\displaystyle \text{Find }f(1).\)

\(\displaystyle \text{Let: }\:f(x) \:=\:ax+b\)

\(\displaystyle \text{Then: }\;f(x) + 2\cdot f(3-x) \;=\;(ax + b) + 2\bigg[a(3-x) + b\bigg] \;=\;\)


\(\displaystyle \text{Then: }\:2\cdot f(3-x) \;=\;4x + 5 \quad\Rightarrow\quad 2\bigg[a(3-x) + b\bigg] \;=\; 4x + 5\)

. . . \(\displaystyle -ax + (6a + 3b) \;=\;4x + 5\)

\(\displaystyle \text{Then: }\;\begin{array}{ccccccccc}-a &=& 4 & \Rightarrow & a &=& \text{-}4 \\ 6a + 3b &=& 5 & \Rightarrow & b &=& \frac{29}{3} \end{array}\)

\(\displaystyle \text{Hence: }\:f(x) \;=\;-4x + \frac{29}{3}\)


\(\displaystyle \text{Therefore: }\;f(1) \;=\;-4(1) + \frac{29}{3} \;=\;\boxed{\frac{17}{3}}\)