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2 How many positive three-digit integers are there in which the sum of its three digits is odd?

An odd sum will occur if we have

odd   even   even          even     odd     even        even    even   odd

( 5   *     5  *     5 )         (4)     *  (5)   *   (5)             (4)  *    (5)  *   (5)

            125             +             100                  +              100 

                                       325

OR

odd  odd    odd

 5  *  5  *     5 

        125     

So...there will be  325 + 125  =   450  three-digit integers

Determine the value of a.

Look at Omi's second diagram, Nickolas.....she decomposes the figure into two triangles each with an area of 1

3.What is the least positive integer with exactly 10  factors?

The integer  is

2^4 * 3  =    48

1 In the land of Noom, all nouns are 4-letter words with consonants at the beginning and end and a repeated vowel (a, e, i, o, or u) in the middle. How many such words are possible?

Assuming that the first and last consonants can be repeated....we have 21choices for each [ 21 consonants]

So....the number of possible nouns =

21 * 5 * 21  =  2205 nouns

However.....if the two consonants must be different, then only

21 * 5 * 20 =  2100 nouns

Since ABCD is a unit square....its height = 1  an d its base = 1

So....its area = 1

And since EFGH  = 1/4 area of ABCD, then its base, FE, just must be 1/4

So.....that means that  BF + EA  = 3/4

So...the combined areas of BFGC  and EAHD must also be 3/4

But FGC  = 1/2 of BFGC   and EAH = 1/2 of EAHD

So...FGC + EAH  = (1/2) ( BFGC + EAHD)  = (1/2)(3/4) = 3/8

So...the area of quadilateral  AFCH  =

area of EFGH   + (area of FGC + area of EAH)   =

       (1/4 )          +      (3/8)        =

      (2/8)           +       (3/8)    =

                 5 / 8               

Connect WY....this will form right triangle  WXY

Since WX = WY....then this will be an isisceles right triangle

And WY  = √[4^2 + 4^2]  = √32  

And angle WYX  = 45

So....in triangle WYZ, angle WYZ =  135 - 45  = 90

So.....triangle WYZ is also a right triangle with WZ being the hypotenuse

So....

a  = √(WZ^2 - WY^2)   = √ ( 9^2  - (√32)^2 ]  = √ [81 - 32]   = √49  =  7

Yeah! It stumped me too!

OK....that one was a little difficult!!!

A pyramid  on the same base with the same height of its associated prism will have a volume of 1/3 of the prism

But the pyramid here has only 1/2 the base of its associated prism

So....its volume  = (1/3(1/2)12   = 2 units^3

So...the volume of the remaining piece  = 

(12 - 2) units^3  =

10 units^3 

Thanks CPhill! You really helped my understand that question!

1.How many distinct odd - 4 digit numbers can be written with the digits 1,2 ,3  and 4 if no digit may be used more than once?

To be odd....the last digit must be odd

So...if the last digit is 1.....we have  3 * 2 * 1  =  6  possibilities

And the same if the last digit is 3

So.....12 distinct odds

2. How many distinct arrangements can be made from the letters in the word "REARRANGE''?

The number of distinct arrangements is given by :

      9!                   

________  =  15120

3! * 2! * 2!

The volume of each cube  = (side)^3  = (1/3)^3  = 1/27 cubic units

The number of cubes  that will fill the prism  =

Vol of prism            4

__________  =    _____  =    4 * 27   =   108 cubes

Vol of cube            (1/27)

If the perimeter of one of the equilateral  triangles is 1, then one side = 1/3

And the area of one of these triangles = (1/2(1/3)^2 * √3/2  =  √3/36 units^2

Also.....the height, h,  of one of these triangles can be found as

tan 60 =  h/ (1/2 base)

√3 = h (!/2 * 1/3)

√3 = h (1/6)

√3/6  = h

And this will also be the height of the middle parallelogram

And the area of a parallelogram  =  base * height

So

Area of parallelogram  = Area of one triangle....so.....

√3/36 =  base * (√3/6)   

1/36 = base * 1/6   

1/6  = base of parallelogram

So....the perimeter of the parallelogram  =

2(1/3 + 1/3 + 1/6)  =

2 ( 2/3 + 1/6) =

2 ( 4/6 + 1/6) =

4 (5/6)

10 / 6 units

Thanks, everyone!

In order I think:

B

D

A

E

C

Looks like the solutions are x = -1, x = 0 and x = 1

2) d

6) c

5) Just follow the same ideas as in 4

4)

\(\text{the primes that can be rolled are }2,3,5,7,11\\ P[\text{roll a prime}] = \dfrac{1+2+4+6+2}{36} = \dfrac{5}{12}\\ E_A = 15 \cdot \dfrac{5}{12} = \dfrac{25}{4}\\ \text{The multiples of 6 that can be rolled are }6,12\\ P[\text{roll a multiple of 6}]=\dfrac{5+1}{36}=\dfrac 1 6\\ E_B = 42 \cdot \dfrac 1 6 = 7\\ E_B = 7 > \dfrac{25}{4} = E_A\)

2) It should be obvious that the number of people in the group must be a factor of 28

3) \(E=2P[\text{odd}]-6P[\text{even}]\\ P[\text{odd}]=P[1]+P[3]+P[5]=\dfrac 3 4\\ P[\text{even}] = P[6] = \dfrac 1 4\\ E= 2\cdot \dfrac 3 4 - 6 \cdot \dfrac 1 4 = 0\)

\(1)~ \text{The possible values are }\\ 6 = (2,4)\\ 8=(4,4)\\ 9=(2,7)\\ 11=(4,7)\)

\(P[6]=\dfrac{\dbinom{1}{1}\dbinom{2}{1}}{\dbinom{4}{2}}=\dfrac 1 3\\ P[8] = \dfrac{\dbinom{2}{2}}{\dbinom{4}{2}}= \dfrac 1 6\\ P[9] = \dfrac{\dbinom{1}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 6\\ P[11]=\dfrac{\dbinom{2}{1}\dbinom{1}{1}}{\dbinom{4}{2}}=\dfrac 1 3\)

\(\text{We want a distribution of }\left(\dfrac 1 3,\dfrac 1 3,\dfrac 1 3\right) \text{ on cone sizes}\\ \text{And choice D is the only one that accomplishes this}\)

Looks like 10

\(\left( \begin{array}{c} x^4-x \\ x^4-3 x^3+3 x^2-x \\ x^4-x^3 \\ x^4 \\ x^4-2 x^3+x^2 \\ x^4+x^3+x^2 \\ x^4-4 x^3+6 x^2-4 x+1 \\ x^4-x^3-x+1 \\ x^4+x^3+x^2+x+1 \\ x^4+2 x^3+3 x^2+2 x+1 \\ \end{array} \right)\)