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By analyzing this question, it’s easy to see the distinguishable fruit is irrelevant here. An orange is the same as an apple. The selection of any fruit creates a fruit basket.

The solution to this is the number of partitions of 15.  

First, use s = v₁t + (1/2)gt² to find v₁, the velocity at s = 1m (g = 9.81 m/s², t = 0.05 s).

Then use v₁²= 2gh to find the distance, h, from the top of the bridge to 1 m above the water (assuming rock is dropped with zero initial velocity).

Then height of bridge = h + 1 m

Thank you Melody!

The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through T).

Each word in their language is 4 letters or less, and for some reason,

they insist that all words contain the letter A at least once.

How many words are possible?

\(\begin{array}{|lr|} \hline \text{words with $1$ letter : } &1 \\ \text{words with $2$ letters: } &39 \\ \text{words with $3$ letters: } &1141 \\ \text{words with $4$ letters: } &29679\\ \hline \text{sum: } & 30860 \\ \hline \end{array} \)

Very impressive Heureka    

5 are not indistinguishable Ginger. They are all different people.

1)
A frog started from the origin of the coordinate plane and made three jumps.
Each time the frog jumped a distance of 5 units and landed at a point with integer coordinates.
How many different possibilities of the final position of the frog are there?

All 264 different possibilities of the final position of the frog:

\scalebox{0.5}{% scale
\begin{tikzpicture}
\draw[thick] (-15,-15) grid (15,15);
% Axes
\foreach \x in {-15,...,-1,0,1,2,...,15} {%
\draw (\x, -.1) -- (\x, 0) node[below left=2pt] {$\scriptstyle\x$};
}
\foreach \y in {-15,...,-1,1,2,...,15} {%
\draw (-.1,\y) -- (0,\y) node[below left=2pt] {$\scriptstyle\y$};
};

\foreach \x/\y in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\x \y,}
{
  \foreach \v/\w in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\v \w,}
  {
    \foreach \r/\s in {5/0,4/3,3/4,0/5,-5/0,-4/3,-3/4,4/-3,3/-4,0/-5/,-4/-3/,-3/-4} %{\r \s,}
    {
           \node[draw,circle,inner sep=2pt,fill] at (\x+\v+\r,\y+\w+\s) {};
    }
  }
}
\end{tikzpicture}
}

The six (6) friends are assigned to one of two (2) classes. Five (5) are indistinguishable and one (1) is distinguishable. Restrictions: The distinguishable friend must have at least one (1) indistinguishable friend as a classmate. 

Remove the distinguishable friend from the set. Distribute the five (5) indistinguishable to the two classes.  Then assign Manoj only to the classes where one (1) or more of his friends are assigned.

\(\left( \binom{c+f-1}{f}\right) \tiny \textbf{ c=number of distinguishable classes & f=number of indistinguishable friends}\\ \left( \binom{2+5-1}{5}\right) * 2 = 12 \tiny \textbf{ Multiply by 2 because each division creates two sets—one for each class. }\\ \)

These 12 sets will range from zero to five and five to zero. There are two (2) arrangements where a class has zero (0) of Manoj’s friends, leaving 10 class arrangements for Manoj to join.   

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This is also an integer partition of five (5) divided into two sets. Each set represents the two classes with five to zero and zero to five friends.  

{5+0}, {4+1}, {3+2}, {2+3}, {1+4}, {0+5}

The resolution is the same as above.

GA

Thank you for keeping us informed :)

The hypotenuse of the triangle is 15cm long.

Calculate the length of the shortest side.

it can be anythign in the range     0

I'm just going to look at one fruit basket, it can have

5apples and 0oranges

5apples and 1oranges

5apples and 2oranges

5apples and 3oranges

5apples and 4oranges

5apples and 5oranges

5apples and 6oranges

5apples and 7oranges

5apples and 8oranges

5apples and 9oranges

That is 9 ways it can have 5 apples

There are 10 ways it can have 4 apples

There are 10 ways it can have 3 apples

There are 10 ways it can have 2 apples

There are 10 ways it can have 1 apples

There are 9 ways it can have 0 apples

So that is

9+10+10+10+10+9=58 ways

If the baskets are different then you will have to double this answer.

Mmm

\((x^2+y^2-z^2)^2-4x^2y^2\\ [(x^2+y^2-z^2)-2xy][(x^2+y^2-z^2)+2xy]\\ [(x^2+y^2-2xy)-z^2][(x^2+y^2+2xy)-z^2]\\ [(x-y)^2-z^2][(x+y)^2-z^2]\\ [((x-y)-z)((x-y)+z)][((x+y)-z)((x+y)+z)]\\ [(x-y-z)(x-y+z)][(x+y-z)(x+y+z)]\\ (x-y-z)(x-y+z)(x+y-z)(x+y+z)\\ \)

Are the fruit baskets identical too?

Not that it makes much difference. You would just have to double the answer if they are not.

How many ways are there to assign each of 6 friends to either the chemistry class or the biology class if one of these six, Manoj, refuses to be in a class without any of his friends?

Yes there are many more than 8 possibilities.

I think I will assign the 5 first and then assign the sixth one later.

number of ways that there can be all 5 in one subject is   1*2 = 2    (because there are 2 subjects)

Manoj would have to be with the others so when he is added there are still only 2 ways

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number of ways that there can be 1 in one subject and 4 in the other in one subject is   5*2 = 10   

number of ways that there can be 2 in one subject and 3 in the other in one subject is   5C2*2 = 10*2=20

So that is 30 ways the 5 can be assigned if they are not all in the same class.

with each of these Manoj can be in either class so when I add him the number of possibilities double.

So that is 60 ways where they are not all in the same class.

--

Add the other 2 and I think there are   62 ways that the children can be placed into subjects where Manoj is not left on his own.

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If you have been given an answer then please let me know what it is. If it is different from mine maybe we can work out why.

Let n be the integer such that \(0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31}.\) What is \(\left(2^n\right)^3 - 2 \pmod{31}\)? Express your answer as an integer from 0 to 30, inclusive.

\(3n\equiv1\pmod{31}\\ 3n\cdot21\equiv1\cdot21\pmod{31}\\ 63n\equiv21\pmod{31}\\ 62n+n\equiv21\pmod{31}\\ \boxed{n\equiv21\pmod{31}} \)

If \(0\le n < 31\ \text{and}\ 3n \equiv 1 \pmod{31},\) n = 21

You can finish this on your own. 

What is the sum of all positive integer solutions less than or equal to 20 to the congruence \(13(3x-2)\equiv 26\pmod 8\)?

Note: you usually can not divide in linear congruences, but since 13 is relatively prime to 8, you can. 

\(13(3x-2)\equiv 26\pmod 8\\ 3x-2\equiv2\pmod8\\ 3x\equiv4\pmod8\\ 3x\cdot3\equiv4\cdot3\pmod8\\ 9x\equiv12\pmod8\\ 8x+x\equiv12\pmod8\\ x\equiv12\pmod8 \)

I hope this helped,

Gavin

What is the residue of \(9^{2010}\), modulo 17?

We can see the pattern:

\(9^1\equiv9\pmod{17}\\ 9^2\equiv13\pmod{17}\\ 9^3\equiv15\pmod{17}\\ 9^4\equiv16\pmod{17}\\ 9^5\equiv8\pmod{17}\\ 9^6\equiv4\pmod{17}\\ 9^7\equiv2\pmod{17}\\ 9^8\equiv1\pmod{17}\\\)

The residues then repeat after that. 

Therefore, since \(2010÷8=251\ R\ 2\), the residue is 13. 

I hope this helped,

Gavin

Note: this is solution is incorrect, since I missed the less than, I thought it was just an equal sign.

We first need to determine the range of for \(a\):

\(1^3=1,\\ 2^3=8,\\ 3^3=27,\\ 4^3=64.\\ \Rightarrow{a=2} \)

\(a=2, \text{since a is even}\\ \Rightarrow2^3+b^2+c=50\\ b^2+c=42\)

Repeating our process, b can be 2, 4, or 6

There is a fixed value for c for each value of b. 

Therefore, there are 3 triples that satisfy \(a^3+b^2+c=50\)

​​​

a)  Making a Venn Diagram, label the diagram.

\(\color{red}{\text{Only telescope club}} \color{black}{|} \color{orange}{\text{Both clubs}} \color{black}{|} \color{yellow}{\text{Only Math}}\)

\(\color{red}{x-y}\color{black}{|}\color{orange}{y}\color{black}{|}\color{yellow}{2x-y}\)

x-y+y+2x-y=3x-y

b)  2x-y+x-y=3x-2y

9^2010 mod 17 = 13

10! base 10 =3,628,800 =205940a in base 11. So, there are no trailing zeros in base 11.

Sorry, can't read the other 2.

Let the 4 kids be: A, B, C, D
A+B=2*47, A+C=2*50, A+D=2*51, B+C=2*53, B+D =2*54, C+D=2*57, solve for A, B, C, D

Solve the following system:
{A + B = 94 | (equation 1)
A + C = 100 | (equation 2)
B + C = 106 | (equation 3)

Subtract equation 1 from equation 2:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+B + C = 106 | (equation 3)

Add equation 2 to equation 3:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+0 B+2 C = 112 | (equation 3)

Divide equation 3 by 2:
{A + B+0 C = 94 | (equation 1)
0 A - B + C = 6 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Subtract equation 3 from equation 2:
{A + B+0 C = 94 | (equation 1)
0 A - B+0 C = -50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Multiply equation 2 by -1:
{A + B+0 C = 94 | (equation 1)
0 A+B+0 C = 50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Subtract equation 2 from equation 1:
{A+0 B+0 C = 44 | (equation 1)
0 A+B+0 C = 50 | (equation 2)
0 A+0 B+C = 56 | (equation 3)

Collect results: A = 44,   B = 50,   C = 56
C +  D =2 x 57
56 + D = 114
D = 114 - 56 = 58

A = 44 and B = 50 and C = 56 and D=58 - which is the heaviest child.

The smallest 3-digit number divisible by 3 and ending in 5 is 105. The largest 3-digit number divisible by 3 and ending in 5 is 975. Each number after 105 must increase by increments of 30 in order to be divisible by 3 and ending in 5.

Therefore: 975 -105 = 870 / 30 + 1 = 30 numbers that are divisible by 3 and end in 5.

There are: [999 - 102] / 3 + 1 = 300 multiples of 3 with three digits.

1 2 3 3 =4!/2! - (3  nPr  1) = 9
1 1 3 6 =+ 4!/2! - (3  nPr  2) = 6
1 1 2 9 =+ 4!/2! - (3  nPr  2) = 6
4!/2! - (3  nPr  1) + 4!/2! - (3  nPr  2) + 4!/2! - (3  nPr  2)=21 - total permutations exluding 1 in "units" position.
(1129, 1136, 1163, 1192, 1219, 1233, 1316, 1323, 1332, 1613, 1912, 2119, 2133, 2313, 3116, 3123, 3132, 3213, 3312, 6113, 9112)=21 permutations.

do you want to know how many or just the list of multiples?

I believe there are 291 3 digit multiples.

The prime factorization of 18 is \(2\cdot3^2\)

Case 1: {1, 2, 3, 3}

If these digits are used, there are 4!/2! combinations. 

Case 2: {1, 1, 3, 6}

If these digits are used, there are 4!/2! combinations. 

Case 3: {1, 1, 2, 9}

If these digits are used, there are 4!/2! combinations. 

There are a total of 4!/2! + 4!/2! + 4!/2! = 36 positive four digit integers. 

However, excluding permutations with ones in the unit digits, we get 36 - 15 = 21

As far as I know this is the best way. 

I hope this helped,

Gavin