Bosco

You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $7$ pieces of candy per scoop to the vat, and another machine that can remove exactly $6$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?  

Well, I say the least possible number number of times is 1 time.  

Turn on both machines.  Machine 1 adds 7, and Machine 2 removes 6,   

then real quick turn off both machines before they have time to cycle again.  

Nobody said the machines have to keep running until all 1000 pieces are gone.    

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Find the discriminant of the quadratic 5x^2 - 2x + 8 - 2x^2 + 6x + 2.   

Combine like terms                              3x² + 4x + 10  

When a quadratic is in the form ax² + bx + c   

the discriminant is (b² – 4ac)   

In this case, that's (4²) – (4)(3)(10)  =  16 – 120  =  – 104  

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In a rhombus, all the side lengths are $12,$ and one of the angles is equal to $90^\circ.$ Find the area of the rhombus.   

If one of the angles is 90^o then all the angles are 90^o    

Given that all the side lengths are the same, namely 12   

That makes the rhombus the special case of being a square     

Area  = 12²  =  144 units²    

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Starting with the A moving one letter at a time vertically, horizontally, or diagonally, how many different paths spell ARCH?   

A can reach 3 R's  

R can reach 3 C's  

C can reach 3 H's  

So there are 3 • 3 • 3 = 27 possible paths   

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I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row, if all three mathematics textbooks must be together?   

Temporarily consider the group of three math books as one unit. 

Then, 1 group + 2 psych + 2 chem can arrange to ₅P₅ = 120 ways. 

But wait.  You could arrange the three math books 6 different ways. 

So all the possible arrangements is 120 • 6 = 720 ways.  

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I have $3$ different mathematics textbooks, $2$ different psychology textbooks, and $2$ different chemistry textbooks. In how many ways can I place the $7$ textbooks on a bookshelf, in a row?  

7 • 6 • 5 • 4 • 3 • 2 • 1  =  5040 ways 

You can make 7 choices for the first book. 

That leaves you 6 choices for the next book. 

And then there are 5 choices for the third book. 

and so on  

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Laverne starts counting out loud by 5's.  She starts with 2.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds 20, Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running? 

Laverne's           Shirley's  

Counting          Summation  

     2                           2  

     7                           9   

   12                         21    

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In a row of five squares, each square is to be colored either red, yellow, or blue, so that no two consecutive squares have the same color. How many ways are there to color the five squares, if there must be at least three yellow squares?  

In order not to have two yellows next to each other,  

the three yellows must occupy squares 1, 3, and 5. 

That leaves squares 2 and 4 to fill, and two colors  

to do it with. There are only four ways. 

Blue Blue     Red Red     Blue Red     Red Blue  

So, the answer is four ways.  

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xy – (x + y) = 39    

when x and y are relatively prime and less than 2  

I'm going out on a limb and saying there aren't any.  

If we hypothesize that the problem meant less  

than 20, and the 2 is a typo, then (11,5) works. 

If we can ignore the non-prime and the limit of 2,    

then two more pairs that work are (6,9) and (21,3).  

They were pretty easy to find by graphing it on Desmos.   

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