Bosco

The prime numbers between 10 and 15, inclusive, are 11 and 13.   

So the largest possible number of students in Jon's class is 13.   

Any number other than a prime number can be divided.   

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The first prime numbers after 100 are   

101, 103, 107, 109, and 113   

So the fifth smallest after 100 is 113.   

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2 times.   

Let the first machine cycle two times and then shut it off, for a total of 10 added.   

Let the second machine cycle three times and then shut it off, for a total of 9 removed.   

10 – 9 = 1 candy left in the bottom    

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Class size is at most 60.  The most students that could not   

be arranged in more than one row would be 59 students.  

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In how many ways can the numbers 1, 2, 3, 4, 5, 6 be arranged in a row,    

so that the product of any two adjacent numbers is at least 5?    

You must have the 1 between the 5 and the 6.   

The rest of the numbers can be anywhere.   

So, consider two cases:   

(1) Permutation of   2  3  4  (5 1 6) ..... ₄P₄  =  24   

(2) Permutation of   2  3  4  (6 1 5) ..... ₄P₄  =  24   

Add them together                 total  .....  48 ways   

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Find the units digit of $13*287$.   

13⁰ = 1            13⁴ = 28561            13⁸ = 615730721               13¹² = 23298085122481   

13¹ = 13          13⁵ = 371293          13⁹ = 10604499373           13¹³ = 302875106592253   

13² = 169        13⁶ = 4826809        13¹⁰ = 137858491849       13¹⁴ = 3937376385699289   

13³ = 2197      13⁷ = 62748517      13¹¹ = 1792160394037      13¹⁵ = my calculator went into scientific notation   

The above tabulation shows that the units digit repeats every four exponents.   

Divide the exponent by 4, disregard the whole number part of the   

quotient, and consider only the remainder part of the quotient.   

Raise 13 to that remainder and see what the units digit is.   

Pick a few examples to demonstrate process.   

13⁶      6 / 4 = 1 R 2      13² = 169      therefore the units digit of 13⁶ is 9.   

13⁷      7 / 4 = 1 R 3      13³ = 2197    therefore the units digit of 13⁷ is 7.   

13¹²    12 / 4 = 3 R 0     13⁰ = 1         therefore the units digit of 13¹² is 1.   

13²⁸⁷   287 / 4 = 71 R 3    13³ = 2197    therefore the units digit of 13²⁸⁷ is 7.   

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Catherine rolls a standard $3$-sided die six times. If the product of her rolls is $600,$ then

how many different sequences of rolls could there have been? (The order of the rolls matters.)   

I can't find on the internet, what the numbers are on a "standard" 3-sided die.  

A couple of sites said the faces are 1, 2, and 3.   

Other places said the faces have other numbers.   

The prime factors of 600 are 2, 3, and 5.  

Of course, 1 is a factor of 600, but by definition 1 is not a prime number.   

If the numbers on the die are 1, 2, and 3, you can't get there from here.   

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On planet Enigma, the residents use a currency called the confusion.  There are only two confusion bills on Enigma, one worth $7$ confusions and the other worth $12$ confusions.  There are also some coins of smaller value, but each weighs over $10$ kilograms, so they are difficult to carry around.  In how many ways can a resident of Enigma use only bills to purchase a toaster that costs 19 confusions?   

I think nobody is answering this problem because it's so obvious that they suspect a trick.   

Well, I'll admit I'm suspicious, too, but I'll take one for the team.  I'm going out on that limb.  

There is only one way to purchase the toaster ... with one 7-confusion bill and one 12-confusion bill.   

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In a store window, there was a box of berries having a total weight of 200 kg.  The berries were 99% water, by weight.  After two days in the sun, the water content of the berries was only 90%, by weight.  What was the total weight of the berries after two days, in kg?   

The water was 99% of 200, so the pulp was 1% of 200 = 2 kg.   

After two days in sun, 90% is water, but the pulp didn't evaporate.   

There's still 2 kg pulp, which now accounts for 10% of the berries.   

Since 10% is 2 kg, the total weight of the berries is now 20 kg.   

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Fill in the blanks to make a quadratic whose roots are $1$ and $-1.$   
x^2 + ___ x + ___   

Since the roots are +1 and –1, the factors are (x – 1) and (x + 1)    

Multiplied together gives x² – 1   

So, the first space contains a     0    

the second space contains a    –1   

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