Bosco

The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?    

x² – mx + 24  =  10    

x² – mx + 14  =  0    

the quadratic factors to  (x – 1)(x – 14)    —>>   x²  – 15x + 14    

                                  or  (x – 2)(x – 7)      —>>   x² – 9x + 14    

so, m could be 15 or 9     two possibilities    

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The product of a set of distinct positive integers is $630$. If one of the numbers is 42, what are the other two numbers?    

630 divided by 42 is 15.  Therefore the other two factors are either 3 & 5 or 1 & 15.    

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You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $52$ pieces of candy per scoop to the vat, and another machine that can remove exactly $39$ 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?    

Both 52 and 39 are multiples of 13.  No matter how many times you add X • (4 • 13) and subtract Y • (3 • 13) the remainder will never get "out of synch" so to speak. The number of pieces left will always be a multiple of 13, including zero.   

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In how many ways can you distribute $8$ indistinguishable balls among    

$2$ distinguishable boxes, if at least one of the boxes must be empty?    

If one of the two boxes is to remain empty, and if all eight balls have to be used,     

then you have to put all eight balls into the other box.  Only two ways to do that.    

Either put the balls in box A and leave box B empty, or vice versa.    

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What is the smallest positive integer n such that \sqrt[4]{675 + n} is an integer?    

I think the problem is asking for the square root of the quantity [ (4) • (675 + n) ]    

If so, I observe that 676 is a perfect square, and so is 4.    

So, we just need to assign n the value 1 to obtain sqrt(2704) which is 52.    

Find all values of c such that 3(2c + 1) = 28*(3c) - 9 + 45(3c) - 18. If you find more than one value of c, then list your values in increasing order, separated by commas.    

3(2c + 1) = 28*(3c) - 9 + 45(3c) - 18    

6c + 3  =  84c – 9 + 135c – 18    

6c – 84c – 135c  =  – 3 – 9 – 18    

– 213c  =  – 30    

         c  =  – 30 / – 213  =  30 / 213  reduces to 10 / 71       

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Find the greatest prime divisor of the value of the arithmetic series    
1 + 2 + 3 + \dots + 200    

The sum of this series is (200 / 2)(1 + 200)  =  (100)(201)  =  20,100    

The factors of 20,100 are 2². 3, 5², 67     (I admit I looked this up.)    

The larges prime divisor of 20,100 is therefore 67.    

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What is the smallest prime divisor of 5^{19} + 7^{13} + 23?    

 23  =  ends with a 3    

5¹⁹  =  ends with a 5    

7¹³  =  ends with a 7     (see below)

Whatever the sum of the three numbers is, it ends with a 5.    

Since the sum ends with a 5, the smallest prime divisor is 5.    

7⁰ = 1         7⁴ = 2,401         7⁸ = 5,764,801    

7¹ = 7         7⁵ = 16,807       7⁹ = 40,353,607    

7² = 49       7⁶ = 117,649     7¹⁰ = 282,475,249    

7³ = 373     7⁷ = 823,543     7¹¹ = 1,977,326,743    

In the value of 7ⁿ as n is raised to successive exponents, the     

units digit cycles 1, 7, 9, 3, repeatedly, so 7¹³ ends with a 7.    

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