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What is the coefficient of x^2 in (x^3 + x^2 + x + 1)(x^4 + 3x^3 + 8x + 7x + 1)?  

                                       Is that 8x a typo that should have been 8x² ?  

To get the product, you have to multiply every term inside the   

first parentheses times every term in the second parentheses.  

You could write it like this  

                                                  x³ • (x⁴ + 3x³ + 15x + 1)       (1)   

plus                                            x² • (x⁴ + 3x³ + 15x + 1)       (2)   

plus                                              x • (x⁴ + 3x³ + 15x + 1)       (3)   

plus                                              1 • (x⁴ + 3x³ + 15x + 1)       (4)   

Let's count up everything that results in an x to the second power.  

From (1) there is no x²   

From (2)                                     1 x²    

From (3)                                   15 x²    

From (4) there is no x²    

Totalling up the x² we have 16x² so the coefficient of the x² term is 16.  

_.

THANK YOU! 53 WAS CORRECT!!!

woops...

Catherine rolls a standard $6$-sided die six times.  If the product of her rolls is $2000,$ then how many different sequences of rolls could there have been?  (The order of the rolls matters.)   

Combinations of 2, 2, 4, 5, 5, 5 and combinations of 1, 4, 4, 5, 5, 5    

I cannot think of any other combinations that would multiply to 2000.  

The number of permutations of 2, 2, 4, 5, 5, 5 is 6! = 720   

The number of permutations of 1, 4, 4, 5, 5, 5 is 6! = 720   

So the total number of sequences would be 720 + 720  =  1440 sequences   

_.

As follows:

@Padewolofoofy:  You forgot to divide by 2^2

What is the smallest positive integer that is the sum of a multiple of $15$ and a multiple of $21$? (Remember that multiples can be negative.) 

                   (15)(3) + (21)(–2)  

                           45 – 42  =  3  

              You can get it another way.  

                   (21)(3) + (15)(–4)  

                           63 – 60  =  3  

_.

Let the boxes be B1 and B2.

It it obvious, that under given condition, EITHER box B1 OR box B2 must be empty.

If box B1 is empty, then all 8 indistinguishable balls are in box B2: so, there is only one such distribution.

If box B2 is empty, then all 8 indistinguishable balls are in box B1: so, there is only one such distribution.

In all, there are 1 + 1 = 2 such distinguishable distributions.

By the quadratic formula, the roots are (-7 + √57)/2 and (-7 - √57)/2.

⍺ = (-7 + √57)/2                β = (-7 - √57)/2

⍺² + β² = [(-7 + √57)/2]² + [(-7 - √57)/2]² = 106 - 14√57 + 106 + 14√57 = 106 + 106 = 212

Converting the first equation to base 10, we have 6A+B+C=6C. Simplifying, we have 6A+B=5C.

Converting the second equation to base 10, we have 6A+B+6B+A=6C+C. Simplifying, we have 7A+7B=7C. Dividing both sides by 7, we have A+B=C.

Substituting C=A+B into the first equation, we have 6A+B+A+B=6(A+B). Simplifying, we have 8A+2B=6A+6B. Combining like terms, we have 2A=4B. Dividing both sides by 2, we have A=2B.

Since A, B, and C are distinct digits less than 6, the only possible values for (A,B,C) are (2,1,3) and (4,2,6). However, C must be less than 6, so the only solution is ABC = 213.

There are two cases to consider:

Case 1: C is on the horizontal line through A or the vertical line through B.

Without loss of generality, suppose C is on the horizontal line through A. If AB is a horizontal segment, then △ABC cannot be isosceles.

Otherwise, AC=AB if and only if C is one of the two points on this horizontal line that are the same distance from A as B is.

The same argument applies if C is on the vertical line through B. Thus, there are at most 2+2=4 such points C.

Case 2: C is not on the horizontal line through A or the vertical line through B.

Then △ABC is isosceles if and only if C lies on the perpendicular bisector of AB. There are 4 such points.

Hence, there are 4+4=8​ such points C.

By the Triangle Inequality,

[|a_1 - a_6| = |a_1 - a_2 + a_2 - a_3 + a_3 - a_4 + a_4 - a_5 + a_5 - a_6|

\le |a_1 - a_2| + |a_2 - a_3| + |a_3 - a_4| + |a_4 - a_5| + |a_5 - a_6|.]

Equality occurs when a2​, a3​, a4​, and a5​ all lie on the line segment joining a1​ and a6​.

Also by the Triangle Inequality,

[|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_4| + |a_4 - a_5| + |a_5 - a_6| + |a_6 - a_{10}| + |a_{10} - a_1|

\ge |a_1 - a_{10}| + |a_{10} - a_1| = 2|a_1 - a_{10}|.]

Equality occurs when a2​, a3​, a4​, a5​, and a6​ all lie on the line segment joining a1​ and a10​.

Thus, if a2​, a3​, a4​, a5​, and a6​ all lie on the line segment joining a1​ and a10​, then

[|a_1 - a_6| = 500 - |a_6 - a_{10}| - |a_9 - a_{10}| - |a_8 - a_9| - |a_7 - a_8|.]

To maximize ∣a1​−a6​∣, we minimize ∣a6​−a10​∣+∣a9​−a10​∣+∣a8​−a9​∣+∣a7​−a8​∣.

The minimum value is 0, which occurs when a6​=a7​=a8​=a9​=a10​.

Thus, the largest possible value of ∣a1​−a6​∣ is 500​.