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The constant term is 672.

a_2+a_8=5, so that forces a_5 to be 5/2. a_4*5/2=5, so a_4 equals 2

Thank you so much!

\(ar^{22}=16\\ ar^{27}=24\)

solve simultaneously to find a and r

etc

You may see this properly depending on what extentions you have on your computer but we do not.

Get rid of the dollar signs.

I do not know if guests answer is right or wrong, I have not checked.

Thanks for answering guest :)

Let

\(f(x)=x^3(ax^2+bx+c)\quad so\\ f(x)+2=x^3(ax^2+bx+c)+2\qquad (1)\\ f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But also}\\~\\ f(x)+2=(x+1)^3(dx^2+ex+f)\qquad (2)\\ \text{By inspection I can see that }f=2\\ f(x)+2=(x+1)^3(dx^2+ex+2)\\ f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\ \text{The x term will be }6x+ex=(6+e)x\\ \text{ but the coefficient of x must be 0 so e=-6}\\~\\ \text{The x squared term will be }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\ \text{ but the coefficient of x squared must be 0 so d=12}\\ so\\ f(x)+2=(x+1)^3(dx^2+ex+f)\\ f(x)+2=(x+1)^3(12x^2-6x+2)\\ \text{expand etc} \)

The logic is good but you need to check for careless errors.

LaTex:

f(x)=x^3(ax^2+bx+c)\quad so\\
f(x)+2=x^3(ax^2+bx+c)+2\qquad (1)\\
f(x)+2=ax^5+bx^4+cx^3+2\\ \text{But  also}\\~\\
f(x)+2=(x+1)^3(dx^2+ex+f)\qquad (2)\\
\text{By inspection I can see that   }f=2\\
f(x)+2=(x+1)^3(dx^2+ex+2)\\
f(x)+2=(x^3+3x^2+3x+1)(dx^2+ex+2)\\
\text{The x term will be   }6x+ex=(6+e)x\\
\text{ but the coefficient of x must be 0 so e=-6}\\~\\
\text{The x squared term will be   }6x^2+3ex^2+dx^2=6x^2-18x^2+dx^2=(-12+d)x\\
\text{ but the coefficient of x squared must be 0 so d=12}\\
so\\
f(x)+2=(x+1)^3(dx^2+ex+f)\\
f(x)+2=(x+1)^3(12x^2-6x+2)\\
\text{expand etc}

What is the value of xyz ?

Hello Guest!

\(xy - z = 15\\ xz - y = 15\\ yz - x = 15 \)

\(z=xy-15\\ z=\frac{15-y}{x}\\ z=\frac{15-x}{y}\)

\(xy-15=\frac{15-y}{x}\\ \frac{15-y}{x}=\frac{15-x}{y}\\ ok\ Max.\)

i got 1/2 wat do i do next?

The fewest number of  handshakes the coach could have pariticipated in is 8.

(a,b) = (-5,2).

(1,  5,  7,  11, 13, 17, 19, 23, 29, 31, 35,  37, 41, 43,  47, 49, 53, 55, 57, 59, 61, 65, 67, 71, 73, 77, 79,  83, 85, 89, 91, 95, 97)==33 integers ==1,665 (their sum)

I don't think that's right

Since f(x) is divisible by x^3, f(x) is of the form ax^5 + bx^4 + cx^3.

You then want ax^5 + bx^4 + cx^3 + 2 to be divisible by (x + 1)^3.  Using long division, you get the equations

-10a  + 6b - 3c = 0

4a - 3b + 2c = 0

-a + b - c + 2 = 0

==> a = 6, b = 16, c = 12

So f(x) = 6x^5 + 16x^4 + 12x^3.

Here is my attempt: 
So suppose the 4 digits are:  _ _ _ _
The first _ could be even or odd.

If it is even then,

let Case (1): EVEN ODD EVEN ODD​

If it is odd then,

let Case (2): ODD EVEN ODD EVEN
Even numbers: 0,2,4,6,8

Odd numbers: 1,3,5,7,9

For Case (1):

Since the first digit _ could be 2,4,6,8 then there are 4 possibilities.

The second digit is odd, so it could be 1,3,5,7,9 then there are 5 possibilities.

The third digit is even, so it could be 0,2,4,6,8 (Notice, the first digit couldn't be 0, as that would make the number "3-digit") then there are 5 possibilities.

The fourth digit is odd, so it could be 1,3,5,7,9 then there are 5 possibilities.

Thus multiplying all the possibilities: 

4*5*5*5 = 500       (*)

For Case (2):

Same method as Case (1), except the first digit is odd so it could be: 1,3,5,7,9 then there are 5 possibilities
5*5*5*5 = 625  (**)

Adding (*) and (**):

Therefore, 500+625 = 1125,  4-digit numbers which satisfy the given condition.

We need n to be even, so the answer is 500.

cos A = 1/3.

The probability is 7/10.

The possible values of AC are 4, 5, 6, 7, 8, 9, 10, 11, and 12.

YY' = 5.

We can note that each repetition of the sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2] has length 24. We do not include the extra 1 at the end to prevent double counting. If we divide 1234 by 24, we get 51 with some remainder. Multiplying 24 by 51, we get 1224. Subtracting this from 1234, we get a result of 10. Thus, our answer is the 10th number in the list: 10.

Suppose we begin by looking at the first 6 numbers. The only numbers that are not divisible by 2 or 3 are 1 and 5. 

Now, you can note that since 6 is divisible by both 2 and 3, this pattern will continue. 

The sum of 1 and 5 is 6. Generalizing this, we can say that the sum of the numbers not divisible by 2 or 3 in the xth group of 6 is 12x - 6. This is obtained by the fact that each number increases by six when moving up a group of 6. 

Now we check how many full groups of six there are from 1 to 99. This gives a result of 16. Therefore, we evaluate \(\sum_{n = 1} ^ {12} 12n - 6\) and get a result of 864. But there is one number we are forgetting: 97, which is greater than 96 and thus not included in our sum. Finally, we add 97 to get our final result of 961. 

(2^1993 + 3)/5 mod 10^10==3829920359 - these are the last 10 digits of the quotient.

Agree with geno3141:

1,234 mod (13 - 1) ==10

Or:

1234 mod (25 - 1) ==10

As you count from 1 to 13 and then back down again, you get this set of numbers:

1-2-3-4-5-6-7-8-9-10-11-12-13-12-11-10-9-8-7-6-5-4-3-2

Then your next set of numbers repeat the above list

1-2-3-4- etc.

There are 24 numbers in each list.

So, how many sets of 24 are in 1234?

Dividing 1234 by 24, we get a quotient of 51 and a remainder of 10.

The quotient of 51 tells me that there are 51 groups of 24 numbers (which I really don't care about)

and the remainder of 10 tells me how far to keep counting (I care about this!)

So what number is the tenth number in the list

     1-2-3-4-5-6-7-8-9-10-11-12-13-12-11-10-9-8-7-6-5-4-3-2 ?

The tenth number in the list is my answer ...

This is obvious if you use the following fact: If m and n are integers, then m - n divides P(m) - P(n).