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So, let us start with the the simple equations...

\(109 mod (d) = 4\)

We know this from the remainder of four from dividing by 109.

To find the possible two digit integers that abide by this, you must take the prime factorization of the greatest integer to follow this rule, giving you

 \(109 - 4 = 105\)

When you take the prime factorization of 105 from above, you get \(3*5*7\)

To find the possible two digit numbers that abide by the rules in the problem, you can get

\(3*5\)

\(5*7\)

\(7*3\)

Or just 15, 35, and 21!

When you take the sum of these, \(15+35 = 50 50+21 = 71\)

And the final answer is 71! Yay, we did it!

Thanks so much! 

Sol: 5x=100%

x=20%, so 20%/5=4%

Debatable

Since 1000! has 164 factors of 7 and 500! has 82 factors of 7,

\(\frac{1000!}{(500!)^2}\) has \(164 - 82 - 82 = 0\)

factors of 7.

In other words, all the factors of 7 cancel,

so the greatest power of 7 dividing \(\frac{1000!}{(500!)^2}\)is 7^0 = 1, 

so \(n = \boxed{0}.\)

Here's one more way using polynomial division

(n + 1)^2                n^2 + 2n  + 1

_______   =          ___________

  n  +  23                 n  +  23

                  n    -   21

n + 23    [   n^2   +   2n     +   1   ]

                  n^2   +  23n

                 __________________

                             -21n   +    1

                             -21n   -  483

                            ___________

                                          484

n -  21    +        484

                     ________

                       n  + 23         

And note that, as GYanggg found, the largest n that makes the last fraction an integer is when n = 461

Thanks Chris!

It's always nice when someone complements your work!

Very nice, GYanggg  !!!!!

This is a permutation

Out of 11 students....we want all the possible orders  in choosing  any 4 of them....this is given by :

P(11,4)   =  7920  different sequences in a week

5x * p  = x/5       divide both sides by 5x

p  =   x / [ 5*5x]       simplify

p  = 1/ 25    =  4/100   =   4%  

(x/(5 x))/5

Cancel terms. (x/5)/(5 x) = 1/25:

 1/25% = 4%

109 - 4 = 105

105 = 3 * 5 * 7

3 * 5 = 15

3 * 7 = 21

5 * 7 = 35

15 + 21 + 35 = 71

See this very nice answer that was  provided by Melody a while back....

https://web2.0calc.com/questions/circles-and-angles

$5 to 2 lbs    as a ratio    is    5/2    (or 10/4    etc)

Hi Rollingblade!

The first step in this problem is to divide the numerator by its denominator. 

We have:

\(\begin{align*} \frac{(n+1)^2}{n+23} &= \frac{n^2 +2n + 1}{n+23}\\ &= \frac{n^2 + 23n - 21n+1}{n+23}\\ &= \frac{n^2 +23n}{n+23} -\frac{21n}{n+23} + \frac{1}{n+23}\\ &=\frac{n(n+23)}{n+23} - 21\cdot \frac{n+23-23}{n+23}+ \frac{1}{n+23}\\ &=n - 21\left(\frac{n+23}{n+23} - \frac{23}{n+23}\right)+ \frac{1}{n+23}\\ &= n - 21 + \frac{21\cdot 23}{n+23}+ \frac{1}{n+23}\\ &=n-21 + \frac{484}{n+23}. \end{align*} \)

This means that n+23 must be a factor of 484 = 2^2 *11^2

The largest factor of 484 is 484.

So n = 484 - 23 = 461.

I hope this helps. 

If x = arctan(-1/5) and x + y = 290 degrees, then cos y

Firstly I find it totally confusing to call angles x and y.

So I am going to replace x with theta and y with alpha.

I am also going to assume that both the angles are positive and between 0 and 290.

\(tan\theta<0\)   therefore theta must be in the 2nd or 4th quadrant but only the second one will work.

\(\alpha=290-\theta\\ cos(\alpha)\\ =cos(290-\theta)\\ =cos290cos\theta+sin290sin\theta\\ =cos70cos\theta  -sin70sin\theta\\ =cos70\times \frac{-5}{\sqrt{26}}  -sin70\times \frac{1}{\sqrt{26}}\\ = \frac{-5cos70-sin70}{\sqrt{26}} \\ \approx 0.5197\)

Density is mass per unit volume, so density = 43.5/50 g/ml = 0.87 g/ml

.

If we start from \(sAk^\alpha=(\delta+n+g)k\) then divide both sides by \(k^\alpha(\delta+n+g)\) we get \(\frac{sA}{\delta+n+g}=\frac{k}{k^\alpha}\)

This is \(\frac{sA}{\delta+n+g}=k^{1-\alpha}\)  Take the \(1-\alpha\) root of both sides to get:  \((\frac{sA}{\delta+n+g})^\frac{1}{1-\alpha}=k\)  which isn't quite the same as your expression for \(k^*\), unless \(k^*=k^\alpha\).

They are the same

the roots of f(x) are the x values that make f(x)=0

The equation of the x axis is y=0

so the roots of the equation y=f(x)  are the x values that make f(x)=0   true

The equation of a circle which has a center at (-5,2) can be written as Ax^2 + 2y^2 + Bx + Cy = 40

 Let r be the radius of the circle.

Find     A+B+C+r

\((x+5)^2+(y-2)^2=r^2\\ x^2+10x+25+y^2-4y+4=r^2\\ x^2+y^2+10x-4y=r^2-29\\ 2x^2+2y^2+20x-8y=2r^2-58\\ A=2,\quad B=20,\quad C=-8\\ 2r^2-58=40\\ r^2-29=20\\ r^2=49\\ r=7 \qquad \text{Since the radius must be positive} \)

You can do the addition. 

By what integer factor must 9! be multiplied to equal 11!?

\(\begin{array}{|rcll|} \hline 9!\times 10 \times 11 &=& 11! \\ 9! \times 110 &=& 11! \\ \hline \end{array}\)

The integer factor is 110.

Let $f(n)$ be the sum of the positive integer divisors of $n$. For how many values of $n$,

where $1 \le n \le 25$, is $f(n)$ prime?

\(\begin{array}{|r|rcr|c|} \hline n & && f(n) & \text{prime} \\ \hline 1 & 1 &=&1 & \\ 2 & 1+2&=&3 & \checkmark \\ 3 & 1+3&=&4 & \\ 4 & 1+2+4&=&7 & \checkmark \\ 5 & 1+5&=&6 & \\ 6 & 1+2+3+6&=&12 & \\ 7 & 1+7&=&8 & \\ 8 & 1+2+4+8&=&15 & \\ 9 & 1+3+9&=&13 & \checkmark \\ 10 & 1+2+5+10&=&18 & \\ 11 & 1+11 &=& 12 & \\ 12 & 1+2+3+4+6+12 &=& 28 & \\ 13 & 1+13 &=& 14 & \\ 14 & 1+2+7+14 &=& 24 & \\ 15 & 1+3+5+15 &=& 24 & \\ 16 & 1+2+4+8+16 &=& 31 & \checkmark \\ 17 & 1+17 &=& 18 & \\ 18 & 1+2+3+6+9+18 &=& 39 & \\ 19 & 1+19 &=& 20 & \\ 20 & 1+2+4+5+10+20 &=& 42 & \\ 21 & 1+3+7+21 &=& 32 & \\ 22 & 1+2+11+22 &=& 36 & \\ 23 & 1+23 &=& 24 & \\ 24 & 1+2+3+4+6+8+12+24 &=& 60 & \\ 25 & 1+5+25 &=& 31 & \checkmark \\ \hline \text{sum } & &&& 5 \\ \hline \end{array}\)

How many perfect squares have a value between 10 and 1000?

\(4^2 = 16 \ldots 31^2=961\)

\(31 - 3 = \mathbf{28}\) perfect squares have a value between 10 and 1000.