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Let's set up for a really cool function. 

First, note that

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2 \)

Now, we use Euler's Totient Function. We have

\(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320 \)

The answer is \(320\)

Thanks! :)

Let's set up for a really cool function. 

First, note that

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2 \)

Now, we use Euler's Totient Function. We have

\(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320 \)

The answer is \(320\)

Thanks! :)

The answer is \(4\)

1*1 = 1

2*2 = 4

Thanks! :)

Let the original amount that Alice had = A   Let the original amount that Bob had = B

1st scenario :  Alice gets n dollars from Bob....her new amt = A + n

Bob loses n dollars.....his new amt times 7 = Alice's new amt

A + n  = 7(B - n)  →    A + n = 7B - 7n  →  A + 8n = 7B    (1)

2nd senario : Bob gets n dollars from Alice....her new amt = A - n

Bob gains n dollars  and alice has 3 times as much as he does

A - n  =  3(B  + n)  → A - n = 3B + 3n  →  A - 4n = 3B  → 2A - 8n = 6B  (2)

Add (1) , (2)

3A = 13B

A / B  =  13  / 3 

Proof :  3A = 13B  →  A = (13/3)B....sub this into  (1)

(13/3)B + 8n  =   7B

8n =(8/3)B    multiply through by 3/8

3n = B

n= B/3

Using (2)

2A - 8(B/3) = 6B

2A = (26/3)B

A = (26/6)B 

A = (13/3)B

A / B  = 13  /  3

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    

Thus, the answer is \(2\)

Thanks! :)

Let's look at the number 512. 

We know that \(512 = 2^9\), which means it has 10 divisors. 
Since it's divisble by 2 and two is the smallest prime, 

our final answer is 2. 

Thanks! :)

Let's first take a look at the fraction 1/17. 

It has 16 repeating digits in the form of \(0. \overline{0588235294117647}\)

We simplfy have to find which digit the 4000th lies on. 

First, let's do 4000/16. 

We get \(4000/16=250\)

So the 4000th digit is the last digit of the repeating decimal, which is 7. 

So 7 is our answer. 

Thanks! :)

the answer is \(-\frac{252}{\left(\sqrt{11}-3\right)^3\left(-3-\sqrt{11}\right)^3}\)

the answer uis \(17n - 8\)

The answer is \(0\)

The answer is \(111\)

THIS is not possible

https://web2.0calc.com/questions/number-theory_66931

First, let's simplify the equation. We have

\(x^2 - 6x + 13\)

Since the constant is equal to uv, we have the equation

\(uv = 13 \\ 2uv = 26\)

The coefficient of x is equal to u+v, so we have

\(u + v = 6\)

Squaring both sides of the equation, we have the equation

\(u^2 + 2uv + v^2 = 36 \\ u^2 + 26 + v^2 = 36 \\ u^2 + v^2 = 10\)

Now, it's time for the fun part of this problem. We can simplify what we want to figure out to

We already know all these terms! We can easily find them! We have

\(u^3 + v^3 = ( u + v) ( u^2 + v^2 - uv) = (6)(10 - 13) = -18 \\ [uv]^2 = [ 13]^2 = 169\)

So, The answer is \(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2 = -18 / 169\)

Thanks! :)

Simplify as

\(10x^2 + 10x - 1\\ ab = -1/10\\ 2ab = -2/10  = -1/5\)

a + b  =  -1      square both sides

a^2 + 2ab + b^2  = 1

a^2 - 1/5 + b^2  = 1

a^2 + b^2  = 1 + 1/5  =  6/5

\((a -4) / ( b -4)  +  ( b -4) / (a -4)  =     [  (a - 4)^2  + ( b -4)^2] ]  ' [ (a -4) ( b-4) ]  = \\ [ a^2 - 8a +16  + b^2 - 8b + 16 ]  /  [ ab - 4(a + b) +16 ]   = \\ [  (a^2 + b^2) - 8(a + b) + 32 ]  / [ ab - 4 (a + b) + 16 ]  =\\ [ 6/5 - 8 (-1) + 32 ]  [ -1/10 - 4 (-1) + 16 ]  =  \\ [ 6/5 + 40 ]  / [ 20 - 1/10] =   20497 / 25\)

The answer is \( 20497 / 25\)

Thanks! :)

Let's write a system of equations to solve this problem. 

Let's let G be Grogg's age and D be the dad's CURRENt age. 

From the first part of the problem, we ge tthe equation

\(9(g-7)=d-7\\ 9(g-7)+7=d\)

From the second part of the problem, we have the equation 

\(6(g-4)=d-4\\ 6(g-4)+4=d\)

Now, since both equations are equal to d, we can set the right side to equal each other. 

Thus, we have the equation

\(9(g-7)+7=6(g-4)+4\)

Now, we can find Grogg's age. We find that Grogg is \(g=12\)

Plugging this value back into the first equation, we get

\(d=9(12-7)+7=9(5)+7=52\)

So Grogg's dad is 52 years old. 

Thanks! :)

MY 600th ANSWER

The answer is \(A, B, C = (1, 1, 1)\)

We get the monstrosity \(10101001010100010111000110010101100011_2\), which when converted to base 10, we get

\(181803967843_{10}\)

This divided by 70 (it would go to scientific notaiton if not) gets us \(229488030591\), 

meaning it is divisible by 7. 

Thanks! :)

Let's create a system of equations to solve this problem .

Let's let A be the number of money Alice has and B be the amount of money Bob has. We have

If Alice receives n dollars from Bob, then she will have 3 times as much money as Bob.

\(A + n  =  3 * (B - n) \)

If she gives n dollars to Bob, then she will have 2 times as much money as Bob.

\(A - n  =  2 * (B + n) \)

Thus, we have the system

\(A + n = 3B - 3n → A = 3B - 4n → 3A = 9B - 12n \\ A - n = 2B + 2n → A = 3B + 3n → 4A = 12B + 12n\)

adding these two equations up, we get

\(7A = 17B → 7A / B = 17 → A / B = 17 / 7 \)

So, the answer is \(17/7\)

Thanks! :)

We can write a handy equation to solve this problem. 

First, let's set a variable. Let's set x as a positive integer. 

Any number congruent to 7 modulo 29 can be written in the form

\(29x+7\) since it is 7 greater than a factor of 29. 

thus, we can write the inequality

\(29x+7\leq 1000\)

Setting the two to equal and solving for x, we have

\(29x+7=1000\\ 29x=993\\ x \approx 34.24\)

Rounding down, we find x is 34. 

So our answer is 34. 

Thanks! :)

the answer is \(34\)

The answer is \(244224 \pmod {64} \equiv 0 \)

The remainder is 10.

The answer is 58.

\((n-3) + (n-2) + (n-1) + n + (n+1) + (n+2) + (n+ 3) = (28/9)(n + 3) \\ 7n = (28/9)(n + 3)\\ 7n = (28/9)n + 28/3 \\ 63n = 28n + 84 \\ 63n - 28n = 84 \\ 35n = 84\\ n = 84 / 35 = 12 / 5 \\ (n - 3) = (12/5 - 3) = -3 / 5\)

This isn't an integer, but

the answer is \( -3 / 5 \)

Thanks! :)