NotThatSmart

We can use long division to solve this problem. We have

As shown, the first digits of 13/29 is \(0.4482758620\)

So \(0.4482758620\) is our answer. 

Thanks! :)

According to the Euclid's algorithm, we start with 2 numbers, a and b. 

We do \(a ÷ b = c\) with remainder R where a is the larger of the two. 

Then, we replace a with b and b with R, and do the same thing. 

We repeat this prcoess until R is 0. 

So, now we do the same with 972 and 1220. 

\(1220 ÷ 972 = 1; R= 248\)

\(972 ÷ 248 = 3 ; R =228\)

\(248 ÷ 228 = 1 ;R= 20\)

\(228 ÷ 20 = 11 ;R= 8\)

\(20 ÷ 8 = 2 ;R =4\)

\(8 ÷ 4 = 2 ;R =0\)

When the remainder is 0, the GCF is the divisor. Thus, 4 is the GCF. 

So our answer is 4. 

Thanks! :)

~NTS

Thanks to you too, CPhill. 

I did not know about Euler's Totient Function until today when we did a problem similar earlier. 

Sure saved me a lot of time on this one! Lol! :)

Thanks! :)

~NTS

The first number above 1000 that is divisble by 47 is 1034. Thus, the smallest number that fits the list is 1037. 

The largest number under 2000 divisble by 47 is 1974, so 1977 is the last number in our list. 

Now, notice that we just have an arithmetic sequences. We have

\(1037, 1084, 1131, ... 1977\)

Subtracting every number in the list by 3, we get

\(1034, 1081, 1128, ...1974\)

Divididing every number in the list by 47, we get

\(22, 23, 24,...42\)

Subtracting 21 from every term in the sequences, we have

\(1,2,3,4...21\)

There are 21 numbers in this list. 

Thus, there are 21 numbers that leave a remainder of 3 when divided by 47. 

So our answer is 21. 

Thanks! :)

~NTS

We can use Euler's Totient Function. 

Let's find all the distinct prime factors of 2835 first. 

The prime factors of 2835 are \(3,5,7\)

Now, we simplfy do \(2835(1-1/3)(1-1/5)(1-1/7)\)

Simplfying this, we have

\(2835 * \frac{2}{3}*\frac{4}{5}*\frac{6}{7} =\frac{2835\cdot \:2\cdot \:4\cdot \:6}{1\cdot \:3\cdot \:5\cdot \:7}=1296\)

So 1296 is our answer. 

I'm not sure if I did this correctly...

Thanks! :)

We want the blanks to be 3 numbers that have a least common multiple of 12. 

We also want them to be as small as possible, but we don't want them to be 1, since that would make this problem impossible. 

The numbers \(2,3,4\) satsify these conditions. We have

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

Converting them to the same denominator of 12, we get

\(6/12+4/12+3/12 = 13/2\)

So 2, 3,4 is our answer. 

Thanks! :)

We can write an equation to solve this problem. 

First, the equation for the number of diagonals in a polygon is \(n(n-3)/2\) where n is the number of sides or vertices. 

Thus, we can write the equation

\(n(n-3)/2=2n\)

Now, we simply solve for n.

\(n^2-3n=4n\\ n^2-7n=0\)

Using the quadratic equation, we get

\(n=\frac{7\pm \sqrt{(-7)^{2}-4\cdot 1\cdot 0}}{2\cdot 1}\\ \)

From this, we get

\(n=7\\ n=0\)

So we have 7 sides. 

So 7 is our answer. 

Thanks! :)

First, let's note that since the area of ABCD is 180, we have the equation

\(AD * AB =180\)

Now, note that the shaded region is a trapezoid with base AB and UT, with height AD. 

We know the length of AB and AD already, and UT is 1/3 the length of DC, so we are all set. 

We can write the equation

\((1/2)(AD)(AB + AB/3) = \\ (1/2)(AD)(4/3)(AB) = \\ (2/3) AD * AB = (2/3)(180) = 120\)

So our answer is 120. 

Thanks! :)