NotThatSmart

Let's focus on the two smallest numbers first. 15 and 24. 

Since N leaves a remainder, we can elmiinate the factors.

Thus, N cannot be \(3, 5, 2, 4, 6, 8, 12\)

This leaves 7, 9, 10, 11, 13, and 14.   Now, let's test out all the other numbers

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

mmm...I don't think there is a solution

Thanks! :)

The first 5 primes are

\(101, 103, 107, 109,113 \)

113 is our answer. 

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Well, let's notice something interesting about this problem. 

The number two has only two residues. 0 and 1. 

If the number is odd, then the residue is 1. If the number is even, then the residue is 0. 

We just have to figure out whether or not \(n \equiv 43 \pmod{165}\)is even or odd. 

We can easily figure this out though. 

Let's note that since n is 43 mod 165, we can write n as

\(165x+43 =n\) where x is a random integer. 

In this case, when x is even, then n is odd. 

If x is odd, then n is even. 

so in turn, both residues are completely possible for the final result. 

Thus, our answer is \(0,1\)

Thanks! :)

Let's work on an equation for this problem. Let's let the missing units digit number be x. 

Since we converting from base 16 (hex) to base 10, 

We have the conversion process

\((7D3x)_{16} = (7 × 16^3) + (13 × 16^2) + (3 × 16^1) + (x × 16^0)\)

Now, calculating the first 3 terms of the numbers, we find that

\((7 × 16^3) + (13 × 16^2) + (3 × 16^1)=32048\)

So, this means that \(32048 + x\) has to be divisble by 25. 

The closest number that satsifies this is \(32050\), which means that x is 2. 

The next closest number is 32075, which would mean x is 27, which is our of range for base 10. 

Thus, 2 is our final answer. 

Thanks! :)

First, let's figure out the slope of the two points. 

Since the slope is \(\frac{y_2-y_1}{x_2-x_1}\), the slope for (3,17) and (82,150) is

\(\frac{150-17}{82-3} = \frac{133}{79}\)

However, notice that between the two points, there are no lattice points. 

This is because following the slope, after (3, 17), the next latice point is

\((3+79, 17+133) = (82, 150)\)

Thus, the two endpoints are the only latice points, there are 2. 

Thanks! :)

Let's write an equation to solve this pronlem. 

First off, let's let x be the tens digit and the other be y. 

From the first of the problem, we have the equation

\(10x+y = 10y+x +72\)

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

\(x=2y-1\)

Solving this system, we get that \(x=17,\:y=9\)

This equation cannot be satisfied. 

From the first equation, we have that \(y=x-8\)

There are only two possibilities. One is where x is 9 and y is 1, so we have 91. 

The other possibility is x is 8 and y is 0. 

Neither of these satisfy the conditions given.

Thanks! :)

First, let's combine all like terms and moving all terms to one side. 

We get the equation

\(6x^{2}-21x-31=0\)

Using the quadratic equation, we find two values of x. we have

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

Obviously, the first term of x is bigger, so taking that, we find our final answer as

\(x=\frac{\sqrt{1185}+21}{12}\)

Thanks! :)

If there are the two roots, we have that (x+1)(x-1). 

This, distributing in, we have

\((x + 1) ( x -1) = x^2 + 0x - 1 \)

Thus, the answer is x^2 - 1. 

Thanks! :)

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

Let's let b be bagels

Let's let t be toast

And let's let m be muffins. 

From the problem, we have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

Now, we want everything to be in terms of bagels. 

From the first two equations, we get the equations

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, plugging these values into the third equation, we get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is 262 cents or about 261 and 2/3 cents. 

Thanks! :)

We can set variables to solve this question. 

Let's let the first term be 2

This means that the last term \( 2 + 5(n-1)  =  5n - 3\)

The sum is  \([ 2 + 5n -3 ] [n/2]  > 100\)

\([ 5n - 1 ] [ n ] > 200 \\ 5n^2 - n > 200 \\ 5n^2  - n - 200 > 0 \)

Since we want the largest possible value, set this to equal 0. We have

\(5n^2 - n -200 =  0 \\ n ≈ 6.4253 \)

n = ceiling [ 6.4253 ]  =  7

The 7th term produces the scream  =  5(7) - 3  =  32

2,  7,   12,  17,   22 ,   27 ,  32

Sum  9    21   38    60     87     119

Thus, the final answer is 32. 

Thanks! :)