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Let's note something important before we begin. 

All the points 5 units away from (2, 7) forms a circle with radius 5 and center of (2, 7). 

A circle with that equation would look something like

\((x-2)^2+(y-7)^2 = 5^2\)

Combining that with the second part of the problem, we have a system with 

\((x-2)^2+(y-7)^2 = 25\\y=5x-28\)

Now, we already have a value for y, and we subsitute that into the first equation to isolate x. 

We find that

\((x-2)^2+(5x-35)^2 = 25\)

Expanding and bringing all terms to one side of the equation, we get a quadratic. We find that

\(26x^2-354x+1204=0\)

This gives us 2 x values that will satisfy thsi equation. We get that

\(x=7,\:x=\frac{86}{13}\)

Now, we plug these two values of x back into the equation to find y, so that we can complete our problem. 

We find that \(y = 7, y= \frac{66}{13}\)

So our final two abswers that satisfy all these conditions given above in the problem are

\((7, 7), (86/13, 66/13)\)

Thanks! :)

5 • 6 • 7 • 8 = 1680.  Then add 1 to it.      

Brandon picked 1681 apples   

check answer   

                          1681 / 5  =  335 R 1   

                          1681 / 6  =  280 R 1   

                          1681 / 7  =  240 R 1   

                          1681 / 8  =  210 R 1   

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The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$   
If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?   

If a number ends with 5, its square will end with 25.  

So the only possible value of the tens digit is 2.   

How to square a number that ends with a 5.   

Separate the 5 out of the number.  

Change that 5 to 25.   

Add 1 to the number that is left. 

Multiply that by the original number.   

Stick the 25 to the end of the product.   

Examples:  

   15²                              45²                              75²                              125²       

   1                  5              4                  5             7                  5              12                  5   

   1                25              4                25             7                25              12                25    

   1 x (1+1)    25              4 x (4+1)    25             7 x (7+1)    25              12 x (12+1)  25    

   1 x 2           25              4 x 5          25             7 x 8           25              12 x 13        25     

   225                              2025                           5625                             15625          

_.

Now, we want to find the ratio PQ/FH. We can find the lengths of PQ and FH by using similar triangles. Let's look at triangle HFG and triangle GNF.

Since HM is parallel to GN, we have angle NFG = angle GFM. Additionally, since FM/MG = 1/2, we have FG = 2GF. Similarly, using triangle HEN and triangle EMF, we find that EH = 3HE.

Now we see that FG:GF = EH:HE = 3:2. Therefore, triangles HFG and GNF are similar.

From here, we know that PQ/FH = QF/FH = HN/HF. By letting HF = a and HN = b, we have a+b = FH. Since HN:NE = 1:3, b/3y = 1/3, b = y. Similarly, we know that a/x = 2x/x = 2, a = 2x.

Now we have PQ/FH = HN/HF = y/(a+b) = y/(2x+y). Let Y = PQ/FH = y/(2x+y). Since y = HN = 1 and x = FM = 1, we have Y = 1/(2+1) = 1/3.

Therefore, the ratio PQ/FH is 1/3.

The value of k is 25.

We have a system of equations to solve. We have

\(N \equiv 2 \pmod{3}, \\ N \equiv 2 \pmod{7}, \\ N \equiv 2 \pmod{10}.\)

Now, let's note something really important. All 3 equations have a remainder of 2. 

Let's anakyze. 

N leaves a remainder of 2 when divided by 3, 7, and 10. 

Therefore, our answer is the LCM of 3, 7, and 10. 

This is because since they all leave the same remainder, the same number must be for all of them. 

Therefore, we have \(LCM[3, 7, 10] = 210\)

Adding 2, we have \(210+2=212\)

So our answer is 212. 

Thanks! :)

We need to solve the system of equations

\(N \equiv 2 \pmod{5}, \\ N \equiv 2 \pmod{7}\)

Well, let's analyze this equation. 

N leaves a remainder of 2 when divided by 5 and 7. 

This means that N is two greater than the LCM of 7 and 5. 

This is because since it leaves the same remainder for both numbers, then it must be the same number. 

Therefore it is the LCM we are looking for. 

We have

\(LCM[5, 7] = 35\)

Adding two, we have \(35+2 = 37\)

thus, our answer is 37. 

Thanks! :)

Let's rewrite the number as \(a18,486.7b\). This number must be both divisible by 9 and 11. For the number to be divisible by 9, the digits must add up to a multile of 9. The sum of all the digits are \(34 + a + b\). The closest multiple of 9 is 36. Therefore, \(a+b \equiv 2\text{ } (\text{mod }9)\), because the sum of any 2 one-digit numbers can't be any greater than 18. For a number to be divisible by 11, the result of alternating adding and subtracting the digits must be divisible by 11 (ex: 121, 1 - 2 + 1 = 0, so 121 is divisible by 11). Therefore, \(a + 1 - 8 + 4 - 8 + 6 - 7 + b = -12+a+b\)

The closest multiple of 11 is -11, so \(a+b\equiv1\text{ }(\text{mod 11} )\). We can now narrow down the possibilities. No two single digit numbers can add up to  more than 18. a + b can now only be 2 and 11 for the first equation, and a + b can now only be 1 and 12 for the second equation. Since none of these are the same, there are no possibilities for the missing digits.

We are starting with this:

\(N \equiv 4 \pmod{6}, \\ N \equiv 4 \pmod{10}, \\ N \equiv 4 \pmod{15}, \\ N \equiv 4 \pmod{24}.\)

We can rewrite this as:

\(N = 6a +4\\N=10b+4\\N=15c+4\\N=24d+4\)

\(N -4 = 6a\\N-4=10b\\N-4=15c\\N-4=24d\)

Therefore, N - 4 is a factor of 6, 10, 15, and 24. To find the minimum value of N-4, we find the least common multiple of 6, 10, 15, 24. We can do this by listing out the factors:

\(6 = 2^13^1\)

\(10=2^15^1\)

\(15=3^15^1\)

\(24=2^33^1\)

\(N-4= \text{lcm}(6, 10, 15, 24) = 2^33^15^1=120\)

Therefore, the least possible value of N is 124.

If a number is written in the form of \(p_1^a p_2^bp_3^c...\), where all p's are prime numners, it will have (a+1)(b+1)(c+1)... factors. Therefore, a terrific number must be writen in the form of either \(p^9\), or \(p_1^1p_2^4\). The smallest possibilities for each respectively are 2^9 or 512 and 3(2^4) or 48. Therefore, 48 is the smallest terrific number.

If you have a number in the form of \(p^n\) where p is a prime number, it will have n + 1 factors. Therefore \(2^9\) or 512 will have 10 factors (1, 2, 4, 8, 16, 32, 64, 128, 256, 512), but 2 will be the only prime number that it is divisible by. This is also true for \(3^9\) (19683), \(5^9\) (1953125), \(7^9\) (40353607), or any number in the form of \(p^9\), where p is a prime number. Therefore, one prime number is the minimum number of factors for a terrific number.

First, let's note that if a number has an odd number of factors, then it must be a perfect square. 

In the problem, it also must be divisible by 8. 

Thus first few that satsify this number is \(16,64,144\)

However, none of them work, as 16 and 64 have too few factors, and 144 has too many. 

The next number is 256. 

Note that 256 has exactly 9 factors of \(1, 2, 4, 8, 16, 32, 64, 128,256\)

Of these 9 factors, \(2\) is the only prime divisor!

So out of the 9, only 1 is prime

So 1 is our answer.

Thanks! :)

First, let's note that if a number has an odd number of divisors, then it must be a perfect square. 

It also must be a multiple of 8.

So we need a perfect square divisble by 8. 

First, we have

\(4*4=16\)

However, 16 only has 5 factors, with \(1,2,4,8,16\)

Next, we have \(8*8=64\). However, 64 only has 7 factors, with \(1, 2, 4, 8, 16, 32, 64\)

144 doesn't work, as it has way too much. 

However, we note that \(16*16=256\)

256 has exactly 9 divisors, of \(1, 2, 4, 8, 16, 32, 64, 128,256\)

Thus, the smallest possible number is 256. 

So 256 is our answer. 

Thanks! :)

There isn't really a model you can follow to solve this problem. 

However, there are multiple numbers that satisfy the condiitions given. 

First, we have

\(244224 \mod 64 = 0\)

Since the final result is 0, then 244224 is an answer. 

Also, let's note that another number, 

\(444224 \mod 64 = 0\) also works. 

So 444224 satisfies the equation as well. 

Thus, the two answers we have are \(444224, 244224\)

Thanks! :)

In the example, the divisors of 100 include 1 and 100.  

Using that standard, the smallest integer would be 4    

because its perfect square divisors are 1 and 4.  

_.

The answer is phi(1200) = 600.

The answer is phi(191) = 172.