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There are 28 numbers in the array that are divisible by p^4.

Carl writes eight consecutive three-digit numbers on a blackboard.  Each number that Carl writes is divisible by 2, 3, 4, or 5.  What is the sum of the digits of the smallest number that Carl wrote?   

The eight numbers could have been 114, 115, 116, 117, 118, 119, 120, & 121    

The sum of 1 + 1 + 4 =  6     

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When a prime is divided by 60, the remainder is a composite number.  When a second prime is divided by 60, the remainder is a prime.  Find the smallest possible value of the second prime.  

2 is the smallest prime.  2 / 60 = 0 R 2 so the remainder is a prime as well.  

However, I think that is not the intent of the problem.  Let's go larger than 60.    

Note the smallest primes – 2, 3, 5, 7, 11, etc – and then add to 60    

                             62 not prime   

                             63 not prime   

                             65 not prime   

                             67 prime and 67 / 60 = 1 R 7 and the remainder is prime as well.    

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Find a six-digit multiple of $64$ that consists only of the digits $2$ and $4$.   

64 x 3,816 = 244,224     

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The remainder of the 1000th number in the fibonacci sequence when divided by 17 is 4

(d-7) = 9 (g-7)     Given

(d-4) = 6 (g-4)      Given       <==== solve for   g= (d+20)/6     <==== use THIS in the first equation 

d-7 = 9 ( (d+20)/6   - 7)       Solve for   d

d-7 = (9d+180)/6 - 63 

d+56 = (9d+180)/6 

6d + 336 = 9d + 180 

156 = 3d 

d = 52 y/o  now

The sum of the digits of the smallest number that Carl wrote is 10.

Let's analyze the problem step by step:

1. Understanding the Given Information:

EFGH is a trapezoid with EF parallel to GH.

P is a point on EH such that EP:PH = 1:2.

Area of triangle PEF = 6.

Area of triangle PGH = 6.

2. Visualizing the Situation:

trapezoid EFGH with point P dividing EH in a 1:2 ratio

3. Using the Ratio of Areas:

Since triangles PEF and PGH share the same height (the perpendicular distance between EF and GH), the ratio of their areas is equal to the ratio of their bases.

So, (Area of triangle PEF) / (Area of triangle PGH) = EP/PH = 1/2.

We know the area of both triangles, so we can confirm this ratio: 6/6 = 1/2.

4. Finding the Area of Triangle EGH:

Since EP:PH = 1:2, we can say that EH = 3 * EP.

Triangles EGH and PEF share the same height, so the ratio of their areas is equal to the ratio of their bases.

(Area of triangle EGH) / (Area of triangle PEF) = EH/EP = 3/1.

Therefore, the area of triangle EGH = 3 * (Area of triangle PEF) = 3 * 6 = 18.

5. Finding the Area of Trapezoid EFGH:

The area of trapezoid EFGH is the sum of the areas of triangles PEF and EGH.

Area of trapezoid EFGH = Area of triangle PEF + Area of triangle EGH = 6 + 18 = 24.

Therefore, the area of trapezoid EFGH is 24.

Jon teaches a fourth grade class at an elementary school where class sizes are always at least 10 students and at most 15.  One day Jon decides that he wants to arrange the students in their desks in a rectangular grid with no gaps.  Unfortunately for Jon he discovers that doing so could only result in one straight line of desks.  What is the largest number of students that Jon could have in his class?    

It will have to be a prime number.  Any nonprime number can be divided into at least two rows.     

The largest prime number that's 15 or less is 13.    

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Thank you very much! I completely understand the problem now.

A baking club wants to form an executive committee. There are $15$ people in the baking club, including Mark. In how many ways can the baking club form an executive committee with $2$ people, not including Mark?   

That depends.  If the two members have different ranks or titles, then there are ₁₄P₂  =  182 ways.    

This way would mean that a pair of members like Jesse and Frank is different from Frank and Jesse.    

If the two members are untitled and otherwise indistinguishable, then there are ₁₄C₂  =  91 ways.    

This way would mean that a pair of members like Jesse and Frank is the same as Frank and Jesse.  

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

First, we know that the two points must be on the equation \(x^2+y^2=25\)

We also know that x must be 4. 

Thus, plugging in x=4 into the first equation, we can find y values. We have

\(4^2+y^2=25\\ y^2=25-16\\ y=\pm\sqrt9\\ y=3, y=-3\)

Thus, the two points A and B are \((4, 3), (4, -3)\)

Since they share an x value, the difference between the y values is the distance. We have

\(3-(-3)=6\)

Thus, 6 is our final answer. 

Thanks! :)