My Adventures as a Math Doctor

A Presentation at the Ohio Council of Teachers of Mathematics Annual Conference

Sharonville Convention Center, Cincinnati, OhioOctober 16, 2015

Mark Jaffee, Annual Conference CoordinatorOhio Council of Teachers of Mathematics26 North Cedar StreetOberlin, Ohio 44074-1530 Telephone: 440-574-1462Email mark.jaffee@ohioctm.orghttp://www.oberlin.net/~markjaffee

My History with the “Ask Dr. Math” Project

Summer, 1995 - I attended a Geometer’s Sketchpad workshop in Philadephia and I was introduced to the Geometry Forum located at Swarthmore College.

Fall, 1995 - I subscribed to a geometry email list through the Geometry Forum and exchanged information and answered Questions.

October 13, 1996 - I received an email from the Math Forum inviting me to apply to “Ask Dr. Math” and I applied immediately. (see attached letter)

October 20, 1996 - A response was received from my application.

November 4, 1997 - My Dr. Math account was created, a user name and password was assigned and information for logging on was provided. Suggestions were made as to which questions to answer and explanation of the “Triage Area” and “Holding Tank” was provided.

January 7, 1998 - I received feedback from my administrator concerning questions that I had answered.

February 3, 1998 - I became a tenured Math Doctor.

How to Become a Math DoctorGo to http://mathforum.org/volunteer/dr.math/signup.htmlYou can choose a sample question and write a reply which gets sent to an administrator. The two of you will have a conversation about how it might be improved. You will then work on live problems, and revise your answers under supervision before sending them out. When you get to the point where your answers are routinely being sent without any changes, you get 'tenure', after which you can correspond directly with the 'patients'.

Interesting ProblemsAs Joesph wrote to Dr. MathOn 01/26/2015 at 18:54:52 (Eastern Time),[Question]An isosceles triangle ABC has its vertices on a circle. If AB=13cm, BC=13cm and AC=10cm. Calculate 1. The height, BM of the triangle.2. The radius of the circle

[Difficulty]Finding the radius of the circle

[Thoughts]I solved for BM And got 12 but since confused on how to solve 4 the radius.

Hi Joesph,

You are off to a good start; I agree with you that BM = 12. Now, here is my suggestion.

You know that the vertices of the triangle are on a circle which means that we have a triangle with a circumscribed circle. Therefore, the center of the circle is located at the intersection of the perpendicular bisectors of the three sides.The perpendicular bisector of AC is the line BM and you already know that the length of the segment BM is 12. Let N be the midpoint of side BC, so if you draw a perpendicular line through BC passing through point N, the line will intersect BM at the center of the circle. Let's call that point, O. Now OB is a radius; let's call its length, r. Draw the radius OC and look at right triangle OMC. You know the length of MC and you can label OM and OC in terms of r.Then use the pythagorean theorem and you can solve for r.Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more.

Thanks for writing to Dr. Math and Good Luck.

As Stephanie wrote to Dr. MathOn 02/03/2015 at 22:24:34 (Eastern Time),[Question]Find the first triple of consecutive odd numbers which are all not primes. Is it true that for each integer k>=2, there are k consecutive odd numbers that are not primes? in fact, it isn't difficult to show that for each positive integer n there are n consecutive integers none of which is a prime. Can you find a proof of this? It follows from this that the answer to the last question is 'yes'.

[Difficulty]I don't understand how the second question works at all, so don't understand the third either--I did try to solve the first one though. Here are my workings: I figured that each triple had to have a multiple of three. This is because the possible values of the first number in the triple are 3n, 3n +1 or 3n +2. If we start with 3n, then our three numbers are 3n, 3n +2, 3n+5.

If we start with 3n+1, then our three numbers are 3n+3, and 5n +5. And if we start with 3n +2, then our three numbers are 3n, 3n +4 and 3n +6.

So, I made a list of multiples of three and tried to find numbers above or below that weren't primes. I did this up to 120, but then realized further than that, I wouldn't be able to tell whether numbers were primes or not because I'm not familiar with them (for instance, the number 169 equals 13^2, but if I didn't know that, I would just assume it was a prime.) Are there any other methods to solve this efficiently?

[Thoughts]

Hi Stephanie,

I came across this problem that you sent in two weeks ago. If you haven't already found a solution, I can help you.

First of all, consider the sequence (n + 1)! + 2, (n + 1)! + 3, (n + 1)! + 4, ... and pick any integer for n. Let's start with n = 4.

Substituting 4 for n, we get 5! + 2 =( 5 x 4 x 3 x 2 x 1) + 2 for the first term. It is obviously divisible by 2.

The second term is 5! + 3 =( 5 x 4 x 3 x 2 x 1) + 3, which is divisible by 3.

The third term 5! + 4 =( 5 x 4 x 3 x 2 x 1) + 4 is divisible by 4, and

The fourth term 5! + 5 =( 5 x 4 x 3 x 2 x 1) + 5 is divisible by 5.

So, here are 4 consecutive integers, none of which is prime.

Now, pick a value for n greater than 4 and you can determine a larger sequence.

Now, if you want to find the 3 smallest consecutive odd integers that are not prime, this method won't help you because it locates a sequence that is not necessarily the smallest.

However, I found the answer fairly easily using the Sieve of Eratasthones. If you don't know what that is, make a list of the first 100 integers. Cross out the 1because it is not prime. Then circle the "2" which is prime, then cross out all the multiples of 2. Circle the "3" because it is prime, then cross out all of the multiples of 3 that haven't already been crossed out. Do the same thing with the 5, then the 7, but when you get to the 11, you should observe that every multiple of 11 up to 100 has already been crossed out.

Next, circle every number larger than 11 that hasn't been crossed out. They are all primes.

It should be easy now to find three consecutive odd numbers that are not prime.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

As Manish wrote to Dr. MathOn 02/19/2015 at 06:50:55 (Eastern Time),[Question]if m^x + m^y +m^z=1 then what is x+y+z=?

[Difficulty]Very confusing to procced

[Thoughts]

Hi Manish,

I suspect that there is more to the problem than what you indicated. For example, if m = 1/2 and x = 1, y = 2, and z = 2, then m^x + m^y +m^z=1 and x+y+z= 5.

But, if m = 1/3 and x = 1, y = 1, and z = 1, then m^x + m^y +m^z=1 and x+y+z= 3.

Or, if m = 1/4 and x = 1, y = 1, and z = 1/2, then m^x + m^y +m^z=1 and x+y+z= 2 1/2.

So, based on what you provided, I suspect that there are infinitely many answers to this problem. Is it possible that you left some information out?

Let me know and if you have some more information about the problem, I'll try to help you find a solution.

What???

As rey wrote to Dr. MathOn 01/27/2015 at 22:31:37 (Eastern Time),[Question]

[Difficulty]

[Thoughts]

Hi Rey,

What is your question?

As Rhea wrote to Dr. MathOn 12/04/2014 at 21:58:19 (Eastern Time),[Question]Find the area of the incenter circle with coordinates A (15, -3), B (-9, 4) and C (-9, -3). My teacher said something about 1/a +1/b+1/c=1/r where a,b, and c are the 3 alitudes of the triangle and r is the radius.

[Difficulty]I am not sure what to do.

[Thoughts]

Hi Rhea,

The way you presented the question, it appears that A,B, and C are points on the circle. Did you mean that A,B, and C are the vertices of the triangle? Let me know. The solution to the problem depends on which of those you meant.

Write back and we will discuss how to solve the problem.

Thanks for writing to Dr. Math and I will be waiting to hear from you.

As rabin wrote to Dr. MathOn 03/13/2015 at 11:46:47 (Eastern Time),[Question]if in triangle pqr, line segment ps is the angle bisector of angle p, pq =12 , pr =15, qr =27 find qs.

[Difficulty]i dont know how to solve it. i know it is a scalene triangle . i dont know where point s is located.

[Thoughts]qr =27. i just guessed that qs = ps and qs = 13.5 and ps =13.5. i am not sure whether this right. if wrong please solve it for me

Hi Rabin,

Are you sure that the information you are giving me is correct. The problem is that since 12 + 15 = 27, pqr isn't a triangle, it is a line segment.

If the information is correct, then p and s are the same point and the answer is 0.

Let me know if you want to change the question.

As Adriana wrote to Dr. MathOn 01/17/2015 at 15:17:02 (Eastern Time),[Question]I have a 99 in a class and got a 67 on a test wwh to my ave? I would like to know if I would still be passing. A 67 is really bad as a test grade

[Difficulty]

[Thoughts]

Hi Adriana,

If the 99 is the average of all of the tests you took before you got a 67 on the last test, then your grade is what you get when you add up all of the test grades, then divide by the number of tests.

If the 99 was determined a different way, you need to talk to your teacher and find out the teacher's method for determining the grade.

Give it a try and if you want to check your answer with me or if you want some clarification about this question, write back and we'll discuss it some more.

Thanks for writing to Dr. Math and Good Luck.

Questions from AdultsAs Kate wrote to Dr. Math

[Question]phrase: "Why was four afraid of one? Because one, two, three."what's about it? may it be play on words?

[Difficulty]

[Thoughts]

Hi Kate,

I don't get it either, but there is another joke that is similar that makes more sense."Why was ten afraid of seven? Because seven, eight (ate), nine."

Is it possible that the individual who told you this joke was a young child who had heard the joke as I told it, didn't understand it, but retold it using the same structure with smaller numbers?

I hope my response helps.

As steve wrote to Dr. Math[Question]I am trying to find a formula that will give me quanties for multiple items in multiple forms, based on sqft. I have to take the area and get sqyds, and get quantitiy of gallons to order to cover that sqyd (based on 7sqyds a gallon) and get lnft based on sqft. I also have to find the converstion rate to make the price at the end add to 13.00 a sqyd.

[Difficulty]if the take off is in sqft, what would the series of formulas be to get the results i need.

[Thoughts]i know sqft to sqyd is divid by 9. i know that 1 gal of glue will cover 7 sqyds. i know the labor rate 35/per hr

Hi Steve,

You have the formulas mostly figured out already.

1. If you know the number of square feet, divide by 9 to get the number of square yards. You figured that out yourself.

2. To get the number of gallons, you know that 1 gallon covers 7 square yards (63 square feet). So 1 square yard can be covered by 1/7 gallon and 1 square foot can be covered by 1/63 gallon.

3. How do you define a linear foot? Different people have different definitions for this.

4. If labor is $35 per hour, that means $1 for 1/35 of an hour and $13 for 13/35 of an hour. So, if it takes 13/35 of an hour to cover one square yard, the charge is $13 per hour.

Let me know if this makes sense and let me know what you mean by determining linear feet from square feet and we will discuss it some more.

Thanks for writing to Dr. Math and Good Luck.

As NVV wrote to Dr. Math

[Question]Nancy left her house to go for a jog down a long hill. Eventually she will turn around and return home. She jogs 20% faster going downhill than uphill, and she wants to go for a 1-hour run. After how long should she turn around and start heading home?

[Difficulty]I feel like there is a piece of information missing. This is my daughter's 6th grade problem of the week.

[Thoughts]

Hi NVV,

All the information you need is there. Think of it this way. If you jog a specific distance twice as fast as normal, your time will be 1/2 as long. If you jog the distance 2/3 as fast, the time will be 3/2 as long. In general, if the rate is multiplied by x, the time it takes will be divided by x (which is the same as being multiplied by the reciprocal of x).

So, going uphill Nancy normally jogs at a certain rate, but going downhill her rate is 20% or 1/5 more; that is, her rate is 1 1/5 or 6/5 as much. Therefore, the time it takes will be 5/6 as long.

Therefore, we have to divide the hour into two parts and the ratio of the shorter part to the larger part must be 5 to 6.

Now, if it takes 5 hours to go downhill and 6 hours to go uphill, the ratio will be 5/6, but the total time will be 11 hours; way too much.

So, let's divide each by 11. 5/11 hour to go downhill and 6/11 hour to go uphill. The ratio is still 5/6 and now the total time is 1 hour.

Now, in an algebra class this problem would probably be solved differently.

Let d = the distance jogged uphill after Nancy turns around.Let r = the rate of speed that Nancy jogs uphill.Let t = the amount of time Nancy jogs uphill.

Therefore, d = rt

Then d = the distance that Nancy jogs downhill before she turns around,(6/5)r = the rate Nancy jogs downhill1 - t = the amount of time that Nancy jogs downhill

Therefore, d = (6/5)r(1 – t)

Since the distance going downhill is the same as the distance going uphill,

rt = (6/5)r(1 - t)

The r factors cancel out and you can solve for t.

Give it a try if you like and you should get the same answer that I got using the other method. If you want some clarification about this problem, write back and we'll discuss it some more.

As Adrian wrote to Dr. Math[Question]Ok. I'm a math tutor working with an Algebra II student and recently one of the exercises was to simplify sqrt(-16).

So, if you follow the Laws of Roots you can say that sqrt(-16) = sqrt(16)*sqrt(-1) = sqrt(16)*i. So, I say that the answer should be +/-4i but the answer book says it is only +4i. I disagree. I've looked on the web and only the + answer is provided.

[Difficulty]I'm not sure why it is only +4i. Can you provide clear reasoning as to why I'm incorrect? Is it because the laws of roots dictate that sqrt of a product of two numbers = product of the square roots of the two numbers only if both numbers are >= 0??

[Thoughts]

Hi Adrian,

I suspect that the issue is this. You are 100% correct that the square roots of -16 are 4i and -4i. However, -16 inside a radical sign, which should be read "radical negative 16" or "root negative 16" has only one solution, by definition.

For example, the square roots of 25 are 5 and-5, but radical 25, by definition, is 5.

I hope this clarifies the issue. Let me know if you want to discuss it some more.

DialoguesAs Isha wrote to Dr. Math[Question]Two long strips of paper are each 2 inches wide, and they overlap so that the angle between the strips is 30 degrees. What is the number of square inches in the area of the parallelogram formed by the overlap?

[Difficulty]Can't figure out base from the given information.

[Thoughts]Area of the parallelogram is base X height. We know the height is 2. How to figure out base? The only thing given is the angle 30 degrees.

Hi Isha,

To solve this problem, it is helpful to know that if you have a right triangle in which one of the acute angles measures 30 degrees, then the hypotenuse will always be twice as long as the side opposite the 30 degree angle.

So, the two strips of paper intersect and form a parallelogram. Let's call it ABCD with angle B measuring 30 degrees. Now, if you draw a segment from C that is perpendicular to side AB, you will create a triangle like the one I described. That should provide you all the information you need to finish the problem.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more.

A B

CD

Hi, I tried it, but could not finish it. I drew a segment from C that is perpendicular to side AB which intersects AB at point x. Using the information you gave me, I found the BX =Sqrt 12. what about AX? I know AX+XB=AB. Once I know AB then AB times height of a parallelogram which is two in this case = area of a parallelogram. So now my question is how to find AX?Thank you for your help.Isha

Hi Isha,You are asking the wrong question. Since both strips have the same width, the four sides of the parallelogram will be equal. So, focus on the triangle BXC. You know that XC = 2 and BX is the square root of 12. But, what you really want to know is how long is BC. Once you know that you have the length of AB because BC = AB.Give it a another try and let me know how that works out.

As Isha wrote to Dr. MathIn the case of AB=BC, BC=4 according to the fact you provided for the hypotenuse of 30-60-90 triangle. So the AB = 4 which is the width of parallelogram and the height is 2, so the area of the parallelogram is 8 sq. in. I am still little confused about what makes AB=BC. If the two strips were perpendicular to each other the square formed by the overlap has the area of 2x2 = 4 sq. in. But for parallelogram, the BC is slanted, so I have hard time understanding AB=BC. Thank you.

Hi Isha,

When you drew the altitude from C to AB, you found that BC = 4. Now draw an altitude from C to AD. Do you see how that results in CD = 4?

Let me know if that helps you understand why the parallelogram is really a rhombus.

As Isha wrote to Dr. Math

I am sorry, but if I draw an altitude from c to AD, where does it intersect? On D? or may be I am not getting this right!!

Hi Isha,

When you drew the altitude from C to AB, you intersected AB at a point you called X. When you draw the altitude from C to AD, the same type of thing should happen. The one end of the altitude is the point C and the other end should be a point Y between A and D such that CY is perpendicular to AD just like CX was perpendicular to AB.

Age: 11TimeStamp: 01/30/2014 at 08:45:04Subject: Thank you (Geometry)To: dr.math@mathforum.orgThreadNum: 940692From: educationid2021@gmail.com (Isha)

I got it now. Thank you so much for your help.

As Isha wrote to Dr. MathOn 01/11/2015 at 16:25:34 (Eastern Time),[Question]The trapezoid is divides into 4 triangles by its diagonals. The areas of two triangles are 9 and 25 (Not consecutive). What is the area of the whole trapezoid?

[Difficulty]I plugged different numbers for the bases and heights of the two triangles. When I tried (3,6) for the 9 triangle, and (5,10) for the 25 triangle (height,base). I used the trapezoid area formula, [1/2 (b1 + b2) * h]. I got 64 as my answer, which is correct. However, when I tried to check my answer, I instead used (2,9) for the 9 triangle and (2,25) for the 2 triangle. This time, I got 68. Why am I getting two different answers? Also, I don't think I am using the best method for this problem. Can you give me a better method to use for this problem.

[Thoughts]See above.

Hi Isha,

I'm glad to hear from you again. I hope you have had a good year. I have a few observations about your trapezoid which I think will help you understand the problem a little more clearly.

First draw a trapezoid ABCD with base(1), AB, then side BC, base(2), CD, then side AD. Make sure AB is a lot longer than CD.Now draw diagonals AC and BD intersecting at point E.

So, let's assume that the area of triangle DEC is 9 and the area of triangle BEA is 25. Since CD is parallel to AB, angle CDE must measure the same as angle EBA, angle DCE must measure the same as angle EAB, and angles DEC and AEB are vertical angles, so they measure the same.

Therefore, the two triangles are similar which means that corresponding sides and altitudes must be proportional. So, when you picked (3,6) and (5,10) as the respective bases and altitudes of the two triangles, you made a good choice because 3/6 = 5/10.

But, 2/9 is not equal to 2/25, so that choice led to the wrong answer. But, suppose you stick with 2 for the height and 9 for the base of the triangle whose area is 9. Then, if b is the length of the base of the other triangle and h is the height of the other triangle, then 2/9 = h/b. Also, (1/2)bh = 25. So, if you solve this system of equations, then use the fact that 2 + h is the height of the trapezoid, you should find out the area turns out to be 64 again.

Give it a try and if you want to check your work with me or if you want some clarification about this problem, write back and we'll discuss it some more.

Thanks for writing to Dr. Math and Good Luck.

Thank you so much for the quick response. I understood the problem and I got 64 again. I hope you had a good year too. Thanks again for your help.

Isha

As Rahul wrote to Dr. Math[Question]A triangle's angles have tangents that are all positive integers. Find the angles of the triangle.

[Difficulty]I don't see how there is enough information in this problem it seems really vague. In my mind this problem has lots of solutions.

[Thoughts]Well since the angles have positive tangents they must lie in the 2nd and 4 quadrents or in degrees must be between 0-90. But any 3 angles between 0-90 that add up to 180 are viable then so how can there be one answer (which I am told there is)? I think that maybe I could find the 3 tangent values and then take the arctan of them but how do I know which they are?

Hi Rahul,

If you are allowed to use a scientific calculator, this problem is fairly simple. You already know that the sum of the measures of the three angles is 180 degrees and each of the angles measures less than 90 degrees.

So, use the calculator to determine the arctan(1), arctan(2), arctan(3), arctan(4), arctan(5), etc... until you see three values whose sum is 180.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more.

(From Dr. Rick)To supplement what Dr. Jaffee said, if you CAN'T use a scientific calculator, you can reduce the problem to a simpler set of choices. Let p = tan(A), q = tan(B), r = tan(C). We want A+B = 180-C; so expresstan(A+B) = tan(180-C) in terms of p, q, and r using trig identities (tangent of a sum in terms of tangents of A and B, and a simple identity for tan(180-C). I obtain a very simple equation in positive integers p, q, and r.

That doesn't mean it's simple to SOLVE, but intuition may tell you that the numbers can't be very large. So (as with Dr. Jaffee's method) test pairs of small numbers to see which work. I solved my equation for p in terms of q and r, then tabulated p for pairs of small q and r to see which gave integer values for p.

PROVING that there is only one solution (counting permutations of p, q, and r as the same) may be difficult, but as the problem seems to imply that there is only one solution, you can stop when you find one.

I shouldn't use a calculator for these problems.Ok so after simplifying as you did (Dr. Jaffee) I got p= (-r-q)/(1-qr)

Not using a guess and check approach (not really impressive...) is there anything else (you can think of) I can do?

Yes, I do. Starting with your equation p= (-r-q)/(1-qr), multiply the numerator and denominator of the fraction by -1 to get p = (r + q)/(qr - 1).

Now, you know that p will be an integer if the denominator of the fraction is 1. So, what must be the value of qr to make the denominator 1? Once you have established that, you can see that there are only two numbers that can be the values of q and r. Substitute them into the equation and you have the value of p.

As venkatesh wrote to Dr. Math[Question]Three tap together drain a complete tank in 5 hours. If first drains this tank in 10 hours alone, then minimum hours taken by second tap will be

[Difficulty]how to solve fast

[Thoughts]

Hi Venkatesh,

If the first tap can drain the tank all by itself in 10 hours, that means that it can drain 1/10 of the tank in 1 hour. So, you should be able to figure out, then, how much of the tank it can drain by itself in 5 hours and how much that leaves for the other two taps to drain.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more.

Age: 16Subject: time and work

solve the problem

Do you mean that you solved the problem? If so, I'm curious what answer you got. Or, do you mean that you want me to solve the problem? Actually, I thought I gave you enough of a suggestion that would enable you to solve it. What have you done so far? How can I help you finish it?

I'll be looking for your response.

i solved the question Let time taken by first, second and third tap are t1, t2, t3Then according to given condition 1/t1 + 1/t2 + 1/t3 = 1/5 (i) and 1/t1 = 1/10So 1/t2 + 1/t3 = 1/5 - 1/10 = 1/10for t3 = Infinite t2(min) = 10

Challenge Problems

As Gail wrote to Dr. Math[Question]I am 11 years old and I do not understand how to show 3/4 divided by 4/8 on a numberline.

[Difficulty]I do not understand how I can show 24/16 if I only go up to 4/8.

[Thoughts]

Hi Gail,

There are several ways to do it. One way uses the fact that 4/8 means the same thing as 4 x 1/8. So, if you locate 3/4 on the number line, you can divide it into 4 equal pieces, with each piece having a length of 3/16. Then, you need to divide 3/16 by 1/8.

Well, you know that if you want to divide a piece by 8, you have to cut it into 8 equal pieces, so if you want to divide by 1/8 you have to do the opposite; that is, create 8 pieces each 3/16 units long. You already have 4 pieces that are each 3/16 units long, so you just have to add on 4 more.

You can see that 3/16 + 3/16 + 3/16 + 3/16 + 3/16 + 3/16 + 316+ 3/16 = 24/16 which is what you said the answer should be.

A second way to do it is a little quicker. It uses the fact that 4/8 =1/2. So, dividing by 4/8 is really the same as dividing by 1/2, which means that you have to make the 3/4 twice as long. If you do that on the number line, you see that you end up at exactly the same place where 24/16 is.

Give it a try and if you want some clarification about this problem, write back and we'll discuss it some more. Let me know if my suggestions made any sense to you or if you would like me to try to explain it a different way.

As mrigang wrote to Dr. Math[Question]a pack of cards contain 1 to n cards. two consecutive numbered cards are removrd from the pack and the sum of remaining cards is 1224. if the smaller of the nos. removed is k then find out k-20

[Difficulty]i am not able to do this question

Hi Mrigang,

Here is how I solved the problem. You know that there are n cards in the deck. Suppose the value of the cards removed from the deck are k and k + 1.

You also know that the sum of the first n counting numbers is n(n + 1)/2.

So, write an equation expressing the fact that the sum of the first n counting numbers minus the sum of the values of the two cards removed is equal to 1224. This should be an equation with the variables n and k and also quadratic.

Treat the n as your variable and the k as a constant and use the quadratic formula to express n in terms of k. Then find the value of k that will make the number inside the radical sign a perfect square.

Once you do that, you can use the quadratic formula again to find the value of n.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more. If you get stuck make sure to show me any work you have done. That will make it easier to help you.

As Ram wrote to Dr. Math[Question]For the quartic equation x^4-18x^3+kx^2+200x-1984=0The product of two of its four roots equals -32.What is the value of coefficient k?

[Difficulty]I tried Vieta's formulas but get a whole bunch of equations. I got cd=62 and ab=-32.... but i dont know how to proceed forward

[Thoughts]

Hi Ram,

I don't know a quick way to solve this problem, but I did find a solution. Here is how I did it.

First, I started solving the problem the same way that you did by calling the four roots a,b,c, and d, then letting ab = -32 and cd = 62.

Now, if we write the equation in the form (x - a)(x - b)(x - c)(x - d) = 0 and multiply the first two factors and then multiply the last two factors, we get

(x^2 - (a + b)x + ab)(x^2 - (c + d)x + cd) = 0, which is equivalent to (x^2 - (a + b)x - 32)(x^2 - (c + d)x + 62) = 0

Next, multiply that out to express the coefficients of x^3, x^2, and x in terms of a,b,c, and d. Then set the coefficient of x^3 equal to -18 and the coefficient of x equal to 200.

If you do this properly, you should be able to multiply the first equation by 62 add it to the other equation and be able to calculate the value of c + d.

Once you know the value of c + d and you already know the value of cd, you can use the quadratic formula or completing the square to evaluate c and d. Warning: They are imaginary numbers.

If you know the value of c + d, it is easy to find the value of a + b and then the individual values of a and b.

Finally,substitute the values of a,b,c and d into the appropriate equation and calculate k.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more. If you have difficulties, it will be easier for me to help you if you show me everything that you tried and where you got stuck.

Thanks for writing to Dr. Math and Good Luck.

As Tushar wrote to Dr. Math[Question]Three girls held a series of swimming races among themselves. They decided to award a positive whole number of points for finishing first, second and third, where the number of points for first was more than the number for second, which was more than the number for third. The same number of points was awarded for the place in each race. There were no ties.

The first race was Breaststroke. Altogether Sarah accumulated 20 points, Michelle 10 points and Catherine 9 points. If Sarah did not win the backstroke race, who did win the backstroke?

[Difficulty]Everything! Please show answer with all working out! Thanks!

[Thoughts]

Hi Tushar,

I figured out an answer to the question, but rather than showing you all of the details and answering it for you, I'll explain the method I used, then give you a chance to figure it out for yourself.

First, I calculated the total number of points that were earned by adding 20 + 10 + 9. Since the total number of points for each race was the same, I divided the sum of all of the points that I just got by 3. Now I had the total number of points for each individual race.

So, all I had to do then was find three numbers which added up to that result such that I could make combinations of them to get 20, 10, and 9.

Once I had that, I could see that I could assign a first, second and third place for each race and the answer was evident.

Give it a try and if you want to check your answer with me or if you want some clarification about this problem, write back and we'll discuss it some more. If you are really stuck show me what work you have done so far and that will make it easier for me to help you.