The expression reads:

(10 + 11) (10 – 11) = 100 + 1!

The challenge is to make the above equation true by either coloring four blue dots or erasing four blue dots.

Hint: the title reads “Keep it 💯” which means to keep it real and be honest.

Watch the solution in the video.

**“Keep It 100” Puzzle: Change 4 Dots To Make This True!**

Thanks to Professor Peer Johannsen at Pforzheim University for creating this puzzle!

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**Answer To Keep It 100 Puzzle: Change Four Dots To Make True!**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

This solution involves the imaginary number i. The answer is to change each 11 into a 1i, which means to erase four dots in total (turning a 1 into an i requires erasing two dots, and we need to do this two times).

The left-hand side then becomes:

(10 + 1i)(10 – 1i) = 100 + 10i – 10i + i^{2} = 100 – i^{2} = 100 – (-1) = 101 = 100 + 1!

So this exactly equals the right hand side.

While the solution is complex, the answer is quite real! (At MindYourDecisions we always keep it 100).

Are there any other solutions?

Thanks to Professor Peer Johannsen at Pforzheim University for creating this puzzle!

]]>In a class of *p* students, the average (arithmetic mean) of the test scores is 70.

In another class of *n* students, the average of the scores for the same test is 92.

When the scores of the two classes are combined, the average of the test scores is 86.

What is the value of *p*/*n*?

Watch the video for a solution.

**The Meanest Test Question Ever?**

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**Answer To The Meanest SAT Test Question Ever**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Suppose *c*_{1} is the sum of scores for the class with *p* students, and *c*_{2} is the sum of scores for the class with *n* students.

**Method 1**

We can work this using the definition of the arithmetic mean. We are given:

*c*_{1}/*p* = 70

*c*_{2}/*n* = 92

(*c*_{1} + *c*_{2})/(*p* + *n*) = 86

We can re-write the equations as follows:

*c*_{1} = 70*p*

*c*_{2} = 92*n*

*c*_{1} + *c*_{2} = 86(*p* + *n*)

We can add the first two equations to get:

*c*_{1} + *c*_{2} = 70*p* + 92*n*

Now we equate this to the third equation to get:

70*p* + 92*n* = 86(*p* + *n*)

16*p* = 6*n*

*p*/*n* = 6/16

And that’s the answer.

But the above equations actually suggest a much quicker way to solve this problem.

**Method 2**

At the start, we know class 1 has an average of 70, which is 86 – 70 = 16 units below the average. So class 1 with *p* students creates a deficit of 16*p* points below the average.

By contrast, we know class 2 has an average of 92, which is 92 – 86 = 6 units above the average. So class 2 with *n* students creates a surplus of 6*n* points above the average.

Since the two classes average out to 86, the deficit has to equal the surplus, meaning:

16*p* = 6*n*

This is the very equation we found in the first method, but we derived it much faster! So we then arrive at the answer *p*/*n* = 6/16 again.

**Source**

Difficult SAT exam questions posted on Quora

https://www.quora.com/What-are-some-of-the-most-difficult-SAT-math-problems

I read this problem even tricked Richard Feynman at first–even the greatest minds can make mistakes!

Can you figure it out? Watch the video for a solution.

**A Puzzle Richard Feynman Missed? Can You Solve It?**

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**Answer To A Puzzle Richard Feynman Missed?**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

There’s a very quick answer to this problem. But first let’s investigate how someone might not solve it. Suppose the circle’s radius is *r*, which means the rectangle would have sides of *r* – *a* and *r* – *c*.

We can then use the Pythagorean Theorem with the hypotenuse of *b* to get:

(*r* – *a*)^{2} + (*r* – *c*)^{2} = *b*^{2}

This can be written as a quadratic equation in *r*:

*r*^{2}(2) + *r*(-2*a* – 2*c*) + (*a*^{2} + *c*^{2} – *b*^{2}) = 0

We can then use the quadratic formula, and then take the positive value of the square root, to get a formula for the radius:

*r* = (*a* + *c* + √(2*b*^{2} + 2*ac* – *a*^{2} – *c*^{2}))/2

This formula will work, but it is quite unnecessary! There is a much simpler formula for the radius of the circle: it is equal to the variable *b*:

*r* = *b*

Why is this? Draw the other diagonal of the rectangle, and notice it is a radius of the circle. Then, since the diagonals of a rectangle are equal to each other, we can conclude the radius of the circle equals *b*. The answer is remarkably simple!

Did you figure it out?

**Source**

Walter Bender’s week 11 puzzles

http://web.media.mit.edu/~walter/MAS-A12/week11.html

Inside rectangle ABCD, there are right triangles AED and BFC as shown in the picture below. Note that point F is located along DE.

If AE = 7, ED = 24, and BF = 15, what is the length of AB?

The original problem was multiple choice, but I am omitting the choices to increase the difficulty. Can you figure it out?

I also learned the exam has 60 problems to solve in 2 hours (an average of 2 minutes/problem). So can you figure it out that quickly? I sure could not!

Watch the video for a solution.

**The Rectangle Side Length Puzzle – Think Inside The Box!**

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**Answer To The Rectangle Side Length Puzzle**

I will mention that this problem was challenging to me as my approaches kept falling short. I first considered finding the altitudes of the triangles, but then I noticed there was overlap between them which was hard to solve. I then considered extending all the other sides of the triangles to a side of the rectangle, but none of those worked. I then searched for similar problems online but did not find any. So I was about to give up and ask for help in a forum online.

But just as I did that, the solution method came to me!

**Solution**

We can calculate AD as the hypotenuse of a triangle with legs of 7 and 24 to get

AD = √(7^{2} + 24^{2}) = 25

(Or we could have remembered the 7-24-25 primitive Pythagorean triple.)

Then side BC is equal to AD, as opposites sides of a rectangle have equal length. So we can solve for CF as the leg of a triangle with a leg of 15 and a hypotenuse of 25.

BC = √(25^{2} – 15^{2}) = 20

(Or we could have noticed BFC is a 5x scaled version of a 3-4-5 primitive Pythagorean triple.)

Next we will draw altitude GF in triangle BFC. We can calculate its length using similar triangles:

GF/20 = 15/25

GF = 12

Then we can calculate CG either using the Pythagorean Theorem, or by noticing GFC is a 4x scaled copy of a 3-4-5 triangle, so then CG = 16, and then BG = 25 – 16 = 9.

Now how to proceed? This is where I needed the insight to “think inside the box.” We should extend GF into an FH in the triangle AED, and FH will be perpendicular to AD.

This creates similar triangles DHF and DEA, from which we get:

HF/16 = 7/24

HF = 14/3

So now we’ve basically solved it!

HF = 14/3 + 12 = AB = 50/3

So that’s the answer!

I do get emails from people who think I can easily solve any problem. That is not the case at all! Even simple problems can give me a challenge. But the lesson is to be persistent and practice to improve your problem-solving ability.

**Thanks to all patrons! Special thanks to:**

Shrihari Puranik

Yildiz Kabaran

Kyle

Share the beauty of math with the world! You can support these videos and posts at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

]]>ABC is a three digit number where A, B, C are non-zero digits.

If ABC = A! + B! + C!, then what is the number ABC?

Can you figure it out? Watch the video for a solution.

**Solve ABC = A! + B! + C! The Factorial Digits Sum Puzzle!**

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**Answer To The Three Digit Factorial Sum Number**

We can start by listing the factorials of the digits 1 to 7.

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

7! = 5040

Since 7! is a four digit number, and ABC is a three digit number, we cannot have 7 as a digit. Furthermore, we can exclude 8 and 9 because 8! and 9! are even larger values.

Could we have 6 as a digit? If we had 6 as a digit, the result would be a number larger than 700, requiring a digit of 7, 8, or 9. But we just excluded 7, 8, and 9, so we also must exclude 6 as a digit.

If ABC only had digits from 1 to 4, the largest possible sum is 4! + 4! + 4! = 72. But this is not a three digit number. Thus, we must have some digit equal to 5.

Since 5! + 5! + 5! = 360 ≠ 555, we can have at most two digits of 5. We can then test the values of 1, 2, 3, and 4 as the third digit, and we can see none of these produce a solution:

5! + 5! + 1! = 241

5! + 5! + 2! = 242

5! + 5! + 3! = 246

5! + 5! + 4! = 264

So we can only have a single digit equal to 4, meaning the largest value is 5! + 4! + 4! = 168. Thus we must have A = 1 (if there is a solution). Now we consider the cases 15B and 1B5. We can check B = 1, 2, 3, 4 for possible solutions.

1! + 5! + 1! = 122

1! + 5! + 2! = 123

1! + 5! + 3! = 127

1! + 5! + 4! = 145

None of these work for 15B, but the very last equation does work for 1B5. So we have the only solution:

145 = 1! + 4! + 5!

I learned from Nestor Abad’s YouTube comment such numbers–where the number equals the sum of the factorial of its digits–are called factorions. There are just a handful in decimal numbers: Factorion

**Sites consulted**

Math StackExchange

https://math.stackexchange.com/questions/2582253/what-three-digit-integer-is-equal-to-the-sum-of-the-factorials-of-its-digits