Category Archives: brain puzzler

July 2017 Brain Teaser Solution

Q: A trader had gold coins but did not tell anyone how many she had.  If the coins are divided into two different sized groups, then 32 times the difference between the two numbers is equal to the difference between the squares of the two numbers.  How many gold coins did she have?

A: The merchant has 32 gold coins.
It is easy to check this… Let’s divide the 32 coins into two unequal numbers, say, 27 and 5. Then, 32 (27 – 5) = (27 x 27) – (5 x 5).
We can also check this by dividing the 32 coins into 30 and 2.
Then, 32(30-2) = (30 x 30) – (2 x 2).

This will work for any two numbers that add to 32.  If we call the two numbers x and y:
32 (x – y) = x^2 – y^2
So x + y = 32 and therefore y = 32 – x
Then we can rewrite the above as:
32(x – (32 – x)) = x^2 – (32 – x)^2
32(x – 32 + x) = x^2 – (1024 – 64 x + x^2)
32(2x – 32) = x^2 – 1024 + 64x – x^2  (the x^2s cancel)
64x – 1024 = 64x – 1024  🙂

 

 

June 2017 Brain Puzzler Solution

Q: Adam, Brian, Cathy, Doug and Eve are at a round table for five when they realize that they are in age order around the table. What are the odds of that?”

A: The odds are 11:1 (against) and the probability is 1/12.

If we imagine that they sat down in age order randomly:
The first person is random.

The second person can sit to the first person’s right or left so there are two slots and 4 people left. (2/4)
The third oldest now has 3 chairs to choose from = 1/3
This leaves two seats for the 4th person to choose from (1/2)
The 5th person has no choice as there is only 1 seat left.
Therefore the probability is:
(1) (2/4)(1/3)(1/2)(1) = 1/12

May 2017 Brain Puzzler Solution

Q: How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

A: 45

With 1 as a middle digit, there are two three digit numbers whose middle digit is the average of the first and last digits:  111 and 210

There are 4 numbers with 2 as a middle digit: 123, 220, 321, 420
There are 6 numbers with 3 as a middle digit: 135, 234, 333, 432, 531, 630
There are 8 numbers with 4 as a middle digit: 147, 246, 345, 444, 543, 642 741, 840
There are 9 numbers with 5 as a middle digit: 159, 258, 357, 456, 555, 654, 753 852, 951
There are 7 numbers with 6 as a middle digit: 369, 468, 567, 666, 765, 864, 963
There are 5 numbers with 7 as a middle digit: 579, 678, 777, 876, 975
There are 3 numbers with 8 as a middle digit: 789 888, 987
There is 1 number with 9 as a middle digit: 999

Total: 2 + 4 + 6 + 8 + 9 + 7 + 5 + 3 + 1 = 45

Another way to solve this problem is to note that the middle digit is half the sum of the first and third digit so the sum must be even to result in a whole number when divided by 2.  To get two numbers to add to an even number they must be both odd or both even. Notice this in the list of 3 digit numbers above!

If both the first digit and the last digit are odd, then 1, 3, 5, 7, or 9 are choices for each of these digits, and there are $5\cdot5=25$ numbers in this case.
If both the first and last digits are even, then 2, 4, 6, 8 are choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are $4\cdot5=20$ numbers here.  25 + 20 = 45

 

 

April and May 2017 Brain Puzzler Solution

Link to the May 2017 Brain Puzzler Solution:
April’s Brain Puzzler
What is the area of the shaded region of the given 8 x 5 rectangle?

[asy]  size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));  label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90));  label("$1$",(8,1/2),dir(0)); label("$4$",(8,3),dir(0));  label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270));  label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180));  [/asy]

 

The area is 6.5 square units.  
Using extra lines to break the shape into triangles is a great way to solve this problem.

https://artofproblemsolving.com/wiki/index.php/2016_AMC_10A_Problems/Problem_11

March 2017 Brain Puzzler Solution

Victoria refuses to sit next to either William or Xavier. Yasmin refuses to sit next to Zack. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?

Let Victoria be V, William be W, Xavier be X, Yasmin be Y and Zack be Z.   
Scenario 1: V sits on an end seat.

Then, since W and X can’t sit next to V, that must mean either Y or Z sits next to V.
After picking either Y or Z, then either W or X must sit next to Y/Z.
Then, the last two people can be arranged in two ways.
Since there are two different end seats that V can sit in, there are a total of .

 Scenario 2: V does not sit in an end  seat.

The are 3 seats for V to choose from
In this case, then only two people that can sit next to V are Y and Z, and they
can be in either order and there are two ways to arrange W and X
 
So the total is 28

Adapted from http://artofproblemsolving.com/wiki/index.php/2017_AMC_10A_Problems/Problem_19