Category Archives: brain puzzler

August 2017 Brain Teaser Solution

August 29, 2017 mathconfidence Leave a comment

Q: With one straight cut a cake can be sliced into two pieces. A second cut that crosses the first one will make four pieces, and a third cut can produce as many as seven pieces. What is the largest number of pieces that you can get with six straight cuts?

A: 22
With 0 cuts, you get 1 big cake
With 1 cut, 2 pieces
With 2 cuts, 4 pieces
For the 3rd cut, if you cut cleverly you can get 7 pieces

click here for more info and nice graphics thanks to the Guardian
http://bit.ly/CuttingthePieGardner

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July 2017 Brain Teaser Solution

July 31, 2017 mathconfidence Leave a comment

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 :)

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June 2017 Brain Puzzler Solution

June 27, 2017 mathconfidence 1 Comment

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

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May 2017 Brain Puzzler Solution

April 26, 2017 mathconfidence 1 Comment

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