Category Archives: brain teaser

September 2020 Brain Puzzler Solution

abcd image002

a has to be either 1 or 2 as multiplying by 4 keeps the number 4 digits.
(if a= 3 then quadrupling that number would be >12000 (5 digits)  )

if a was 1 then d would have to be 4 so our number abcd would be:
1bc4 and its reverse would be 4cb1 which cannot be as an integer multiplied by 4 has to be even.

Therefore a has to be 2.
Making d = 8.

When multiplying 2bc4 x 4  to be 8cb2, b and c cannot be too large as multiplying by 4 would then make the thousands place >8.

For example if b = 3 then even the lowest # in the 2300s,  2300 x 4 = 9200  would be too large as a = 9.

So b must be 0, 1 or 2
b cannot be 0 as 20cd x 4 = dc02 and multiples of 4 always with the last two digits as a multiple of 4.
b cannot = 2 since a is already 2 so b must = 1!

c cannot be 1, 2, or 8 as they have already been used

So abcd is now 21c8 x 4 = 8c12
If c = 0, 2108 x 4 does not end in 12
If c = 3, 2138 x 4 does not end in 12
If c = 4, 2148 x 4 does not end in 12
If c = 5, 2158 x 4 does not end in 12
If c = 6, 2168 x 4 does not end in 12
If c = 7, 2178 x 4 DOES end in 12
If c = 9, 2198 x 4 does not end in 12
c cannot be 1, 2, or 8 as they have already been used

2178 x 4 = 8712

 

 

August 2020 Brain Puzzler Solution

Question

If the probability of a false positive of a medical test (could be COVID-19) is 3%, what percent of people who test positive would actually have it?

Scenario 1: The disease is present in 1 person in 1000

Scenario 2: the disease is present in 1 person in 5


Answer

Scenario 1: 3.2% and Scenario 2: 89.3%

We need to compare the actual positives with the sum of the true positives and false positives.  In Scenario 1, if 1 in 1000 (.001) have it then 999 out of 1000 (.999) do not.

In Scenario 2, if 1 in 5 (.2) have it then 4 out of 5 (.8) do not.

 

Desmos False PositiveScenario 1:

comparing the true positives with the total positives  (true positive plus false positive)

approximately 3.2%.

Scenario 2:

comparing the true positives with the total positives (true positive plus false positive)

approximately 89%

June 2020 Brain Puzzler Solution

Q: What is the median of the following list of numbers: (There are 4040 numbers)

1, 2, 3…2020, 1^2, 2^2, 3^2…2020^2?

A: 1976.5

In order to put these into order from least to greatest, we have to figure out how many of the square(d) numbers are less than 2020:

 1^2 (1) – 10^2 (100) is 10 numbers
11^2 (121) – 20^2 (400) is 10 numbers
      21^2 (441) – 30^2 (900) is 10 numbers
        31^2 (961) – 40^2 (1600) is 10 numbers
        41^2 (1681) – 44^2 (1936) is 4 numbers

There are 4040 numbers altogether so we need to know the 2020th and 2021st numbers and get the mean of those two numbers to find the median.
Inserting the 44 perfect squares above into the first 2020 numbers would make 2020 the 2064th number so we need to subtract 43 and 44 from 2020 to find those two numbers (the 2020th and 2021st).
By the way, the nearest perfect square under 2020 is 1936 which (luckily!) is too far away for us to think about!

2020 – 43 = 1977 and 2020 – 44 = 1976

Therefore the median of the list of numbers above is 1976.5.

May 2020 Brain Puzzler Solution

Q: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

A:  112 numbers
There are four numbers that end with zero that satisfy this condition: 120 240 360 and 480.   Each of these numbers can only be rearranged four ways (as a leading zero is not allowed) making 16 combinations.

Each of the numbers below can be arranged into 6 distinct numbers.  For example, 213 can be 123 132 213 231 312 321.  Below are 16 numbers and each can be rearranged in 6 ways.  Therefore there are 96 three-digit numbers that satisfy this condition that do not have a zero in it.

The total is 96 + 16 = 112.

213

315

324

417

426

435

519

528

537

546

639

648

657

759

768

897