# 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

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.

Scenario 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%

# July 2020 Brain Puzzler Solution

A: 333

The clocks in the first row represent 9 + 9 + 3
The calculators in the second row are each 10 (also 1234 added together is 10)
Each lightbulb in the third row is 15 (15 + 15 – 15) = 15.
Which means each ray above the lightbulb =3
So in the last row:
9 + 9 x (12 + 12 + 12)
9 + 9(36)
9 + 324
333

10/31/20 update from Colin Porteus — thanks for your sharp eyes!!

Hello Robin, can you check the last row on this one please? I get 9 + 9 x (12 + 16+ 16)?? The bottom two globes have 4 threads each x the 4 light marks makes them 16 each?? So consequently I get a result of 405.
Thank you

# Homeschooling Math Ideas Part 1

By sheer coincidence, my very first tutoring family were homeschoolers back in 2000 (www.homeschoolnyc.com).    Whatever this Fall may bring, students may be learning from home at least part of the time.  I will be sharing online resources  biweekly.

Estimation 180 has photos for estimation: “tasks that make mathematical reasoning accessible to students and enjoyable.”

Visual Patterns asks visitors to “Click on a pattern to see a larger image and the answer to step 43. What is the equation?”

New York Times What’s Going on in This Graph? is a colorful ‘real-world’ Math resource sure to spark discussion (and even debate!)

Delta Math also has online practice although mostly for middle and high schoolers

# 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