All posts by mathconfidence

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%

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.
Please comment/share ideas on what has been helpful

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!)

Khan Academy and their Get Ready for Grade Level for all grade levels

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.