What a kid learning math can look like – percents

My son was confused today by a percent problem from Art of Problem Solving’s Algebra book. I liked the problem a lot and thought talking through it would be really helpful for him.

Here’s the introduction to the problem plus some initial steps – he is really confused about how to get going with the problem. His initial thought is to combine equal amounts. That gets us going, but he struggles to see where to go after that:


At the end of the last video we were still a little stuck. He knows that we need more than 1 cup of the mixture with more fat and less than one cup of the mixture with less fat. Our first try here is a new split – 1.5 cups and 0.5 cups. This mixture turns out to have too much fat, but it does help us understand the problem a bit better.


To finish up the conversation today we summarized what we’d done so far. That gave him the idea to try a split of 1.25 and 0.75. This split got us to the right answer. We’ll revisit this problem again to see how algebra can help us solve it.


The distribution of the last digits of consecutive twin primes is strange

I don’t quite know what I seeing with distribution of the last digits of consecutive twin primes. The data is in the last tab on the right, “Improved Twin Prime Sheet”, here:

My google doc with twin prime counts – see the “Improved Twin Prime Sheet” tab

Walking through that sheet:

The sheet has data about the last digits of consecutive twin primes. There are 9 possible sets of last digits from { (1,3) (1,3) }, { (1,3) (7,9) }, . . . { (9,1) (9,1)}.

(1) The top part in blue is just background data. I’m looking at twin primes in sets of 1 billion primes. So, in column E, for example, between the 2nd and 3rd billionth primes, there are 165.7 million twin primes, the 3 billionth prime is 71.8 billion, and most of the rest is just some log calculations. The last line tells you how many twin primes you’d see for each pair of consecutive digits if all 9 pairs were equally distributed.

(2) The next part in yellow tells you how many consecutive twin primes with each particular last digit pair happened in that set of 1 billion primes.

(3) The next part in green is the cumulative data.

(4) The next chart down in yellow is where things get interesting. This chart looks at the difference between the actual distribution and the equal distribution. I wanted to compare the numbers here to the Ln(Ln(x)) / Ln(x)^2 term because this term is the important term in the recent prime distribution paper by Robert Lemke Oliver and Kannan Soundararajan.

Evelyn Lamb’s article about the new result about last digits of prime number

I computed the value of the LN(LN(X) / LN(x)^2 term on line 13 of the spreadsheet.

(5) The next chart in red shows the difference between the LN(LN(X) / LN(x)^2 term and the chart above. You can see that this first term in the approximation nearly gets the distribution of last digits of consecutive twin primes following a (9,1) twin prime perfectly.

However, the error terms for twin primes with last digits of (3,1) and (7,9) are much larger. These larger terms have left me scratching my head because some of these error terms appear to be increasing at the same rate as the count of twin primes. That growth would suggest that last digits of consecutive twin primes have a different behavior than last digits of consecutive primes.

For example, it would suggest that as you head down the number line to infinity the ratio of the number of consecutive twin primes with last digits { (1,3) (1,3) } to the number of consecutive twin primes with last digits ( {9,1} {9,1}) is less than 1. That seems odd to me.

Unfortunately my little twin prime counting program is really slow and I can only count 1 billion per day. I’m very interested to see what the counts and the error terms look like as the program counts higher and higher.