How Many Pokémon do you need to Catch ‘Em All in Pokémon Go?

The easy answer is 151. Obviously.

Let me rephrase the question. What’s the maximum number of Pokémon you need to catch in Pokémon Go to catch them all?

Given how rare some Pokémon are and how candies allow Pokémon to evolve I wanted to investigate the worst case scenario. Let’s say you never caught a Pokémon that’s in an evolved form already. Let’s say you never hatched any eggs. And also, let’s say that restrictions such as continent specific Pokémon or some being unreleased isn’t an obstacle for you.

If you caught only unevolved forms and worked your way up from there just how many would you need to catch them all? Also, how many candies would it take?

1. Completing the Pokedex.

I’ll quickly review the math I used to figure this out.

There are 25 Pokémon that don’t evolve at all. So our count starts at 25 and you would get 3 candies each for each one caught.

Next there are 40 Pokémon that only have one evolution and 37 when you exclude Eevee (we’ll get to that in one second). You need 50 candies to evolve any of these Pokémon. To evolve one you would need to catch 13 of them, receiving 39 candies and another 11 candies from trading that many in. This would take 481 Pokémon and 1,850 candies.

Now for Eevee. Eevee has the 3 evolutions which only require 25 candies each to evolve to. Better yet, all of the evolutions all come from Eevee candy. So, when you evolve your first one, then you actually have 2 candies leftover that can be applied to the next one (capture 7 and trade in 6 for a total of 27 candies). You would need to capture 20 Eevees and use 79 candies.

Lastly, there are 16 Pokémon that evolve twice and require 125 candies total to make that happen. You would need to capture 32 then trade 29 for the total of 125 candies. This adds up to 512 Pokémon necessary and 2,000 candies.

Edit: Realized after first posting that there are more special cases for Magikarp-Gyarados, Pidgey-Pidgeot, Weedle-Beedrill, and Caterpie-Butterfree. These adjustments are factored in below.

Our totals then are 1,072 Pokémon needed to catch ’em all. They would use up 4,193 candies.

This is definitely the hardest path to take, which no one would recommend as its the absolute maximum requirements. This would complete your Pokedex and as proud as the professor would be there is still a little bit more to do if you’re truly a diehard fan.

2. Capturing Every Pokémon.

Since you would need to evolve Pokémon to fill the Pokedex you would lose their unevolved forms. If you wanted to own all 151 entirely and completely then you would need some extra ones.

That means one extra Eevee. You also wouldn’t need to catch any more Pokémon with a single evolution, because we caught 13, but only traded in 11. One of the remaining two of them stays unevolved and the other evolves.

For the double evolutions we traded in 29 of the 32 captured, leaving you with 3 remaining. One of them is evolved to the highest form, one stays in the lowest form, and the other would need to evolve just once. This would take 25 candies and another 7 Pokémon captured for each.

All in all, our new totals for being a Pokémon Master.

Pokémon: 1,177

Candies: 4,635

3. Time Factor

Going this far with the data there was one more thing I wanted to find out – how long would it take to catch them all. Given a quick estimate of 30 seconds to catch each Pokémon and requiring 1,177 Pokémon total it would take at least 9.8 hours. This is just for capturing them, it doesn’t take into account how long you need to be out there searching for the rare ones.

Let me know what you think! Also, let me know if there’s any errors in the math or special cases like Eevee that I need to take into account.



3D Nonagram Puzzle 9x9x9

My first time attempting to solve a nonagram was part of the British Spy Christmas Puzzle. The first challenge was to complete a nonagram which revealed a QR code that led to further questions.

British Spy Christmas Puzzle

A nonagram is basically a grid puzzle, where the numbers along each column and row reveal what cells are shaded. A sequence of 3 – 2 – 3 – 1, means that a set of 3 cells are shaded, then 2, then 3, and finally 1. What the numbers do not reveal is how many unshaded cells are in between the shaded ones.

Recently, I advanced this style of puzzle into the third dimension. The puzzle below is a 9x9x9 cube, comprised of rows, columns, and diagonal lines. Each of the 9 squares are slices that have been cut from the cube. Follow the instructions on the document to try to solve it.

3D Nonagram Puzzle – 9x9x9

Feel free to message me for the solution:


Good luck!

The Google Weighing Problem Extension

When Google is interviewing potential employees they are given a series of seemingly outrageous questions to test their problem solving ability. How many piano tuners are in the world? How would you design an evacuation plan for San Francisco? Why are manhole covers round?

The full list of revealed questions can be found on this website:

A functional knowledge of mathematics is needed to know how to approach these problems, and some are rooted in famous math historical questions. The piano tuners question is a Fermi Problem, named after Enrico Fermi, who posed many estimation problems with little information and had startlingly accurate results. (I plan to simulate a Fermi Problem in a future post). On another note, I once attended a seminar at a math conference where the guest speaker spoke entirely about the manhole problem and its extensions.

The problem I would like to look at today from Google is the following:

  • You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you find the ball that is heavier by using a balance and only two weighings?

I highly encourage to spend some time attempting to solve it before reading onward – its challenging, but can be done with some thought.

Here’s how I went about solving it:

  1. Start with weighing 6 of the balls – 3 on each side of the scale.
  2. If the scale is perfectly balanced, then weigh the remaining 2 and find the one that is heavier.
  3. If the scale was not balanced, then take the 3 balls from the heavier side.
  4. Weigh 2 of them.
  5. Either the heavier one will be identified, or if they are balanced then the remaining one is heaviest

I kept thinking about this problem for another week and wanted to investigate further. What is the maximum number of balls where a heavier one could be identified in two weighings? What about three weighings?

My approach was to solve for a certain number of cases and see if a pattern would appear that would predict the number of balls versus weighings needed.

Some cases were trivial, such as with 1, 2, or 3 balls. At 4 for the first time you would need to do a second weighing. 5, 6, 7, 8, and 9 followed suit. To describe it, with 9 balls your first weighing would be between two sets of 3 balls. The outcome (the scale revealing the heavier set or balancing to reveal it was the unmeasured ones) would whittle down the possibilities to 3 balls. The second weighing would determine which one it was.

At 10 balls you finally need a third weighing, and the reason why ties into the groupings. I realized that a key component to this problem is that the process of weighing separates the balls into three groups. Thus, the number of weighings increases each time the number of balls surpasses a power of 3. Below is a table that shows select cases that demonstrate this transition.

# of Balls # of Weighings
3 1
4 2
8 2
9 2
10 3
26 3
27 3
28 4

Using this information I was able to find an equation to help illustrate this. By inputting the number of balls, b, you can find the maximum number of weighings, w, required to single out a heavier one.

Untitled picture

And using it, I could find out the number of weighings for ridiculous numbers of balls.

# of Balls # of Weighings
100 5
12345 9
999999 13
1000000 13

The Greatest Tile on 2048

For the past few weeks I have been completely addicted to the game of 2048. 2048 has a simple design and concept, but it has had amazing popularity, last week being #1 free app on Apple Appstore. The game involves sliding tiles together, starting with 2s. Each time two identical tiles collide then they join together to form a new one that is twice the value of the previous. So, two 2s slide together to form a 4. Two 4s make an 8…so on and so forth.

This type of doubling is exponential growth and can be modeled by the formula f(x) = 2x.

Exponential growth tends to explode rapidly out of control. The main examples used to describe it is a population of multiplying bunnies, or having your allowance doubled each day for a month starting at a penny (you become an instant millionaire!) Suddenly the 2s and 4s tiles are sliding together to form 8s, 16s, 32s, 64s, 128s, 256s, 512s, 1024s, and 2048s. Experienced players know that the game is not over at 2048, despite being the name of the game. I’ve gone as far as 4096 after an unhealthy and concerning amount of time trying.

Is there a limit to how high the tiles can get? The answer is yes. Given the allotted space on the 4×4 grid the tiles are eventually constricted from doubling any higher. The largest possible tile is 131,072 and is achieved by creating a long string of progressively smaller numbers and randomly receiving a 4 for the final one ( I wonder what color that one would be? ) Take a look at this link to understand just how much work would go into achieving this score :

131,072 is 217. 17 is not a “random” number, it is derived from the fact that there are 16 spaces on the board. Technically, this should mean the answer is 216, but this is not always the case because the game is programmed so that a 2 or 4 could appear in the final cell. Since 4 is 2 squared, it raises the maximum from 216 to 217.

From this we can determine what the maximum tile would be for larger game boards, say instead of a 4×4 that it was 5X5. A 5X5 board would have the largest tile be 226 or 67,108,864, again depending on the last one being a 4 and not a 2. Looking in the other direct, a 2×2 board can only have a maximum tile of 25 or 32. This depends on the last tile being randomly generated as a 4, but it tends to be a 2 so personally I have never made it past 16. This game can be played here:

Since 2048 is 2 to the eleventh power, a more politically correct game would only have 10 tiles (perhaps arranged as a 2×5 rectangle). An incredible number of variations to 2048 exist with different board sizes, styles, numbers, etc. For a look at some of the many variations of 2048 and the classic game itself go to this link:

A Palindromic Week

A palindrome is a group of numbers or letters that can be written the same way backwards. For example, the word radar is a palindrome, because reversed it is still radar. Every so often the date is a palindrome. One of the best examples was on November 11, 2011 (11/11/11), but the 22nd of that month would have been a palindrome too (11/22/11).

This entire week of 4/13 – 4/19 has an amazing quality: that every single day is a palindrome giving us a palindromic week. Take a look:

4-13-14 – 41314

4-14-14 – 41414

4-15-14 – 41514

4-16-14 – 41614

4-17-14 – 41714

4-18-14 – 41814

4-19-14 – 41914                                                  

                The following day, April 20th, breaks this pattern (42014). The only other palindrome days we will have this year are 4/10, 4/11, and 4/12. This decade is full of palindromic weeks – 2012, 2013, 2014, 2015, 2016, and 2018.

                Granted, these are not entirely true palindrome days, because the year is abbreviated. So far in the 21st century we have had 7 days that were entirely palindromes.

October 2, 2001 – 10022001

January 2, 2010- 01022010

January 10, 2011 – 11022011

November 2, 2011 – 11022011

February 10, 2012 – 2102012

March 10, 2013 – 3102013

April 10, 2014 – 4102014

                Palindrome days are occurring much more frequently lately, because we are in the early part of the 21st century. After the year 2040 these dates become much rarer (in fact there are only 6 entire palindrome days after 2039), because the years do not “flip” as easily. While palindromic days will still happen, having an entire week won’t be possible.


For more cool calendar dates check out this link:

The Abbott and Costello Math Method

Famous comedic duo Abbott and Costello have a routine in which Costello performs clearly incorrect math, but somehow continues to arrive at the same answer. Abbott is baffled to see how Costello’s outrageous claim proves to be true time and time again. Before going any further take a look at their skits, the first showing how 25 divided by 5 is 14, and the second how 7 x 13 = 28.

Costello shows that 25 divided by 5 is 14 by addition, multiplication, and by division. However, he performs each of these operations incorrectly, because of how he handles the ones and tens places. When adding the five 14s together, he counts each number individually: 1 + 4 + 1 + 4 + ….. instead of using their place values ( 10 + 4 + 10 + 4…). When multiplying Costello similarly does 14 x 5 as 4 x 5 added with 1 x 5. The correct way is 4 x 5 added to 10 x 5. The same holds for division, where his method disregards place values.

What is astounding about this incorrectness is that he always arrives back at the same conclusion that 5 x 14 = 25, by those three operations. However, he makes the same mistake in each operation, which is the reason why it always works out the same way. In the second skit, Costello uses the same flawed methodology to show that 7 x 13 = 28. This led me to wonder if these select examples were special and unique, if other numbers would have a similar outcome, or if all numbers would behave this way.

After some experimenting with 16 x 6 (which equals 42 by the Costello Method) among other number combinations, the results continued to hold true. The numbers selected for the skit are more impressive than some that could be chosen. Evidence shows that if you commit to the Costello Method for addition, multiplication, and division then you can usually manipulate your way to the same answer each time. Two exceptions would be if there is a 0 in the ones place, or if you are multiplying by 1.

Example: Consider 3 x 20. By ignoring place value this adds as 2 + 0 + 2 + 0 + 2 + 0 = 6, which is so wrong that it is just goofy. It would take clever misdirection to even try to pull off that 6 divided by 3 is 20.

Example: 1 x 14 by the Costello Method = 5. The addition step for 1 x 14 is adding 14 to nothing, and I would love to see Costello try to explain that one.

There are certainly more exceptions that could be considered, such as if both numbers were each one digit, if it was being multiplied by zero, etc.

So – why does this happen?

Consider the case of 29 x 3.

29 x 3 would be 2 x 3 + 9 x 3 = 33. By adding this it would be 2 + 2 + 2 + 9 + 9 + 9 = 33. If I rewrite the addition step in this way 3(2) + 3(9) then it becomes clear how the mistake in the multiplication example leads you astray in both cases. Again, you are multiplying 3 to both digits, regardless of their place value.

As a teacher, these videos are among my favorites for the shock value and then the discussion about where Abbott and Costello made their mistakes.