How many quadratics factorise?

Write down a quadratic—any quadratic you like, but let’s say it should have integer coefficients between 0 and 20. What is the probability that it factorises?

What I really mean is will it factorise ‘over the integers’. So
\[x^2 + 5x + 6 = (x+2)(x+3)\]is in, but
\[x^2 + 2x + 2 = (x + [1-\mathrm{i}]) (x + [1 + \mathrm{i}]) \quad \text{and} \quad x^2 – 2 = (x-\sqrt{2})(x+\sqrt{2})\]are out.

To make it simpler, we will look for quadratics of the form
\[x^2 + bx + c\]where $b$ and $c$ are both positive. Try extending it to negative coefficients yourself afterwards!

Let’s plot a graph of $c$ against $b$, and colour in the values where $x^2+bx+c$ factorises. We’re going to colour these in with a 1×1 box where the bottom-left corner is at the relevant coordinate.

Quadratics of the form x² + bx + c which factorise, for b,c ≤ 20

Quadratics of the form $x^2 + bx + c$ which factorise, for $b,c \leq 20$.

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Review of Elastic Numbers

I spend a lot of time solving maths puzzles. Many of my favourites appear in Chalkdust and on my website. But there is a problem with spending so much time doing puzzles: its not very easy for me to find new and interesting puzzles any more.

I was therefore pleased to hear that Daniel Griller—author of the Puzzle Critic blog, a great source of less well-known puzzles including this gem—was releasing a book of original puzzles. Elastic Numbers (Amazon UK, US) is this book, and boasts 108 puzzles. These puzzles are sorted into four sections by difficulty: bronze (easiest), silver, gold and diamond (hardest).

I highly recommend the bronze and silver puzzles to teachers, who will find a collection of well posed questions they can give to students to make them think about common school topics. However, these puzzles don’t offer much challenge to the seasoned puzzler, and although many are neat they feel a little unspectacular.

But the slight disappointment I was feeling about the book immediately disappeared when I flicked forwards to the gold and diamond puzzles. These puzzles will make you immediately reach for the nearest pen and paper and getting solving. With so many good puzzles in these sections, its hard to pick favourites, but the following puzzle stood out (so it’s perhaps not surprising that this puzzle is the source of the title of the book):

Elastic numbers

Source: Elastic Numbers by Daniel Griller (obviously)
A two-digit number $ab$ ($a$ and $b$ are the two digits of the number; the number is not $a$ multiplied by $b$) is called elastic if:

  1. Neither $a$ nor $b$ is zero.
  2. The numbers $a0b$, $a00b$, $a000b$, … made by putting any number of zeros between $a$ and $b$ are all multiples of the original two-digit number $ab$.

Find three elastic numbers, and explain why they are elastic. 

As any mathematician will be able to spot, Elastic Numbers is typeset in $\mathrm{\LaTeX}$. I greatly approve of this and the pretty equations it gives (we use $\mathrm\LaTeX$ for Chalkdust too), although this leaves the book looking more like a puzzle dictionary than a fun puzzle book that you might give straight to the kids. But to puzzlers like me, this doesn’t matter: the best thing about a puzzle is the new and exciting mathematical situation it gives you to investigate. And this book is packed full of mathematical excitement. And on that note, I’m off to work out where Evariste is standing…

Where is Evariste?

Source: Elastic Numbers by Daniel Griller (obviously)
Evariste is standing in a rectangular formation, in which everyone is lined up in rows and columns. There are 175 people in all the rows in front of Evariste and 400 in the rows behind him. There are 312 in the columns to his left and 264 in the columns to his right.

In which row and column is Evariste standing? 


On the cover: dragon curves

Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper makes. If you folded the paper $n$ times, then the edge will make an order $n$ dragon curve, so called because it faintly resembles a dragon. Each of the curves shown on the cover of issue 05 of Chalkdust, and in the header box above, is an order 10 dragon curve.

Left: Folding a strip of paper in half four times leads to an order four dragon curve (after rounding the corners). Right: A level 10 dragon curve resembling a dragon

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Seven things you didn’t notice in Issue 04

With just a few days to go until we launch issue 05, we thought it’d be fun to share a few bits and pieces that we hid around issue 04. If this gets you excited for issue 05, why not come to the launch party on Tuesday?!


Since we published the horoscope in issue 03, scorpions have been running around all over Chalkdust HQ. Three of them managed to sneak into issue 04.
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The Chalkdust Christmas card

This post was part of the Chalkdust 2016 Advent Calendar.

Recently, some of you may have received a Chalkdust Christmas card. If not, it’s not because we hate you, it’s just that we couldn’t find your address… Unless we hate you, in which case it is because we hate you.

The card initially looks very boring: it is just a grid of squares with “Merry Christmas” written below it. Definitely NOT HOT… But there’s more. There’s a puzzle inside that leads you to add some colour to the squares to reveal a Christmassy picture.

Without giving any more away, here is the puzzle. If you’d like to give it to someone as a Christmas card (or just want to actually be able to colour it in), you can print and fold this lovely pdf.

Christmas Card 2016

The grid (click to enlarge)


  1. Solve the puzzles below.
  2. Convert the answers to base 3.
  3. Write the answers in the boxes on the front cover.
  4. Colour squares containing a 1 green. Colour squares containing a 2 red. Leave squares containing a 0 unshaded.


  1. The square number larger than 1 whose square root is equal to the sum of its digits.
  2. The smallest square number whose factors add up to a different square number.
  3. The largest number that cannot be written in the form $23n+17m$, where $n$ and $m$ are positive integers (or 0).
  4. Write down a three-digit number whose digits are decreasing. Write down the reverse of this number and find the difference. Add this difference to its reverse. What is the result?
  5. The number of numbers between 0 and 10,000,000 that do not contain the digits 0, 1, 2, 3, 4, 5 or 6.
  6. The lowest common multiple of 57 and 249.
  7. The sum of all the odd numbers between 0 and 66.
  8. One less than four times the 40th triangle number.
  9. The number of factors of the number 2756×312.
  10. In a book with 13,204 pages, what do the page numbers of the middle two pages add up to?
  11. The number of off-diagonal elements in a 27×27 matrix.
  12. The largest number, $k$, such that $27k/(27+k)$ is an integer.

Advent puzzles III

This post was part of the Chalkdust 2016 Advent Calendar.

Welcome to the twelfth day of the 2016 Chalkdust Advent Calendar. Today, we have another puzzle for you to enjoy, plus the answer to the puzzles from 06 December.

Today’s puzzle is taken from Daniel Griller‘s talk at the MathsJam conference earlier this year.

Odd factors

Pick a number. Call it $n$. Write down all the numbers from $n+1$ to $2n$ (inclusive). Under each of these, write its largest odd factor. What is the sum of these odd factors?

Now for the solutions to the puzzles from 06 December.

Digital sums

Source: Advent calendar, day 6
When you add up the digits of a number, the result is called the digital sum.

How many different digital sums do the numbers from 1 to 1091 have?

As this puzzle is part of a larger advent calendar (with prizes!), I’m not going to give you the answer here!


Source: nrich Secondary Advent calendar, day 10
You are given the numbers 1,2,3,4,5,6 and are allowed to erase one. If you erase 5, the mean of the remaining numbers will be 3.2. Is it possible to erase a number so that the mean of the remaining number is an integer?

If you are given the numbers 1,2,3,4,…,$N$, can you erase one number so that the mean of the remaining numbers is an integer?

For the first part, erasing 6 will leave numbers that sum to 15, with a mean of 3.

For the second part, if $N$ is even, erasing $N$ from the list 1,2,3,4,…,$N$ will leave numbers that sum to $\tfrac12N(N-1)$. $N$ is even, so $\tfrac12N$ is an integer; therefore $N-1$ is a factor of the sum, so the numbers have an integer mean.

If $N$ is odd, removing the middle number from the list leaves an integer mean. I’ll let you work out why this is and will return in a few days with the answer to today’s puzzle…