Exploding Dots

5.8 Solutions

As promised, here are my solutions to the question posed.


1. \(2783 \div 23 = 121\).

2. \(3900 \div 12 = 325\).We need some unexplosions along the way. (And can you see how I am getting efficient with me loop drawing?)

3. \(46632 \div 201 = 232\).


5. We have \(2789 \div 11 = 253\) with a remainder of \(6\). That is, \(2789 \div 11 = 253 + \dfrac{6}{11}\).

6. \(4366 \div 14 = 311 +\dfrac{12}{14}\).

7. \(5481 \div 131 = 41 +\dfrac{110}{131}\).

8. We see two groups of two at the hundreds level (all the dots in the blue loops would unexplode to make two-hundred blue sets of two at the ones level), one group of two at the tens level (all the dots in the green loop would unexplode to make ten green groups of two at the ones level), and three orange groups of two at the ones level. That makes for \(213\) groups of two.

9. In the picture for \(404 \div 3\) we see two leftover dots unaccounted for.

So \(404 \div 3\) equals \(134\) with a remainder of \(2\).

Note: We could regard this remainder as “two dots still to be divided by three,” and so write

\(404 \div 3 = 134 +\dfrac{2}{3}\).


10. We certainly see one group of five right away.

Let’s perform some unexplosions. (And let’s write numbers rather than draw lots of dots. Drawing dots gets tedious!)

We see \(61230 \div 5 = 12246\).


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