Earlier today I set you the following puzzles, adapted from Math Games with Bad Drawings by Ben Orlin.
1. Five nice dice
A roll of five dice produces the numbers 1, 3, 4, 5 and 6.
i) By combining the numbers using +, –, x and ÷, what is the closest you can get to the target of 55?
ii) What is the largest whole number you can reach?
iii) What is the largest whole number you can reach using only subtraction? (tip: it’s a positive number.)
iv) What is the largest whole number you can reach using only division?
Clarification: By combining the numbers I mean creating an expression using addition, subtraction, multiplication and division (and as many brackets as you like.) For example, using the five numbers above, you could combine them in this way: ((5x6)+ 4) ÷ (3 – 1), which is equal to 17. You can use as many, or as few, plus, minus, times and division signs as you like. You must use all five numbers exactly once.
Solutions
i) We can hit it exactly: (5 x 4 x 3) – 6 + 1 = 55
ii) 6 x 5 x 4 x (3 + 1) = 480
iii) 6 – (1 – 3 – 4 – 5) = 17
iv) 3/(((1/6)/5)/4) = 3*4*5*6 = 360
2. Five nicer dice
You roll five standard dice.
i) What is the probability the five numbers can be combined to reach 0?
Solution
100%. If any two dice are identical, multiply their difference (e.g., 5 – 5) by the sum of the other dice. If all five dice are different, then find a trio that form zero (1 + 2 – 3 or 5 + 1 – 6 or 4 + 2 – 6) and then multiply this by the sum of the other dice.
ii) What’s the probability that the maximum result is obtained simply by multiplying them all together?
Solution
The maximum can always be obtained this way as long as there are no 1’s present (for example (1+2) x 3 x 4 x 5 is larger than 1 x 2 x 3 x 4 x 5) so this is equivalent to rolling no 1’s. That has probability of (5/6)5 , or approximately 40%.
iii) What’s the probability that the maximum result is obtained uniquely by multiplying them all together?
Solution
There can be at most a single 2, because with multiple 2’s present, the maximum can be attained by using addition as well (for example (2+2) x 3x3x3 = 2x2x3x3x3). Thus, either there must be no 1’s or 2’s, which has probability (4/6)5, or no 1’s and a single 2, which has probability 5(1/6)(4/6)4. These sum to approximately 30%.
iv) What is the probability that the five random dice cannot be combined to reach any of the targets from 33 to 99?
Solution
One way this failure can occur is if all the dice are 1’s and 2’s. This has a probability of (2/6)5, or approximately 0.4 per cent.
The failure also occurs with the following combinations of dice. (In brackets I have listed the number of ways to rearrange that particular set of dice)
1 1 1 1 3 (5) , 1 1 1 1 4 (5) , 1 1 1 1 5 (5) , 1 1 1 1 6 (5) , 1 1 1 2 3 (20), 1 1 1 2 4 (20), 1 1 1 2 5 (20), 1 1 1 3 3 (10), 1 1 2 2 3 (30).
There are 120 total combinations, so total probability of all of these combinations is 120(1/6)5 , or approximately 1.5 per cent. Thus, the probability of being unable to form any of the numbers from 33 to 99 is just under 2 per cent.
To read more about Math Games with Bad Drawings, see my previous post.
I hope you enjoyed the puzzles. I’ll be back in two weeks.
Thanks to Ben Orlin. Math Games with Bad Drawings is out now.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
I’m the author of several books of puzzles, most recently the Language Lover’s Puzzle Book. I also give school talks about maths and puzzles (online and in person). If your school is interested please get in touch.
On Thursday 21 April I’ll be giving a puzzles workshop for Guardian Masterclasses. You can sign up here.
