number puzzles

it may be helpful to have numbers (e.g. on card) to move around

there is an added dimension/extension of proving various statements, usually connected to the sum of the numbers used and those numbers that are in more than one line

Graeme Brown did some Excel versions of four of these puzzles (the first three and hollow triangle (i))
these can be found at nrich 5512

penguins on ice

life is now complete...

number braids

an article by Richard Bennett appeared in the June 1976 edition (075) of Mathematics Teaching, published by the ATM

he called them 'number plaits' rather than braids

the 'triangle totals' must be in smallest to largest order going from left to right

a variety of 'triangle totals' are (usually) possible

why is there no variety of 'triangle totals' for the 1 to 6 case?














and of course you could carry on...

6 totals occur due to the braids...

mobile inequalities

the idea for this task came from Ian Sugarman who has worked with Melvyn Rust on NumberGym software

see also Paul Salomon's work on imbalancing on Lost in Recursion

fiveways

use the digits 1 to 9, once only so that
all five lines (of three squares) have the same total

can you find the one way to make lines total 13?

there is also only one way to make lines total 17

there are four ways (ignoring rotations etc.)
to make the lines sum to 12 and 18

totals of 14 and 16 can also be made in 4 different ways (each)


you can fairly easily prove that totals of 11 or lower and 19 or higher are impossible

15 has four solutions as well (I think...) although it appears that there should be more...

although a sytematic approach can be helpful, it is fairly easy to find solutions by playing around

the task provides practice in adding digits repeatedly
and results can be collected as a class

what is probably not obvious to many students is the complementartity: once you have one solution you have another solution by using individual complements to 10 (which is why the sets of results for a particular total are symmetrical)

4 key positions can determine the rest of the grid - it might be convenient to have these as the four corners

the digits in the top right and bottom left hand corner positions are key: they need three ways to combine with different pairs of digits to make the target total

olympic rings

using all of the digits 1 to 9 can students arrange these in the nine spaces of the Olympic rings so that the totals in the circles are the same?

this task is made more interesting by there being at least three substantially different answers and you can prove that the lowest possible total for the circles is 11 and the highest is 15

you can also prove that the sum of the overlap numbers must be a multiple of 5 and that circle totals of 12 and 15 are impossible

eight odds

add 8 odd numbers to get a total of 20

six of the 11 solutions are shown

you might consider the numbers of 1s or the options for the high numbers, e.g. you can't use 19, 17 or 15

try to find the other solutions

[Moscow puzzles 49]

deletions

choose any number (circle it) and cross out the other numbers in the same row and column as this

then do this again

and do it a third time

then put a circle round the remaining number and add up the four numbers

do this a couple of times

what happens?

why does it happen?

[ idea from Mathematics in School, January 1989 ]

unlucky sum

using the digits 1 to 9, each used once only, can you make all (four) of the lines of 3 squares sum to 13?

this task can involve some thinking, after an initial exploration - helped by having digits to move around

the sum of 1 to 9 is 45 and four lines of 13 equal 52 so the additional 7 must come from those squares counted twice
there is only one way for the three digits to be placed and the rest follows


ignoring swapping the digits around at the ends there is one solution to this task

exploring other line totals is an obvious development:
there are five ways for line totals of 14 and 16
one way with a line total of 17
other line totals are impossible - easily proved for a line total of 15

daisy

this problem appeared in one of Brian Bolt's books*
it's promoted a keen interest in getting from 1 up to the highest number
but it's daunting to check repeatedly...

the task is to try to be able to make all numbers from 1 up to as high a number as you can

you put any six numbers in the regions (yes, there could be repeats)

you can only add up numbers in regions that are adjacent (have a common border)

you can only add in a number from a region once

for example, with a 1 in the centre:

a fairly low
highest number
but you can make all the numbers from 1 up to 25

note that 10 is not 4 + 6 because those regions are not adjacent










another example, if 18 is in the middle and 1 , 2 , 4 , 5 , 5 round the outside (in this order) then you can make from all numbers from 1 up to 35

there's an interesting consideration about whether to place a small number in the centre (for versatility) or a large number (to get to bigger numbers)

in the solution section of Brian's book it gives 43 as a highest total
this was beaten by trainee teacher in 1996 who got up to 45
then this record was beaten, also in 1996, by an 11 year old who spent a lot of time on it at home - who got from 1 to 46

nrich ran this activity on their website and 46 was the highest number submitted

the easier-to-play-around-with versions are also quite interesting:

for:

• 2 pieces, the highest is 3
• 3 pieces, the highest is 7
• 4 pieces, there are two different solutions that give from 1 to 13
• 5 pieces, the highest is 19
• 6 pieces gives 1 to 27
• 7 pieces, I make 1 to 33
• 8 pieces, I make 1 to 41

then I gave up...

* 'a mathematical pandora's box', by Brian Bolt

right triangle using 1 to 7

use 1 to 7, once only, so that the lines have the same total 

prove that the line totals cannot be below 12 or above 15 

prove that a line total of 14 is impossible

there are two solutions for sum = 13 and one for sums of 12 and 15 

students can be asked to find all of the solutions, once they have luxuriated in the glory of finding one solution of course...

hard haitch using 1 to 7

use 1 to 7 so that the (four) lines of three all add up to the same number

prove that 4 must go in the middle and the only possible total is 12


there are three solutions but these (maybe trivially) involve switching numbers within the lines of three

Zorro using 1 to 7

use digits1 to 7 (once only) so that the lines of three have the same total

what totals are possible and how many variants are there? 

prove that the middle must be even

prove that 7 and 6 must not be in the same line









ignoring unimportant swaps, show that there are: 

• two solutions with 2 in the middle
• three with 4 in the middle 
• two with 6 in the middle

prove that only totals of 11, 12 and 13 are possible

why is a total of 14 impossible?

this is the same as a task called 'Haitch'
Graeme Brown produced an Excel interactive version to download for nrich here

                                                                                                                               

7 regions

place all the digits 1 to 7, one in each region
so that the three circles all have the same total

I'm fairly confident that there are 18 solutions altogether and these are not too hard to find:

• 1 way totalling 13
• 3 ways totalling 14
• 2 ways totalling 15
• 6 ways totalling 16
• 2 ways totalling 17 
• 3 ways totalling 18 
• 1 way totalling 19

why is there a symmetry to these numbers of possibilities?

proofs:

• 1 must go in the centre for totals of 13 
• 7 must go in the centre for totals of 19

impossibility proofs:

• you cannot make totals greater than 19 
• you cannot make totals less than 13
• circles cannot total 16 with a 4 in the centre

place letters (a to g) in the regions, going left to right and then down
prove that:

• a + b = g + f
• b + c = d + g
• c + f = a + d
• b + e + g (etc) must be even

sum 46

place all of the the first ten odd numbers in the little triangles so that the three big triangles all add up to 46

some aspects of this problem can lead to proof: that the central little triangle must be 19 when the big triangle totals are 46; that there are only two (essentially) different solutions, because 17 has to go with either (9 and 1) or (7 and 3)

other totals are possible:  
there are two different solutions with big triangle totals of 34, 38 and 42
the middle numbers for these are 1, 7 and 13 respectively
you can prove that 3 must be a factor of (the middle number minus 1)
 
a lot of thinking can determine which numbers have to go with which other numbers
it's also good practice at addition, which could be done using cards with dot arrangements for those who might find this helpful

subsets

test out this rule:

for any 5 integers you can always find three of them 

that add up to a multiple of 3

the numbers can be as easy or tricky as you like (and could be negative)

e.g. 7   6   4  14  6

6 + 14 + 4 = M(3) and also:
6 + 7 + 14 = M(3)

a proof can be developed from a consideration of remainders after dividing by 3 (i.e. mod 3 arithmetic) and a consideration of the various options

also prove that in any 3 integers you can find two that sum to an even number

in any 6 integers you can find four that sum to a multiple of 4

two reversal problems