"Write four digits in a 2x2 square": Maths problem from Year 5. What is it called?

[John Bibby]

Has anyone come across this problem? It looks good and has quite deep ramifications - even vector spaces etc if you want to take it that far. Good for teacher-training, I suspect.  However, our Year 5s were more concerned with learning to add.

* Write four digits in a 2x2 square e.g.     34

                                                                 56

(In case this has not come out correctly, I mean write "34" immediately above "56")

* Now write down and add the four two-digit numbers that you get from this square:  34+56+35+46=???

* What is the largest total you can get?  What is the smallest?

* Find four digits that give the total 100. Try other totals.

* How many such sets of four digits are there?

* Extend .....!

Please try this on your friends! I welcome your comments.

JOHN BIBBY

[John.Mason]

Productive Differences

A slight variant that is particularly good for showing the effectiveness of ‘tracking arithmetic’ is to use the same two by two grid. Add the products along the rows; add the products down the columns; calculate the difference. Your answer in John’s example will be 3.(his first row is 3 4 and second row is 5 6).

Now find four numbers for which the same operation will give you, say, 7. In essentially how many different ways can this be done?

To use ‘tracking arithmetic’, carry out the operations but DON’t Actually DO any arithmetic calculations. Then smile.

Now multiply the two digit number formed by the top row by the two digit number formed by reversing the bottom row. Also multiply the two digit number formed by the first column, and the two digit number formed by the second column reversed. Take their difference. Smile even more broadly!

I don’t know of any generalisation to more digits.

JohnM

[John.Mason]

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[John.Mason]

[Peter Farrell]

Hi, John,

Nice challenge! (Python and) I came up with 9 groups of numbers, the first of which was 11 and 57. 11 + 57 + 15 + 17 = 100

def bibby():

    nums = [1,2,3,4,5,6,7,8,9]

    for i in nums:

        for j in nums:

            for k in nums:

                for m in nums:

                    total = 10*i + j + \

                            10*k + m + \

                            10*i + k + \

                            10*j + m

                    if total == 100:

                        print(i,j,k,m)

Reminds me of something from Project Euler, a website I highly recommend for math and programming challenges.

Peter Farrell

[Christian Baune]

It gives:
ab
cd

And translates to:
10a+b+10c+d+10a+c+10b+d
Which simplify to:
20a+11b+11c+2d

I guess that the digits should be distincts otherwise 9 everywhere and one zero everywhere would give respectively max and min.
With distincts:
max: a=9; b=8; c=7; d=6 (b and c can be swapped)
For 100
a=[0,4]
b=c (mod 2)
d=/=c (mod 2)

We can take an intermediate step. Let e be the sum of b and c.
e is a multiple of 11.
20a+11e+2d=100
Now, we know that 20a+2d is always even.
So, we can keep: 22,44,66 and 88.
20a+2d=22
a=1; d=1; e=0 (-> b=0; c=0)
20a+2d=44
a=1; d=12 (-> d>9)
a=2;d=2 (-> a=d)
20a+2d=66
a=3;d=3 (-> a=d)
20a+2d=88
a=4;d=4 (-> a=d)

We shown that with distinct numbers, this problem has the empty set as solution. (Eg. "no solution")

If the same number is allowed, we have the candidates eliminated above. With the cross product of sum of 2 multiples of 11 to N.
As "a=b" is a prerequisite (as shown above) we can proceed like this:
{...},{(0,4),(1,3),(2,2),(3,1),(4,0)},{...}
Cardinal is N+1 for each set.
3+5+7+9=24

If duplicates are allowed, there's 24 sums.
We can consider removing sums where operands are a permutation too.
For even sets, it is cardinal/2 for odd sets, (cardinal+1)/2.
2+3+4+5=14

This fully discuss the given problem ;-)

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