Schools Mathematics Grand Challenge
Week four's Puzzles
Problem 10:
The tenth problem was:
Three women go into a bread shop one after another. The first woman buys
half of the total number of loaves in the shop and an extra half loaf (So
if there were 21 loaves in the shop, she buys 10.5 + 0.5 = 11 loaves).
The second woman buys half of the loaves left in the shop and an extra
half loaf. The third woman then goes in and buys half of the remaining
loaves and an extra half loaf. When the third woman leaves the shop,
there are no loaves left. How many loaves were in the shop to start with?
The solution was:
The answer is 7. Let x,y,z be the number of loaves bought by the first,
second and third woman respectively. The total number of loaves in the
shop is then x+y+z. The first woman takes half of the total number of
loaves in the shop plus an extra half loaf:
x = (x+y+z)/2 + 1/2
At this point, there are (x+y+z)/2  1/2 loaves left in the shop.
Now the second woman comes in and takes half of what is left plus an
extra half loaf:
y = (x+y+z)/4 + 1/4
At this point, there are (x+y+z)/4  3/4 loaves left in the shop.
Finally, the third woman comes in and takes half of what is left plus
an extra half loaf:
z = (x+y+z)/8 + 1/8
After she leaves there no bread left. This means that
(x+y+z)/4  3/4  (x+y+z)/8  1/8 = 0
or
(x+y+z)/8  7/8 = 0
or
(x+y+z) = 7.
Thus the first woman takes 4 loaves, the second woman 2, and the third woman 1.
Problem 11:
The eleventh problem was:
John, Mary, Thomas and Sheila all have mobile phones with the same
company, so all four of them have different numbers. One day, while
quite bored, they decide to play the following game. First, each of them
subtracts the numbers of the other three people from their own number,
then multiplies the three resulting differences together, and finally
divides 1 by the result.
So, for instance, if the numbers of the four friends were J (for John's
number), M (for Mary's number), T (for Thomas' number), S (for Sheila's
number), then John would calculate
1 divided by the product (J  M)x(J  T)x(J  S),
while Mary would calculate
1 divided by the product (M  J)x(M  T)x(M  S).
If John, Mary, Thomas and Sheila add up the four answers obtained in
this way, what will the sum be?
The solution was:
The sum is zero (0). To see this, let us introduce the shorthand notation
a := (JM)
b := (JT)
c := (JS)
Note that we can write all other differences in terms of a,b, and c, as follows:
(MJ) = a
(MT) = ba
(MS) = ca
(TJ) = b
(TM) = ab
(TS) = cb
(SJ) = c
(SM) = ac
(ST) = bc
The sum can then be written as
sum = 1/(abc) + 1/((ab)(ac)(bc))*[(bc)/a + (ac)/b + (ab)/c]
The term in square brackets [] can be rewritten as
[...] = 1/(abc)*{(bc)bc + (ac)ac  (ab)ab}
The term in curly brackets {} can be rewritten as
{...} = (ab)(ac)(bc)
Collecting terms we get
sum = 1/(abc) + 1/((ab)(ac)(bc))*[{(ab)(ac)(bc)}/(abc)],
or simply
sum = 1/(abc)  1/(abc)
In other words, we find that sum = 0.
Problem 12:
The twelveth problem was:
The number 14 is the smallest whole number with the following two properties:
 It is bigger than 10;
 It is smaller than the sum of the squares of its digits  (1 x 1) + (4 x 4) = 1 + 16 = 17 > 14.
What is the largest whole number satisfying these two properties?
The solution was:
The answer is 99.
It is easy to see that there is no 4digit number with these properties,
because with 4 digits the sum of squares can never exceed 4*9*9 = 324,
while by definition all 4digit numbers are greater than or equal to
1000, which is strictly greater than 324. A similar argument applies
for all numbers with more digits.
What about 3digit numbers? Can we find a
number with 3 digits, d1, d2, and d3, such that
d1*100 + d2*10 + d3*1 < d1*d1 + d2*d2 + d3*d3?
Or equivalently, can we find d1, d2, d3, such that
d1*(100  d1) + d2*(10d2) + d3*(1d3) < 0?
The answer is no. Remember d1 > 0 (when d1=0 we have a 2digit number!), we have that
d1*(100  d1) ≥ 99
d2*(10  d2) ≥ 0
d3*(1  d3) ≥ 72
When we add it all up, we find that
d1*(100  d1) + d2*(10d2) + d3*(1d3) >= 27
Since 27 is strictly greater than 0, we conclude that there exists no
threedigit number with the above mentioned properties. Now consider the
case of twodigit numbers. We try 99 and find that it works (because 99 <
81+81). Since 99 is the largest 2digit number and since we know that
there is no number with 3 or more digits that satisfies the desired
properties, 99 must be the large whole number that does.
