Schools Mathematics Grand Challenge
Week three's Puzzles
Problem 4:
The solution was:
The key to this problem is to spot that
 750 is (3/4) x 1000
 If we group the digits of the long number in pairs
92 8X 8Y 80 76 72 68 64 60 56 52 48 44 40 36 32 28
all of the individual pairs are multiples of 4 (no matter what multiple
of 4 X or Y is).
This means that we can write the product as
(92 8X 8Y 80 76 72 68 64 60 56 52 48 44 40 36 32 28) x (3/4) x (1000)
which is equal to
(23 2W 2Z 20 19 18 17 16 15 14 13 12 11 10 9 8 7) x 3 x (1000)
where W and Z must be 0 1 or 2.
Thus the product is
69 6A 6B 60 57 54 51 48 45 42 39 36 33 30 27 24 21 000
(A and B must be 0 3 or 6), and the middle digit is 4 (the 19th digit).
Problem 5:
The solution was:
John throws away the earlier sock until he finally chooses two consecutive
socks of the same colour. The sock colours can only alternate for as
long as there are still both blue and black socks in the drawer. Thus,
the longest possible sequence in which he picks alternating colours is
blue black blue black blue black blue black blue black blue black blue blue
In this case he would have thrown away 12 socks so the answer is 12.
(Note that if he started with a black sock, he would only have had to
throw away 11 socks)
Problem 6:
The solution was:
As you all have difference user codes, this problem is slightly different
for everyone taking part. Let's illustrate how to solve it with the
user code 8D2E.
In Hex the digits and letters represent the numbers from 0 to 15 as follows
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
A 10
B 11
C 12
D 13
E 14
F 15
So
8D2E = (14 x 1) + (2 x 16) + (13 x 256) + (8 x 4096) = 36142
and
36142 = (13 x 2780) + 2
so the remainder when we divide by 13 is 2.
Note that the answer can be different for different usercodes.
Also, if you divided by 13 hex, which is 19 decimal, that's OK,
we accepted either answer.
