Download INTERMEDIATE SOLUTIONS 1996

INTERMEDIATE SOLUTIONS 1996
1. A train takes 45 seconds to cross a bridge 450 metres long. It then
takes 15 seconds to move past a thin lightpost. How long is the train
and at what speed is it travelling?
Solution.
When passing the thin lightpost (assumed to be of zero width), the
train travels a distance equal to its length, in 15 sec. In crossing the
bridge, it travels the length of the bridge, plus its own length, in 45
sec. Therefore the train takes 45 − 15 = 30 sec. to travel the length
of the bridge which is 450 m. Hence the train’s speed is 450
= 15m/s.
30
Its length is its speed multiplied by the time taken to travel its length,
that is 15 × 15 = 225 m.
2. When Nick was as old as Olga is now, their aunt was one year younger
than the sum of Nick and Olga’s current ages. How old was Nick when
his aunt was his current age? (You may assume that their aunt is older
than both Nick and Olga, and that Nick is older than Olga.)
Solution.
Let Nick’s age be n, Olga’s age be o and the aunt’s age be a, (all in
years). Then the time when Nick was as old as Olga is now was (n − o)
years ago, so that their aunt’s age at this time was a−(n−o) = a−n+o.
It was also at this time that the aunt’s age was 1 less than the sum of
Nick’s and Olga’s current ages - that is, n + o − 1. Hence we have
a − n + o = n + o − 1 ⇒ 2n − a = 1
The time when their aunt was Nick’s current age was (a − n) years ago,
so Nick at this time was n − (a − n) = 2n − a = 1 year old.
3. Prove that the product of four consecutive natural numbers, plus 1, is
a perfect square.
Solution.
Let x be the smallest of four consecutive numbers, which would then
be x, x + 1, x + 2, x + 3. Then their product, plus 1, is
x(x + 3)(x + 2)(x + 1) + 1 = (x2 + 3x)(x2 + 3x + 2) + 1
= ((x2 + 3x + 1) − 1)((x2 + 3x + 1) + 1) + 1
= ((x2 + 3x + 1)2 − 12 ) + 1 = (x2 + 3x + 1)2 ,
which is a perfect square.
4. Show that the last digit of each term in the sequence 1, 1 + 2, 1 + 2 +
3, 1 + 2 + 3 + 4, . . . is ultimately periodic.
Solution.
Experimentation leads us to believe that the last digit of each term in
the sequence forms a pattern which repeats itself every 20 terms:
1, 3, 6, 0, 5, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, |1, 3, 6, . . .
However simply listing the first 20, 40 or even 500 terms does not
guarantee that the sequence will remain periodic thereafter. We require
a proof! For convenience, let an denote the nth term of the sequence
1, 1 + 2, 1 + 2 + 3, 1 + 2 + 3 + 4, . . .
(so, for example, a50 is the 50th term). We then observe that an+1 =
an + 1 for all positive integers n.
In particular,
an+20 =
=
=
=
=
an+19 + (n + 20) = (an+18 + (n + 19)) + (n + 20)
((an+17 + (n + 18)) + (n + 19)) + (n + 20)
···
an + (n + 1) + (n + 2) + · · · + (n + 19) + (n + 20)
an + 20n + 210.
The last digit of both 20n and 210 is 0, so that the last digit of an+20
is the same as that of an . This shows that the last digit of each term
of the sequence is indeed periodic.
5. You are given a cardboard cubical box without a lid. (Its sides and
bottom are squares of unit area.) Making as many straight line cuts
as necessary, first cut it so that it can be opened out into a connected
planar shape, (in plain language, cut the box so that it can be folded
out flat), and then cut it into three pieces that can be put together to
make a square of area 5.
Solution.
There are several valid solutions to this question - here is one:
A
D
B
E
C
H
F
G
Let ABCDEFGH be the cubical box, where face ABCD is missing.
Make cuts along the edges BF, EF, DH and GH. The box can then be
flattened to look like the net below.
B1
F1
A
D1
E
H1
G
F2
B2
C
H2
D2
(Here B1 and B2 came from the one point B in the box, and similarly
for the other points.) Then cut along D1 F1 and B2 H2 , to give three
pieces. These can be arranged to give the square shown below.
(It can then be easily checked that all side lengths and angles of this
figure are equal, so the construction indeed works!)
6. A computer printed out two numbers: 21995 and 51995 . How many digits
in all were printed?
Solution.
First write the two numbers in standard index form:
21995 = a × 10x ; 51995 = b × 10y
where x and y are whole numbers and 1 ≤ a < 10; 1 ≤ b < 10.
Clearly both a and b are strictly greater than 1, as neither 21995 nor
51995 are powers of 10. 21995 thus has x + 1 digits, 51995 thus has y + 1
digits. Therefore (x + 1) + (y + 1) = x + y + 2 is the number of digits
printed in total. Multiplying the top two equations gives 21995 ×51995 =
a × 10x × b × 10y , that is, 101995 = ab × 10x+y . This equation tells
us that the product ab must be a power of 10. But the conditions
1 < a < 10; 1 < b < 10 derived earlier imply that 1 < ab < 100, so ab
can only equal 10. Hence
101995 = 10 × 10x+y = 10x+y+1 .
Therefore 1995 = x + y + 1, and so the total number of digits printed
is x + y + 2 = 1996.
7. A positive integer is said to be happy if one can choose a number of
digits from its decimal notation such that their sum equals the sum
of all the remaining decimal digits. Find the smallest happy number
a such that a + 1 is also happy. Is it possible that there exists three
successive happy numbers?
Solution.
We can split the digits of any happy number into two non-empty groups,
each having equal sum. Hence the sum of all the digits of a happy
number is twice that of one of the groups, so is an even number.
If a and a + 1 are consecutive happy numbers which only differ in the
last digit, then one of the numbers will have an even sum of digits, and
the other will have an odd sum. But this is not possible from the above
statement, so a and a + 1 differ in more than 1 digit - in particular, a
must have a last digit of 9.
Clearly, no single digit number can be happy, as it cannot be split into
two non-empty groups. The only two-digit happy numbers are those
with equal digits, and 99 is the only one having a last digit of 9, but 100
is not happy. Therefore we need consider at least three-digit numbers
to find the first instance of two consecutive happy numbers.
In a three-digit happy number, the sum of two of the digits must equal
the third. As the last digit of a is 9, either the first two digits add to
9, or 9 plus one of the other digits must equal the third. We readily
see that 909 is the only happy number in the second case, but 910 is
not happy, so we reject this possibility. Hence writing the digits of a
as x, y, 9 (in order) we have x + y = 9. But x > 0 as a is a three-digit
number, so y < 9. So the digits of a + 1 must be x, y + 1, 0 and as we
cannot have x + (y + 1) = 0, it follows that x = y + 1. Solving these two
equations gives x = 5, y = 4. Hence 549, 550 are the only two successive
three-digit numbers, and we conclude that a = 549.
The condition that the smaller of two successive happy numbers must
have a last digit of 9 means that three successive happy numbers cannot
exist, as we would need two successive numbers to have a last digit of
9, which is impossible.
8. Two circles S1 and S2 touch each other externally at a point F . A line
L is tangent to both S1 and S2 at points A and B, respectively, and a
line parallel to L is tangent to S2 at C. Prove that points A, F, C are
collinear. (Collinear means they lie on a straight line).
Solution.
Let O1 be the centre of S1 , O2 be the centre of S2 . Then O1 , F and O2 are
collinear, as the line tangent to S1 and S2 through F is perpendicular
to both O1 F and O2 F and so 6 O1 F O2 = 90◦ + 90◦ = 180◦ .
Let the line tangent to S2 at C be M. Then as O1 A is perpendicular
to L, O2 C is perpendicular to M, and L and M are parallel, it follows
that AO1 and CO2 are parallel.
Hence 6 AO1 O2 = 6 CO2 O1 (alternate angles).
4O1 AF and 4O2 CF are both isoceles as two of their sides are radii
of S1 and S2 respectively.
Hence
6
1
(6 O1 F A + 6 F AO1 ) (as 6 O1 F A = 6 F AO1 )
2
1
1
=
(180◦ − 6 AO1 F ) = (180◦ − 6 AO1 O2 )
2
2
1
1
◦
=
(180 − 6 CO2 O1 ) = (180◦ − 6 CO2 F )
2
2
1
=
(6 O2 F C + 6 F CO2 ) = 6 O2 F C (as 6 O2 F C = 6 F CO2 ).
2
O1 F A =
This, together with the fact that O1 , F and O2 are collinear, shows that
the two angles are vertically opposite, so A, F, C are collinear.