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BMOS MENTORING SCHEME (Senior Level)
October 2011 (Sheet 1)
Solutions
These solutions represent one way of solving the questions. They are not the only way,
and just because your solution is different does not mean that it is worse or wrong. You
will get most from the solutions if you have at least attempted the questions. Only then
will you appreciate the difficulties that a problem posed and how they were addressed by
the solution. Relevant theorems have been identified where possible, and references to
further information on the Web have been given. If you are still unclear, ask your mentor.
c
A. Rzym & J. Cranch UKMT
2011.
1. For any positive integer n, define n! to be the product of all positive integers not exceeding
n, i.e. n! = 1 × 2 × · · · × (n − 1) × n.
Exactly 9! numbers in the list start with ‘1’ (because the remaining digits 0,2,3,4,5,6,7,8,9
can be arranged in 9! ways). The same applies to numbers starting with ‘2’,‘3’,‘4’ etc.
But 2.9! < 1000000 ≤ 3.9!, so we conclude that the millionth number in the list starts
with ‘3’. Furthermore, it occupies position 1000000 − 2.9! within the numbers in the list
starting with ‘3’.
There are exactly 8! numbers starting with ‘30’ (because the remaining digits can be
arranged in 8! ways) and the same applies to numbers starting with ‘31’, ‘12’ etc.
But 6.8! < 1000000 − 2.9! ≤ 7.8!, so we conclude that the millionth number in the list
starts with ‘37’. Furthermore, it occupies position 1000000−2.9!−6.8! within the numbers
in the list starting with ‘37’.
Repeating this argument, we conclude that the millionth number in the list is ‘3782915460’.
2. For positive x, y we have
(x − y)2 ≥ 0
2(x2 + y 2 ) ≥ (x + y)2
2
1/2
x + y2
x+y
≥
.
2
2
Also,
√
√
( x − y)2 ≥ 0
x+y
√
≥
xy.
2
Replacing x and y by 1/x and 1/y in the above and rearranging gives
√
xy ≥
1
x
2
+
1
y
and combining these three results gives
2
1/2
x + y2
x+y √
≥
≥ xy ≥
2
2
1
x
2
.
+ y1
Supported by the Man Group plc Charitable Trust
This is the RMS-AM-GM-HM (root mean square – arithmetic mean – geometric mean –
harmonic mean) inequality in two variables.
(a) Writing x = |CB| and y = |AB| (and noting that by Pythagoras we have x2 + y 2 =
|AC|2 ) then applying the RMS-AM inequality gives
1/2
x+y
x2 + y 2
≥
2
2
|AC|
|CB| + |AB|
√
≥
2
2
from which the result follows. (b) Setting x = a + 1/a and y = b + 1/b, the RMS–AM inequality implies
(a + a1 )2 + (b + 1b )2
2
Now setting x =
1
a
1/2
≥
a+
1
a
+b+
2
1
b
=
1+
1
a
2
+
1
b
.
(1)
and y = 1b , the AM-HM inequality gives
1
a
+
2
1
b
≥
2
=2
a+b
and substituting this into (1) gives
(a + 1/a)2 + (b + 1/b)2
2
1/2
≥
1+4
.
2
Squaring gives the result we seek. 3. Let No denote the number of permutations of the letters where the letter ‘o’ is in the
correct place (all other letters may or may not be in the correct place). Similarly, Nne
denotes the number of permutation of the letters where ‘n’ and ‘e’ are in the correct place
(all other letters may or may not be in the correct place) etc. In particular, N represents
the number of permutations of the letters without constraint.
Obviously, N = 6!. N includes permutations where exactly 1 letter is in the correct place.
We can correct for this by subtracting No + Nr + Na + · · · + Ne .
The quantity N − (No + · · · + Ne ) counts ‘words’ where exactly 2 letters are in the correct
place – they are counted once in N and subtracted twice in (No +· · ·+Ne ). We can correct
for this by adding Nor + Noa + · · · + Nge , to give N − (No + · · · + Ne ) + (Nor + · · · + Nge ).
Continuing this argument we arrive at the solution
N − (No + · · · + Ne ) + (Nor + · · · + Nge ) − (Nora + · · · + Nnge ) + · · · =
6
6
6
N−
No +
Nor − · · · =
0
1
2
6
6
6
6! −
5! +
4! − · · · = 265.
0
1
2
2
4. Applying the Law of Sines to triangle ACD we have
CA
AD
=
.
sin ADC
sin ACD
Similarly applying the Law of Sines to triangle BCD we have
CB
DB
=
.
sin BDC
sin BCD
However ∠ACD = ∠BCD, and ∠ADC = 180 − ∠BDC therefore sin ADC = sin BDC.
Combining these relations with the equations above, the result follows. 5. There is unlikely to be anything special about the number 2001 in the context of this
problem, and therefore we will solve the more general problem:
‘How many solutions are there (as a function of m, n) to the equation
x1 + x2 + · · · + xm = n
where the xi and n are all non-negative integers (and m is positive)?’
Denote this number by f (m, n). For small values of m, n we can construct a table by
hand:
0
1
1
1
1
f (m, n)
1
2
3
4
1
1
2
3
4
2 3
1 1
3 4
6 10
10 20
We first note that each term is an entry in Pascal’s Triangle. We next note that the
North–East diagonals
of this table are the rows of Pascal’s Triangle, so further guess that
f (m, n) = m+n−1
. It is then just a matter of matching the correct element in Pascal’s
·
Triangle with the correct term in the North–East diagonal to guess f (m, n) = m+n−1
.
m−1
How do we prove this? Clearly, if there is only one variable (i.e. m = 1) then there is
only one representation, i.e.
f (1, n) = 1
∀n.
(2)
If there are m variables, then xm may take any of the values 0, . . . , n, for which there
are f (m − 1, n), f (m − 1, n − 1), . . . , f (m − 1, 0) representations (respectively) for the
remaining variables. Thus
f (m, n) = f (m − 1, 0) + f (m − 1, 1) + · · · + f (m − 1, n).
(3)
Equations (2) and (3) are sufficient to define f (m, n). Does our guess satisfy them? If
f (m, n) = m+n−1
then f (1, n) = n0 = 1.
m−1
Also
f (m − 1, 0) + f (m − 1, 1) + · · · + f (m − 1, n) =
m−2
m−1
m−1
+
+ ··· +
m−2
m−2
m+n−2
3
m−1
and the latter can readily be shown to be to equal m+n−1
by induction (show, explicitly,
that the claim is true for m + n ≤ 2, assume that it is true for m + n ≤ k and show that
it is true for m + n = k + 1).
Thus our guess satisfies both equations, and
therefore the number of solutions to the
m−1
equation in the question is given by m+2000 . 6. A move on a 3 × 3 square can be positioned in any of (8 − 2)2 = 36 locations, and a
move on a 4 × 4 square can be positioned in any of (8 − 3)2 = 25 locations. A repeat
application of a given move is the same as no application at all, therefore the total number
of operations is given by which of the 36 3 × 3 moves and 25 4 × 4s moves are used, i.e.
261 possibilities. By contrast, there are 264 possible initial colourings of the chessboard.
The latter exceeds the former, and therefore there must exist an initial colouring which
cannot be transformed into an entirely white chessboard. 7. Do there exist powers of 3 which end in the digit ‘1’ ? Modulo 10 we have
30 ≡ 1,
31 ≡ 3,
32 ≡ 9,
33 ≡ 7,
3k+4 ≡ 3k · 34 ≡ 3k .
Thus every power of 3 of the form 34k ends in ‘1’.
Do there exist powers of 3 which end in the digits ‘01’ ? Clearly, they must also be of the
form 34k , and modulo 100 we have:
30
34
38
312
316
3k+20
≡
≡
≡
≡
≡
≡
1,
81,
81 × 81 ≡ 61,
61 × 81 ≡ 41,
41 × 81 ≡ 21,
3k × 21 × 81 ≡ 3k .
Thus every power of 3 of the form 320k ends in ‘01’.
Do there exist powers of 3 which end in the digits ‘001’ ? Clearly, they must also be of
the form 320k , and modulo 1000 we have:
30
320
340
360
380
3k+100
≡
≡
≡
≡
≡
≡
1,
((35 )2 )2 ≡ (2432 )2 ≡ 492 ≡ 401,
401 × 401 ≡ 801,
801 × 401 ≡ 201,
201 × 401 ≡ 601,
3k × 601 × 401 ≡ 3k .
Thus every power of 3 of the form 3100k ends in ‘001’.
For powers of 3 ending in ‘2001’ we proceed similarly. Mod 10000 we have:
30 ≡
3100 ≡
3200 ≡
3300 ≡
3400 ≡
3k+500 ≡
1,
(((35 )5 )2 )2 ≡ ((243)5 )2 )2 ≡ (9443)2 )2 ≡ (249)2 ≡ 2001,
2001 × 2001 ≡ 4001,
4001 × 2001 ≡ 6001,
6001 × 2001 ≡ 8001,
3k × 8001 × 2001 ≡ 3k .
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Thus every power of 3 of the form 3100+500k ends in ‘2001’.
Let log x and {x} denote the logarithm to the base 10 of x and the fractional part of x,
respectively.
If x begins with the digits ‘2001’ then log 2.001 ≤ {log x} < log 2.002. Thus the question
can be rephrased: does there exist an integer k such that:
log 2.001 ≤ {log 3100+500k } < log 2.002?
(4)
Now z = log 3100+500k = log 3100 +500k log 3, but log 3 is irrational, therefore the fractional
part of z is never equal for distinct values of k. Thus (by the pigeonhole principal) two
k’s can be found such that the fractional parts are arbitrarily close, and therefore a k
can be found such that (4) is satisfied, and therefore a power of 3 can be found such the
decimal representation begins and ends with 2001. References
Derangement:
http://en.wikipedia.org/wiki/Derangement
Arithmetic Mean – Geometric Mean Inequality:
http://en.wikipedia.org/wiki/Inequality of arithmetic and geometric means
Law of Sines:
http://en.wikipedia.org/wiki/Law of sines
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