ALL THE TRICKS I KNOW
ROBERT C. RHOADES
Abstract. The Putnam exam is hard. Often it is hard to even start thinking about the
problems on the exam. The goal of this talk is to give you some tools that will help to make
many of the problems on the Putnam easier.
1. Advice from the best
During the summer of 2005 I worked with Po-Ru Loh of Caltech University. Po-Ru
participated in the IMO three times for the USA. He won a silver medal and two gold
medals. He also placed 6th on the 2003 Putnam exam. I asked him for tricks or advice that
I should give you and this was his response
“One thing that I can recommend is calculating out simple cases and looking for patterns
on problems where you don’t know where to start. It may sound like an obvious thing to
do, but I’ve found it easy to fall into the trap of just staring at a problem for a long time
and not getting anywhere, when doing out some examples would have made it clear what to
do. One other thing is just making sure you get points for the problems you’ve solved – as
we both know, the Putnam grading is extremely harsh toward unclear solutions.”
I agree completely with his advice. Keep these things in mind.
2. General Advice
(1) The problems are arranged from easiest to hardest. So try the A1 and B1. Even
if you don’t think you know how to do it, try it. Putnam problems are written to
be attractive. It is easy to look at a B5 or A6 and think the problem is cool and
interesting and then spend 3 hours on it without getting anywhere. Spend sometime
working on the A1!
(2) If you solve a problem spend an extra 10 minutes writing it up PERFECTLY. The
graders have to grade over 3500 students papers and they do it in 2 days. If they see
something that they don’t think is right they just give it a 0 and move on. I am sure
the graders appreciate a well-written solution and are much more likely to award a
high score to it, then a solution that may have all the ideas but is unclear. Also in
the last 15 minutes of the exam go back and read over all your solutions again and
again. Do not lose points for stupid reasons!
(3) Get sleep the night before the exam. It is tempting to try to solve old problems
the night before. You are much better off getting sleep. It will keep you more alert
during the exam.
(4) Do a lot of old problems. The only way to really learn how to do problems is by
working on old problems. Do as many of the old Putnam exams as you can. You
can find the statements and solutions to old problems online. Here is a good website:
http://www.unl.edu/amc/a-activities/a7-problems/putnam/
The author thanks Po-Ru Loh for useful conversation.
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ROBERT C. RHOADES
If you need some motivation check out the winners of the 2004 exam. Look at the
top ten teams.
(5) Don’t get frustrated. The exam is hard. Some years the exam will not suit you.
Work hard and you will succeed.
3. No Square is Negative
The basic fact is that if x is a real number then x2 ≥ 0. This may sound dumb but it can
be useful.
Exercise Let a be a real number, prove that 4a − a4 ≤ 3.
Solution One approach to this problem is to define a function f (x) = 4x − x4 and then use
calculus to show that it has a maximum at x = 1 and that in that case f (x) = 3. Or you
can reduce this inequality to the following trivial inequality
(a2 − 1)2 + 2(a − 1)2 ≥ 0.
The next one is slightly harder and you need one more idea in addition to the fact that
no square is negative to prove the result.
Exercise Determine whether there exists a one-to-one function f : R → R with the property
that for all x,
1
f (x2 ) − f 2 (x) ≥ .
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Solution The first thing to try is just to plug in some numbers and try to calculate what
f is for these numbers. The first numbers that you should always try are x = 0 and x = 1.
If we can calculate what f must be for these numbers we gain some information about
f . Notice we have f (0) − f 2 (0) ≥ 1/4. So moving everything to the right side we have
0 ≥ f 2 (0) − f (0) + 1/4 = (f (0) − 1/2)2 . Since no square is negative f (0) = 1/2. Similarly
f (12 ) − f 2 (1) ≥ 1/4 and we see that f (1) = 1/2. Therefore no such f can be one-to-one.
The extra idea in this problem was that 0 and 1 are the only numbers such that they equal
their own square. This is a good fact to know as well.
Try the following problems. All of these as well as the ones above where taken from
Mathematical Olympiad Challenges by Andreescu and Gelca.
Exercise Determine f : N → R such that for all k, m, n one has
f (km) + f (kn) − f (k)f (mn) ≥ 1.
Exercise Let a, b and c be real numbers such that a2 + c2 ≤ 4b. Prove that for all x ∈ R,
x4 + ax3 + bx2 + cx + 1 ≥ 0. Hint: Write the expression as a sum of two squares and x2 times
an expression in a, b, and c that must be positive.
ALL THE TRICKS I KNOW
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4. Telescoping Sums and Products
The two following relations are known as the formula for the telescoping sum and the
formula for the telescoping product.
N
X
(f (k) − f (k − 1)) = f (N ) − f (1).
k=1
N
Y
f (k + 1)
k=1
Exercise Prove
f (k)
=
f (N + 1)
.
f (1)
PN
k!k
< 1.
(N + 1)!
k=1
Solution We evaluate the sum by rewriting k!k = k!(k + 1 − 1) = (k + 1)! − k!.
Exercise Evaluate
Q∞
n=2 (1
− 1/n2 ).
Solution Begin by just writing the product from 2 up to N and then we let N go to infinity
to obtain the desired product. (This is a good technique when evaluating infinite sums and
products.) We have
N µ
Y
n=2
1
1− 2
n
¶
=
N µ
Y
n=2
=
¶µ
1
1+
n
¶
N
N
Y
n−1 Y n+1
n=2
=
1
1−
n
n
n=2
n
1 N +1
·
.
N
2
Taking the limit as N → ∞ gives the desired result.
Exercise Compute
Pn
k=1
k!(k 2 + k + 1).
Exercise Prove the double inequality
√
√
√
√
1
1
1
1
2( n + 1 − m) < √ + √
+ ··· + √
+ √ < 2( n − m − 1).
m
n
n−1
m+1
Exercise Evaluate
∞
Y
n3 − 1
n=2
n3 + 1
.
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ROBERT C. RHOADES
4.1. Partial Fraction Decomposition. Look up this trick in a calculus book or online.
The idea is the following: you have a rational function with the degree of the polynomial in
the numerator less than the degree of the polynomial in the denominator and you want to
split it apart and write it as the sum of rational functions that are “simpler”. For example
consider
1
(3x − 2)(3x + 1)
and we want to write it as
1
A
B
=
+
(3x − 2)(3x + 1)
3x − 2 3x + 1
where the A and B are just numbers. Making a common denominator on the right and
setting the two sides equal we see that we must have 1 = A(3x + 1) + B(3x − 2). If this is
going to be true for all x then we must have the following equations satisfied
3A + 3B =0
A − 2B =1
Solving these we find that A = 1/3 and B = −1/3.
This trick can often be used with the telescoping trick to evaluate sums.
Exercise Sum
Exercise Sum
P∞
j=1
1/(3x − 2)(3x + 1).
∞
X
n=1
2n + 1
.
n(n + 1)(n + 2)
5. Binomial Coefficients
Find a combinatorics book or a book on problem solving, like Larsen’s Problem Solving
Through Problems and look up the basic facts about binomial coefficients. The basic facts
can go a ¡long
¢ way!
¡ ¢
Fact: nk = nk n−1
k−1
¡ n¢
Exercise Prove that the numbers 2k for k = 1, 2, . . . , 2n − 1 are all even and that exactly
one of them is not divisible¡ by
¢ 4.
Fact: Let p be a prime pj ≡ 0 (mod p) for j = 1, . . . , p − 1.
Exercise Prove this fact and then prove that for any prime p, the number
by p2 .
6. Taylor Series
Recall the Taylor series for a function f centered at 0 is given by
f 00 (0) 2
f 0 (0)
x+
x + ...
f (x) = f (0) +
1!
2!
¡2p¢
p
−2 is divisible
ALL THE TRICKS I KNOW
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The Taylor series for (1 + x)α is often useful.
x2
x3
+ α(α − 1)(α − 2) + . . .
2!
3!
In the special case that α is a positive integer this will be a finite series and we have
µ ¶ µ ¶
µ ¶
n
n
n n
n
(1 + x) =
+
x + ··· +
x .
0
1
n
(1 + x)α = (1 + αx + α(α − 1)
Exercise Show that
µ ¶
µ ¶
µ ¶
n
1 n
n+1 1 n
+ · · · + (−1)
.
=
−
k
2 2
n n
1
n
X
1
k=1
[Hint: Use the Taylor series for 1−(1−x)n
and integrate from 0 to 1.]
x
Another common Taylor series is the one for the geometric series
1
= 1 + x + x2 + x3 + . . . ,
1−x
which is valid for |x| < 1.
Exercise Use the geometric series to derive the Taylor series for log(1 − x) for |x| < 1. [Hint:
integrate the geometric series.]
Exercise Look up the Taylor series for ex . Use the fact that eiθ = cos(θ) + i sin(θ) to derive
the Taylor series for sin and cos.
Exercise Show that the power-series representation for the infinite series
X xn (x − 1)2n
n!
n≥0
cannot have three consecutive zero coefficients. [Recall the Taylor series for ez .]
Exercise Evaluate
lim
³ ex
´
+ x2 ((1 + 1/x)x − e) .
x→∞
2
[Hints: Use the Taylor series for ez . Also use the fact that ab = eb log(a) .]