Difference between revisions of "2011 AIME I Problems/Problem 7"
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− | <math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>. Notice that, also, <math>m^{a_2011}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_2011}</math> equal to <math>1-(\frac{1}{m})^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-(\frac{1}{2})^x</math> by doing <math>\frac{1/2}+\frac{1/4}+...</math> (the simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\ | + | <math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>. Notice that, also, <math>m^{a_2011}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_2011}</math> equal to <math>1-(\frac{1}{m})^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-(\frac{1}{2})^x</math> by doing <math>\frac{1/2}+\frac{1/4}+...</math> (the simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times. |
==Solution 1== | ==Solution 1== |
Revision as of 02:09, 7 February 2016
Problem 7
Find the number of positive integers for which there exist nonnegative integers
,
,
,
such that
Solution
. Now, divide by
to get
. Notice that since we can choose all nonnegative
, we can make
whatever we desire. WLOG, let
and let
. Notice that, also,
doesn't matter if we are able to make
equal to
for any power of
. Consider
. We can achieve a sum of
by doing
(the simplest" sequence). If we don't have
, to compensate, we need
's. Now, let's try to generalize. The "simplest" sequence is having
times,
times,
. To make other sequences, we can split
s into
s since
. Since we want
terms, we have
. However, since we can set
to be anything we want (including 0), all we care about is that
which happens
times.
Solution 1
Let . The problem then becomes finding the number of positive integer roots
for which
and
are nonnegative integers. We plug in
and see that
. Now, we can say that
for some polynomial
with integer coefficients. Then if
,
. Thus, if
, then
.
Now, we need to show that for all
,
. We try with the first few
that satisfy this.
For
, we see we can satisfy this if
,
,
,
,
,
,
,
, because
(based on the idea
, leading to a chain of substitutions of this kind)
. Thus
is a possible value of
. For other values, for example
, we can use the same strategy, with
,
,
,
,
and
, because
.
It's clearly seen we can use the same strategy for all
. We count all positive
satisfying
, and see there are
Solution 2
One notices that if and only if there exist non-negative integers
such that
.
To prove the forward case, we proceed by directly finding . Suppose
is an integer such that
. We will count how many
, how many
, etc. Suppose the number of
is non-zero. Then, there must be at least
such
since
divides all the remaining terms, so
must also divide the sum of all the
terms. Thus, if we let
for
, we have,
Well clearly,
is greater than
, so
.
will also divide every term,
, where
. So, all the terms,
, where
must sum to a multiple of
. If there are exactly
terms where
, then we must have at least
terms where
. Suppose there are exactly
such terms and
for
. Now, we have,
One can repeat this process for successive powers of
until the number of terms reaches 2011. Since there are
terms after the
th power, we will only hit exactly 2011 terms if
is a factor of 2010. To see this,
Thus, when
(which is an integer since
by assumption, there are exactly 2011 terms. To see that these terms sum to a power of
, we realize that the sum is a geometric series:
Thus, we have found a solution for the case
.
Now, for the reverse case, we use the formula Suppose
has a solution. Subtract 2011 from both sides to get
Now apply the formula to get
where
are some integers. Rearranging this equation, we find
where
. Thus, if
is a solution, then
.
So, there is one positive integer solution corresponding to each factor of 2010. Since , the number of solutions is
.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
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