AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
.
be an interior point of the acute triangle 
with 
so that 
The point 
on the segment 
satisfies 
the point 
on the segment 
satisfies 
and the point 
on the line satisfies 
Let 
and 
be the circumcenters of the triangles 
and 
respectively. Prove that the lines 
and 
are concurrent.
such that:
for all 
.
, define 
as the product of the digits of (in decimal). Find all positive integers such that 
.
consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
be a function satisfying the following conditions:
, if 
and 
, then 
;
is surjective.
. be a triangle with 
and 
. Let and be the midpoints of 
and 
respectively. The point is inside of 
such that 
is equilateral. Let 
and 
. Prove that 
.
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size 
, the blue tiles are of size and the black tiles are of size 
. Let 
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
. such that the following statement holds: Suppose real numbers 
, 
, 
, 
, 
, 
, , 
satisfy 
for all 
. Then there exists 
, 
, , 
, each of which is either 
or 
, such that![\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]](http://latex.artofproblemsolving.com/6/5/8/658ddc5e8626e7d5ef29809c8c3ac641e3a87757.png)
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
is not a multiple of 
for every 
satisfies
is not a multiple of 
for every satisfies
is not 
and not 
. Now, let us define the function F(n) = 
. We can see that F(
-j) = F(j). From this we can see that 
mod 
. We can also see that 
since 
. From this we can see that 
mod . Now we notice that 
mod and 
mod . Thus we now need only to show that 
mod 
and 
mod 
. To prove this we need only to show that 
mod 
mod . However, from Vieta's relatoins we can see that 
can be rewritten as 
for some integer x. So we know want to show that 
mod . However, we notice that 
is odd which means that there cannot be any factors of 2 in the denominator for 
to be divisible by and this is contradiction. Thus, the problem is proved.
is clearly odd, we need to show that 
is not 1 modulo 
, if 
. We can do this by writing this as the product of 
for 
. Now, choose 
such that 
is maximized: it is easy to check that is unique. From here, after canceling powers of 2, it is easy to see that each term except than the one indexed by is modulo 
, and it is also easy to check that the exceptional one is not 
. Thus is not 1 modulo , so if 
certainly it is not 1 modulo either. If 
, on the other hand, we only have one factor in the product, so the fraction is equal to modulo , so we are done in this case as well.
be a positive integer. Show that the numbers![\[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\]](http://latex.artofproblemsolving.com/2/a/3/2a3629f4a907a9be48ba5faa50dfbf067252d3d6.png)
to 
, 
, 
, 
, 
in some order.
leaves distinct remainders .
. We can take 
. Using the binomial identities we get,
. , it is handy to use kummer's theorem . we will find 
.
then using kummer';s theorem we get 
. Note that , if k is odd then 
. Thus , in this case will divide 
. which are even. is even then 
. Now , we choose a 
such that 
is minimal. Then we will be able to take 
common , and the left term will be odd . so there will be a term where 
such that , 
. Now from earlier observation we see that , more the 
less will be . here we have , 
. Suppose that 
. Obviously 
between 
and 
such that 
. Contradicting the minimality of 
. So we are done! 
?? I'm dying to understand this, especially kummer's theorem !??
, but it doesn't divide 
, so it is equivalent to 
for defined values of 
. Now observe that for 
, it suffices to show that 
:
the assertion for . We use induction. The result is true for 
, and . We prove it for larger values., observe that each term in 
, the values of the binomial coefficients, are distinct mod . When we use the above relations, we obtain a 
, after using the lemma. How could there be a duplicate in this set? Well, if there is one, it would have to be a number from , and then the lemma would produce another value. These two values would somehow be the same if the duplicate occurred. We show that this cannot happen. Observe that this implies that in , there are two numbers such that, when summed, give 
. First, the sum is obviously 
, which each element in has only one pair for: one of its neighbors. But the sum, as shown, isn't , so we are done.
and 
satisfying 
for any given odd number 
. Note that![\[\binom{2^n-1}{2k} = \prod_{i=1}^{2k} \frac{2^n-i}{i} = \left( \prod_{i=1}^{k} \frac{2^{n-1}-i}{i} \right) \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right) = \binom{2^{n-1}-1}{k} \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right).\]](http://latex.artofproblemsolving.com/c/7/4/c744238836dac515fd6d7f389453430ea373a741.png)
We can safely take the left and right ends of this equality modulo , whence 
. Now suppose that 
so that![\[(-1)^i\binom{2^{n-1}-1}{i} \equiv (-1)^j\binom{2^{n-1}-1}{j} \pmod{2^n}.\]](http://latex.artofproblemsolving.com/6/4/8/648124658e052da4c4987e51d11e4c7a9f5838aa.png)
Some casework:
and 
are of the same parity, then the above binomial coefficients are also congruent modulo 
, so by hypothesis either 
or 
. Parity forces . and are of different parity, we have![\[\binom{2^{n-1}-1}{i} \equiv -\binom{2^{n-1}-1}{j} \pmod{2^n}. \quad (\heartsuit)\]](http://latex.artofproblemsolving.com/c/d/3/cd35334cf34a06cb9e44b29d07ad442a19f52f18.png)
WLOG is even. This congruence 
holds modulo , but we can also check that![\[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} = -\binom{2^{n-1}-1}{i+1} \pmod{2^{n-1}}\]](http://latex.artofproblemsolving.com/e/4/7/e47698232e09b48461e0bb3715bd19fa2654add8.png)
![\[\implies \binom{2^{n-1}-1}{j} \equiv \binom{2^{n-1}-1}{i+1} \pmod{2^{n-1}}.\]](http://latex.artofproblemsolving.com/5/3/f/53f89a3596447715490b6db29207b3ada78f4214.png)
By hypothesis, this forces either 
or 
, but and have different parity so we must have . Now examine . Rewrite the right-side binomial coefficient to get![\[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} \pmod{2^n} \]](http://latex.artofproblemsolving.com/9/c/f/9cf9800d485480d478e6678d3f0163fae5351cf6.png)
![\[\implies \frac{2^{n-1}-(i+1)}{i+1} \equiv -1 \implies 2^{n-1} \equiv 0 \pmod{2^n},\]](http://latex.artofproblemsolving.com/d/b/4/db4c8c165d1c1b5af7580e6eaaec48097b165dfd.png)
which is absurd. So this case is impossible.
for 
are distinct odd numbers modulo , which easily converts to the desired claim.
is odd for any integer . Then we show that![\[ \binom{2^n-1}{a} \not\equiv \binom{2^n-1}{b} \pmod{2^n} \]](http://latex.artofproblemsolving.com/5/6/1/56162549a82a689a04162f92e9f2b24aa69b115f.png)
for any 
. It is easy to see that these two claims put together give the result.![\[ \nu_2 \left( \binom{2^n-1}{a} \right) = s_2(a) + s_2(2^{n-1}-1-a) - s_2(2^n-1) \]](http://latex.artofproblemsolving.com/9/7/c/97cf447a5b95831b36550e0f1974e4bf13a37ad5.png)
Since no carries occur, this is just zero.![\[ \prod_{k=b+1}^a \frac{2^n - k}{k} \not\equiv 1 \pmod{2^n} .\]](http://latex.artofproblemsolving.com/6/3/8/63873b8fbdf0699d1dba36478d7a73cae8304c6a.png)
Let 
. Denote by 
the unique integer in that range where the maximum is taken.Clearly, 
. Let 
. Then
At the last step, we make use of the fact that 
is odd. Since 
, it is impossible that![\[ 2^{n-M} - 1 \equiv \pm 1 \pmod{2^{n-M + 1}}. \]](http://latex.artofproblemsolving.com/6/f/9/6f911358e21a04b7b280e73dee2716bcf9d09f50.png)
Since![\[ \binom{2^n-1}{a} = \binom{2^n-1}{b} \prod_{k=b+1}^a \frac{2^n - k}{k} \]](http://latex.artofproblemsolving.com/a/b/9/ab99e765d9edbb80d05cb4ff69d1da5c95cadfb3.png)
we are done.
.
is odd for 
.
denote the sum of the digits of 
written in binary. It is well known that ![\[v_2\left(\binom{a+b}{a}\right) = s_2(a+b)-s_2(a)-s_2(b).\]](http://latex.artofproblemsolving.com/0/6/b/06b06c267662dab5dd4a4607bda6fb1c0a6ac508.png)
Using this, we see that ![\[v_2\left(\binom{2^n-1}{k}\right) = s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k).\]](http://latex.artofproblemsolving.com/9/2/5/9254a96e9d7a786f02dfec062be74e2a02973923.png)
Since 
is ones written in binary and , we see that the set of ones in the binary representation of 
is exactly the complement of those in the binary representation for , so in fact ![\[s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k)=0.\]](http://latex.artofproblemsolving.com/c/f/d/cfd98e0e55f5d5b8b1fc43ba24f55c3b82bc11cb.png)
This proves the claim. 
, with 
trivial. Now suppose the problem is true for . Before proving 
, we need to prove the following intermediate claim.![\[(-1)^a\binom{2^n-1}{a}\]](http://latex.artofproblemsolving.com/2/9/3/29313d350adb54ac6cd45ca30c0572165ac35311.png)
generate all the odd residues modulo as varies from 
to .![\[(-1)^{a_1}\binom{2^n-1}{a_1} \equiv (-1)^{a_2}\binom{2^n-1}{a_2}\pmod{2^{n+1}}\]](http://latex.artofproblemsolving.com/a/4/b/a4b46fa14e72f3eea7797beece8b2603e5ae0454.png)
implies that 
. The inductive hypothesis implies that in fact, 
generates all the odd residues mod as varies from to 
, and a similar statement holds if we replace 
with 
. Thus, if and have the same parity, we are done trivially by simply looking modulo . is even and is odd. Note that ![\[\binom{2^n-1}{a_2}=\frac{2^n-a_2}{a_2}\binom{2^n-1}{a_2-1},\]](http://latex.artofproblemsolving.com/b/0/d/b0dd0aacf234fe2a70972ff50f40c219911177c2.png)
and that ![\[\frac{2^n-a_2}{a_2} = \frac{1}{a_2}2^n - 1 \equiv 2^n-1\pmod{2^{n+1}},\]](http://latex.artofproblemsolving.com/d/0/6/d060946879f974f8a2f753550fc2f5e0a1bd5e04.png)
so ![\[\binom{2^n-1}{a_2}\equiv (2^n-1) \binom{2^n-1}{a_2-1}\pmod{2^{n+1}}.\]](http://latex.artofproblemsolving.com/0/e/7/0e7fb998dadc6c2959412322f2ae2afc56d381ce.png)
Taking our original equation mod combined with the above implies that ![\[\binom{2^n-1}{a_1}\equiv \binom{2^n-1}{a_2-1}\pmod{2^n},\]](http://latex.artofproblemsolving.com/d/4/1/d417f42cd36a574470196b61133d2c115635a52c.png)
and since both and 
are even, the inductive hypothesis implies that 
. However, this them implies ![\[\binom{2^n-1}{a_2}\equiv (1-2^n)\binom{2^n-1}{a_2-1}\pmod{2^{n+1}},\]](http://latex.artofproblemsolving.com/2/0/9/209f583df914f5dada911c0cf33956cabbb609a8.png)
which is a clear contradiction as we can cancel out the binomial coefficients since they are odd. This is the desired contradiction, so and must in fact have the same parity, which we already resolved. This completes the proof. ![\[\binom{2^{n+1}-1}{2a} = \frac{\prod_{i=0}^{a-1}(2^{n+1}-1-2i)}{\prod_{i=0}^{a-1}(1+2i)}\binom{2^n-1}{a}\equiv (-1)^a\binom{2^n}{a}\pmod{2^{n+1}},\]](http://latex.artofproblemsolving.com/b/6/4/b64421642da038f73f72f49041bff9d452ae3a9f.png)
so ![\[\binom{2^{n+1}-1}{2a}\]](http://latex.artofproblemsolving.com/a/1/a/a1a76efbe64c635d87de368280fb4627d854597d.png)
generates all odd residues mod , which completes the induction, and thus the proof.
where 
hence 
meaning that 
which means that 
and therefore is odd for all 
.
, assume for the sake of contradiction otherwise.
among 
such that 
as odds have inverses.
.
and 
, then 
for odd positive integers 
such that without loss of generality 
, then as 
, 
as well and hence 
is among the integers achieving a higher , as desired. 
and consider the positive integer with maximal 
among 
for which 
because we are only considering 
. Then all but one of the fractions are of the form 
for odd 
and 
meaning that 
for some odd which therefore means that 
as 
we must have that 
which because is odd implies that 
contradicting the fact that , as desired. 
is odd for all 
. This follows by Lucas' Theorem because 
, and thus subtracting any integer from it does not induce any carrying in base two.
for 
. Divide both sides of the congruence by , which is valid as the number is odd: 
We induct on , with the stronger hypothesis that this expression cannot be congruent to 1 modulo , and furthermore that it cannot be congruent to modulo unless 
. The base case 
can be checked manually, and furthermore, notice that any two consecutive terms 
if is odd, and we may reduce to the case if is even, so the base case holds.
such that the expression is congruent to or 1 modulo . Because 
for all , we may remove terms on the numerator and denominator that are both odd and cancel out a factor of 2 from the remaining terms: 
for some positive integers 
. Notice that 
and obviously 
, so we may apply the inductive hypothesis to obtain 
If there is only one term in the reduced expansion that is congruent to 
, notice that 
so the quotient cannot be congruent to either. This yields a contradiction, so the inductive step is complete.. It follows they must be congruent to 
in some order, as required.
, so 
. Now by induction we prove that 
, base case evident, with inductive step following 
similarly we deduce 
. The problem is finished by inductive hypothesis of only odd residues appearing in tandem with noting that 
, which follows from basic algebra.
we have
Then,![\begin{align*}
{2^{n}-1 \choose j} - {2^{n}-1 \choose i}
& = \frac{(2^{n}-1)! ((2^{n}-1-i)!i! - (2^{n}-1-j)!j!)}{i! j! (2^{n}-1-i)! (2^{n}-1-j)!}\\
& = {2^{n}-1 \choose i} \cdot \frac{i!(2^{n}-1-j)! [(2^{n}-i-1) \cdots (2^{n}-j)-j \cdots (i+1)]}{j! (2^{n}-1-j)!} \\
& = {2^{n}-1 \choose i} \cdot \frac{i!}{j!} \cdot \left(\prod_{k = i+1}^{j} (2^{n}-k) - \prod_{k = i+1}^{j} k \right).
\end{align*}](http://latex.artofproblemsolving.com/b/0/2/b021fdccee7fe14229f97f39919cd4174aceee70.png)
Notice that 
is even, so we can rewrite the first product as 
So, when considering the number of factors of 
the expression is just
has a negative -adic.
Consider the largest such that there exists an 
with 
and 
We claim that only such one can exist. Considering the smallest such 
if there are more possible values, then 
must work as well, as it clearly has -adic equal to 
But, then, if both of these numbers are in the range ![$[i+1,j],$](http://latex.artofproblemsolving.com/6/7/4/6741e8e769f95c70c99dcb9ff9f5f5c22abc49dc.png)
then so is 
which has -adic at least 
a contradiction. So,
and thus we have proven the desired. 
and so
are 
and 
is true. Now, assuming the statement for 
is true, we will prove it for 
Note that 
Hence, if we define 
it suffices to prove that there don't exist such that 
Note that for even 
this is equivalent to the previous sequence, and for odd it is 
By induction hypothesis, we are done.
be a positive integer. Show that the numbersare congruent modulo to , , , , in some order.
with 
and 
. Note that obviously works, so . Firstly, we see that for each integer in ![$[0,2^n - 1]$](http://latex.artofproblemsolving.com/5/7/7/577b912f0ea8c24b70c4e401c23f703c0160cf63.png)
, 
is odd by directly taking 
: by Legendre's it becomes 
and then since you can't have any carries, this result is .
of ![\[X = (a+1) (a+2) \cdots b - (2^n - (a+1))(2^n - (a+2)) \cdots (2^n - b)\]](http://latex.artofproblemsolving.com/a/a/9/aa9d41b96ea78a18926ed6a541ab6f78ac25601b.png)
as 
has a of 
. can be rewritten as ![\[ (a+1) (a+2) \cdots b - ((a+1) - 2^n)((a+2) - 2^n) \cdots (b - 2^n) \]](http://latex.artofproblemsolving.com/2/8/b/28b07bef0ed77bece0541f5631cf9e8c1748f240.png)
when treated as a polynomial in . Also let be the integer in ![$[a+1, b]$](http://latex.artofproblemsolving.com/1/e/3/1e333034c46b049c72210780a17c69e1788a94e2.png)
with maximal (exists because if there were two or more, take the closest two and then average them). that has a minimal is equal to 
., just replace the with some 
that is not included in the term, and we decrease the , so it does not have a minimal . Therefore, the term must have exactly one copy of , after which we find that the term is equal to 
for some integer ![$i \in [a+1,b]$](http://latex.artofproblemsolving.com/f/5/8/f589561233b4268a84cd7c9495a5127dac4396dc.png)
, so we need to maximize 
and this holds uniquely at 
. 
![\[\nu_2(X) = \nu_2\left (2^n \cdot \frac{(a+1)(a+2) \cdots b}{k} \right) > \nu_2( 2^n \cdot (a+1)(a+2) \cdots b) = n + \nu_2((a+1)(a+2) \cdots b) ,\]](http://latex.artofproblemsolving.com/1/3/6/1365db569ea031d2c83eaf39a41f983d2b41b752.png)
absurd (where the inequality follows from the fact that 
are of the same parity, so there has to be an even number in ).
for 
,
modulus 
modulus 
.
will become something of the form 
which is clearly easier to reduce 
,
for all 
,
(as in any of the other terms, one of the factors of 
where ![\[\nu_2(X) = \nu_2(2(a+1)(a+2) \cdots b) = \nu_2((a+1)(a+2) \cdots b) + 1 < \nu_2((a+1)(a+2) \cdots b) + n ,\]](http://latex.artofproblemsolving.com/3/f/5/3f57ce3cbdefe8e87a79fdf371bc593ac2304a49.png)
absurd.