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.
+1 w
, 
and 
be positive real numbers such that 
. Prove the inequality
, the number 
is the product of at least 
(not necessarily distinct) primes.
x square composed of 
unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement 
.
be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement than of arrangement . Prove that arrangement can be obtained from 
by performing a number of switches, defined as follows:
have the property that both![\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]](http://latex.artofproblemsolving.com/d/e/c/dec527c4673470e7023d6aed00fac90268bd2fcc.png)
are consecutive perfect squares.
is the incenter and 
is the circumcenter. The line 
intersects the lines 
at points 
respectively. Let 
be the intersection of 
and 
. The points 
and 
are defined similarly. The incircle of 
is tangent to sides at points 
respectively. Let the lines 
and 
intersect at points 
respectively. Prove that the circles with diameters 
and 
have a common point.
, 
, 
, and 
lie on a line in that order, and points 
and 
are located outside the line such that 
, 
and 
. Let the circumcircles of triangles 
and 
intersect at points 
and 
, and the circumcircles of triangles 
and 
intersect at points 
and 
. Prove that the lines 
and 
pass through the midpoint of segment 
.
, 
, 
, 
of real numbers satisfies![\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]](http://latex.artofproblemsolving.com/4/d/a/4da3b88a20c42c1798141b8db086de341cfb9a67.png)
for every positive integer . Prove that there exists a real number such that 
for every positive integer .
be positive real numbers such that 
. Find the minimum value of the following sum :
knowing that the denominators are positive real numbers.
is inscribed in a circle whose center lies inside it.
, then the diagonals of the quadrilateral are perpendicular.
such that the following two conditions hold: and satisfying 
, We have 
for all real numbers 
.
and 
, there exist real numbers and , such that and 
.
be the least positive integer for which 
is divisible by 
. Find the number of positive divisors of .
don't solve for 
.
for some m, and then I continued on with LTE.
from 
and 
from doing some stuff with Binomial Expansion. I just guessed 
and 
.
.
is 
you can check that 
.
implying 
.
.
though
.
Similarly, for 
we have 
The condition 
is slightly trickier because we cannot apply LTE directly. However, notice that 
Let 
since 
and 
This means that 
.
giving 
.
.
and 
divide 
.
Therefore, 
.
Therefore, 
.
to be divisible by 
,
,
.
Therefore, 
.
![\[\nu_3(149^n - 2^n) = \nu_3(149-2) + \nu_3(n) = 1+\nu_3(n) \implies 3^2 \mid n\]](http://latex.artofproblemsolving.com/f/8/3/f8381b8946dd9d55048ed5d956793476b8b8b639.png)
![\[\nu_7(149^n-2^n) = \nu_7(149-2) + \nu_7(n) = 2+\nu_7(n) \implies 7^5 \mid n\]](http://latex.artofproblemsolving.com/7/f/3/7f391510cb699426b44162291c4c28387cced1d2.png)
![\[149^n-2^n \equiv 2^n(2^n-1) \implies 4 \mid n\]](http://latex.artofproblemsolving.com/4/b/6/4b607f7ee983cf393d6c4e66923523098f39f9fc.png)
,![\[\nu_5((149^4)^k-(2^4)^k) = \nu_5(149^4-2^4) + \nu_5(k) = 1+\nu_5(k) \implies 5^4 \mid k\]](http://latex.artofproblemsolving.com/0/2/c/02c8e1b5d299a15a57eba3dbb15264e694fcbda4.png)
![\[2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5 = 378157500 \implies \boxed{270}.\]](http://latex.artofproblemsolving.com/2/4/0/240dc1bad853b1d3575fca2980240d1bd7b407ca.png)
case first. We must have 
for ![\[v_5(149^{4k}-2^{4k}) = v_5(149^4-2^4) + v_5(k) \implies v_5(k) \ge 4.\]](http://latex.artofproblemsolving.com/f/a/0/fa026fe79e3f54a763a2ec19158d13e7170fdb60.png)
and 
,![\[v_3(149^n-2^n) = v_3(149-2) + v_3(n) \implies v_3(n) \ge 2.\]](http://latex.artofproblemsolving.com/a/6/d/a6d1100ef3589c853d7ce2ea974af0ffabcf551a.png)
![\[v_7(149^n-2^n) = v_7(149-2) + v_7(n) \implies v_7(n) \ge 5.\]](http://latex.artofproblemsolving.com/2/e/5/2e5b227bb43d2840354ca2a8ea5da499dae38df0.png)
.
if 
if and only if 
is the smallest number and it have 270 divisors
and 
.
check, we get that 
.
gives that 
.
from which 
satisfies which has 
divisors. 
.
.
case, note that 
,
.
,
.
,
.
,
.
.
and 
such that 
and 
and 
,
.
.
.
.
and 
.
.
and 
.
,
factors.
so 
and
so 
by LTE.
Now, notice that 
since otherwise the number of powers of 
Now, if 
we have that
Now, notice that 
while
Thus, we have that
so 
Thus, 
so our answer is 
.
.
,
.
.
won't work so we check 
.
Since 
this means that 
.
which has 
and 
,
we notice that 
thus LTE cannot be applied hence, we notice that 
fo
w
is any positive integer,
as 
and 
hence no. of positive divisor is 
Therefore, 
.
Therefore, 
.
Therefore, 
.
Apply LTE to find that 
is a multiple of 
.
so the Lifting the Exponent lemma yields
Also, we have 
so applying the Lifting the Exponent lemma again gives
Finally, note that the order of 
modulo 
so it follows that 
after which the Lifting the Exponent lemma yields
Hence, combining all of the previous results yields 
so the smallest possible value of 
which has
positive divisors. 
? how is the 4 coming?