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
are such that 
, find the minimum value of expression 
of prime positive integers 
such that number 
equals to a square of an integer.
be real number greater than 
. Let![\[S=\log_a {bc}+\log_b {ca}+\log_c {ab}\]](http://latex.artofproblemsolving.com/5/9/e/59e1a33017116963165f744b49b8515f126c64ef.png)
.
for which:
for all 
.
, with 
, is inscribed in circle 
and point 
lies inside triangle 
. Rays 
and 
meet again at points 
and 
, respectively. Let the line through parallel to 
intersects 
and 
at points 
and 
, respectively. Prove that quadrilateral 
is cyclic if and only if 
bisects 
.
such that 
, 
, and 
are all equal to
, 
be real numbers, such that 
. Prove that 
.
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be the incircle of triangle 
. Line 
is parallel to side 
and tangent to . Line 
is parallel to side 
and tangent to . It turned out that the intersection point of and lies on circumcircle of 
, let![\[a_n = \frac {(n + 9)!}{(n - 1)!}.\]](http://latex.artofproblemsolving.com/f/d/5/fd5a5ee0252b440fd5fcd9694ac0b00de61b83b2.png)
denote the smallest positive integer for which the rightmost nonzero digit of 
is odd. The rightmost nonzero digit of is
.
, we must have an odd number. This further implies that 
, the function is LTE. For powers of two, a product of 10 consecutive numbers will divide at least 
by tracking factors 
. Intuitively, to minimize , we need one of the these ten factors to be a power of 
, so that another factor will be congruent to 
. Note that to satisfy 
, we cannot have any factor divisible by 
. Also, 
, and to minimize , we make 
. Quick checking ensures that this satisfies our two LTE functions. Dividing 
by 
and taking 
, we get 
. Now, simple inspection shows that the last digit of 
must be 9, since 
and all odd last digits multiplied by 256 yields different remainders . Hence, the answer is 
.
if 
.
factors of 2. But this product must also contain at most two multiples of 5, including one that is only once divisible by 5. For the rightmost non-zero digit to be add, the product must by divisible by 5 at least at many times as it is divisible by 2, hence, there must be a multiple of 
.
factors of 2, which means that, since it contains more factors of 2 than of 5, the rightmost non-zero digit is even. However, the product
factors of 2 and the same number of factors of 5. This is the smallest such product with the desired property. To determine the rightmost non-zero digit, we divide out the extra 8 zeros and, since we know this digit is odd, take modulo 5 (as oppose to mod 10):
simply goes away thanks to Fermat's Little Theorem. Since this quantity is 4 mod 5, it must be that the last non-zero digit was 9, making E the answer.
must necessarily be divisible by is 2*4*2*8*2
.
,
is the highest power of 10 that 
(or its multiple).
,
.
.