Clements MA
Problem sources:
1-4: Math Prize for Girls 2009
5-6: AoPS
Solution to rt3 problem:
Advanced Number Theory? Handout Solutions
10/02/13
Clements MA
Advanced Number Theory? Handout Solutions
Solution to 2009^4 problem:
10/02/13
Clements MA
Solution to the problem with x^x:
Advanced Number Theory? Handout Solutions
10/02/13
Clements MA
Advanced Number Theory? Handout Solutions
10/02/13
Solution to abc problem:
27000001 problem:
Solution.
Note that
Plugging in
so the answer is
yields
. From trial and error, we see that
.
,
Clements MA
Advanced Number Theory? Handout Solutions
10/02/13
Solution to the 888 problem:
Solution.
Let
end in
digits of
. Since we are considering congruence modulo
. Therefore, we can assume that
where
are digits. Modulo
We want to find digits
such that
must be congruent to
must end in
. This means that
must be congruent to
First consider the case
so
ends in
Next, consider the case
so
ends in
Therefore, the answer is
,
, we have
the last digit
digit
, we only care about the last three
, so suppose
ends in
modulo
, which forces
must end in
modulo
, which forces us to have
. Now, note that the last
or
. Then, we want
or
. The smallest possible
. Then, we want
.
.
to end in
, and hence
, and hence
. First, note that
. Plugging this in, we see that
or
in this case is thus
to end in
. The smallest possible
,
in this case is thus
.
,
.
Clements MA
Advanced Number Theory? Handout Solutions
10/02/13