Difference between revisions of "2013 AMC 10B Problems/Problem 25"
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<cmath>(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)</cmath> | <cmath>(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)</cmath> | ||
− | Since <math>N_6</math> has at least <math>3</math> digits, <math>(b_1,b_2)=(0,0)</math> doesn't work. Furthermore, when <math>b_1=5</math>, <math> | + | Since <math>N_6</math> has at least <math>3</math> digits, <math>(b_1,b_2)=(0,0)</math> doesn't work. Furthermore, when <math>b_1=5</math>, <math>216b_1</math> exceeds <math>1000</math> which is not possible as <math>N</math> is a three digits number, thus <math>(b_1,b_2)=(5,0)</math> won't work as well. |
Since we know that <math>a_i<5</math>, for each of the other ordered pairs, there is respectively one and only one solution <math>(a_1,a_2,a_3,a_5)</math> that satisfies the equation | Since we know that <math>a_i<5</math>, for each of the other ordered pairs, there is respectively one and only one solution <math>(a_1,a_2,a_3,a_5)</math> that satisfies the equation |
Revision as of 19:12, 7 August 2019
- The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25, so both problems redirect to this page.
Problem
Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer
. For example, if
, Bernardo writes the numbers
and
, and LeRoy obtains the sum
. For how many choices of
are the two rightmost digits of
, in order, the same as those of
?
Solution 1
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that
also that
Substituting these equations into the question and setting the units digits of and
equal to each other, it can be seen that
, and
, (otherwise
and
always have different parities) so
,
,
,
Therefore, can be written as
and
can be written as
Just keep in mind that can be one of five choices:
or
, ;
Also, we have already found which digits of
will add up into the units digits of
.
Now, examine the tens digit, by using
and
to find the tens digit (units digits can be disregarded because
will always work)
Then we take
and
to find the last two digits in the base
and
representation.
Both of those must add up to
()
Now, since will always work if
works, then we can treat
as a units digit instead of a tens digit in the respective bases and decrease the mods so that
is now the units digit.
Say that (m is between 0-6, n is 0-4 because of constraints on x)
Then
and this simplifies to
From careful inspection, this is true when
This gives you choices for
, and
choices for
, so the answer is
Solution 2 (Shortcut)
Notice that there are exactly possible values of
. This means, in
, every possible combination of
digits will happen exactly once. We know that
works because
.
We know for sure that the units digit will add perfectly every added or subtracted, because
. So we only have to care about cases of
every
subtracted. In each case,
subtracts
/adds
,
subtracts
and
adds
for the
's digit.
As we can see, there are cases, including the original, that work. These are highlighted in
. So, thus, there are
possibilities for each case, and
.
Solution 3
Notice that ranges from
to
digits and
ranges from
to
digits.
Then let ,
denotes the digits of
,
, respectively such that
Thus we have
Now we are given
Moving the
to left, we have
Since ,
determine the unit digits of the two sides of the congruence equation, we have
. Thus,
canceling out
, we have
Thus
is a multiple of
and the possibilities are
.
In addition, since is a three digits integer, the only values for
are
, or otherwise
will exceed 1000.
Now going back to our original equation
Since
,
Since the left hand side is a multiple of , then so does the right hand side. Thus
.
Since we already know that ,
, from where we also know that
is a multiple of
.
For , there is a total of 7 ordered pairs that satisfy the condition. Namely,
Since has at least
digits,
doesn't work. Furthermore, when
,
exceeds
which is not possible as
is a three digits number, thus
won't work as well.
Since we know that , for each of the other ordered pairs, there is respectively one and only one solution
that satisfies the equation
Thus there are five solutions to the equation. Also since we have 5 possibilities for , we have a total of
values for
.
~ Nafer
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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All AMC 10 Problems and Solutions |
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