2013 AIME I Problems/Problem 11
Contents
Problem 11
Ms. Math's kindergarten class has registered students. The classroom has a very large number,
, of play blocks which satisfies the conditions:
(a) If ,
, or
students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers such that when
,
, or
students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of satisfying the above conditions.
Solution
Solution 1
must be some multiple of the
of
,
, and
; this
is hereby denoted
and
.
,
,
,
,
,
,
,
,
, and
all divide
, so
We have the following three modulo equations:
To solve the equations, you can notice the answer must be of the form where
is an integer.
This must be divisible by
, which is
.
Therefore, , which is an integer. Factor out
and divide to get
.
Therefore,
. We can use Bezout's Identity or a Euclidean algorithm bash to solve for the least of
and
.
We find that the least is
and the least
is
.
Since we want to factor , don't multiply: we already know that the prime factors of
are
,
,
, and
, and since
is prime, we have
.
Solution 2
Note that the number of play blocks is a multiple of the LCM of ,
, and
. The value of this can be found to be
. This number is also divisible by
,
,
,
,
,
,
,
,
, and
, thus, the three numbers
are
.
Thus, when taken mod
,
,
. Since
is congruent to
mod
and
mod
, and congruent to
mod
, the number
must be a number that is congruent to
mod
,
mod
(because
is a multiple of
, which is a factor of
that can be divided out) and cause
to become
when multiplied under modulo
.
Looking at the last condition shows that mod
(after a bit of bashing) and is congruent to
mod
and
mod
as previously noted. Listing out the numbers congruent to
mod
and
mod
yield the following lists:
mod
:
,
,
,
,
,
,
,
,
,
,
...
mod
:
,
,
,
,
,
,
,
,
,
,
,
,
...
Both lists contain elements where
is the modulo being taken, thus, there must be a solution in these lists as adding
to this solution yields the next smallest solution. In this case,
is the solution for
and thus the answer is
. Since
is prime, the sum of the prime factors is
.
Solution 3
It is obvious that and so the only mod
number of students are
. Therefore,
. Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two equations together, then we get
. Divide by
. Note that since the RHS is
, and since
is
, then
, where
is some nonnegative integer, because
must be
.
This reduces to . Now, take out the
With the same procedure,
, where
is some nonnegative integer.
You also get , at which point
.
cannot be equal to
. Therefore,
, and we know the prime factors of
are
so the answer is
.
Solution 4
We start by noticing that for some integer
in order to satisfy the first condition.
Next, we satisfy the second condition. Since must leave a remainder when dividing
, they are not divisors of
. Thus, we can eliminate all
s.t.
which leaves
. Thus,
. Now, we seek to find the least
which satisfies this set of congruences.
By Chinese Remainder Theorem on the first two congruences, we find that (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on
and
we find that
.
Thus, the minimal value of is possible at
. The prime factorization of this minimum value is
and so the answer is
.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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