1993 AIME Problems/Problem 4
Problem
How many ordered four-tuples of integers with
satisfy
and
?
Contents
Solution
Solution 1
Let so
. It follows that
. Hence
.
Solve them in terms of to get
. The last two solutions don't follow
, so we only need to consider the first two solutions.
The first solution gives us and
, and the second one gives us
.
So the total number of such quadruples is .
Solution 2
Let and
. From
,
.
Substituting ,
, and
into
,
Hence,
or
.
For , we know that
, so there are
four-tuples. For
,
, and there are
four-tuples. In total, we have
four-tuples.
Solution 3
Square both sides of the first equation in order to get and
terms, which we can plug
in for.
We can plug in for
to get
on the left side, and also observe that
after rearranging the first equation. Plug in
for
.
Now observe the possible factors of , which are
.
and
must be factors of
, and
must be greater than
.
work, and yields
possible solutions.
does not work, because if
, then
must differ by 2 as well, but an odd number
can only result from two numbers of different parity.
will be even, and
will be even, so
must be even.
works, and yields
possible solutions, while
fails for the same reasoning above.
Thus, the answer is
Solution 4
Add the two conditions together to get . Rearranging and factorising with SFFT,
. This implies that for every quadruple
, we can replace
,
, etc. and this will still produce a valid quadruple. This means, that we can fix
, and then just repeatedly add
to get the other quadruples.
Now, our conditions are and
. Replacing
in the first equation, we get
. Factorising again with SFFT gives
. Since
, we have two possible cases to consider.
Case 1: ,
. This produces the quadruple
, which indeed works.
Case 2: ,
. This produces the quadruple
, which indeed works.
Now, for case 1, we can add to each term exactly
times (until we get the quadruple
), until we violate
. This gives
quadruples for case 1.
For case 2, we can add to each term exactly
times (until we get the quadruple
). this gives
quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence possible quadruples.
Solution 5
Let . From the equation
, we have
so
and
. We then have
Since
,
, or
. Since the prime factorization of 93 is
, we must either have
and
, or
and
. We consider these cases separately.
If , then
,
, and
. Thus
can be any integer between 95 and 499, inclusive, and our choice of
determines the four-tuple
. We therefore have
possibilities in this case.
If , then
,
, and
. Thus
can be any integer between 35 and 499, inclusive, and our choice of
determines the four-tuple
, as before. We therefore have
possibilities in this case.
Since there are 405 possibilities in the first case and 465 possibilities in the second case, in total there are four-tuples.
Solution 6
Assume , and
. This clearly satisfies the condition that
since (
) . Now plug this into
. You get
Since (as given by the condition that
),
and
and
are integers, there are two cases we have to consider since
. We first have to consider
, and then consider
.
In the first case, we get and in the second case we get
. Now plug these values (in separate cases) back into
. Since the only restriction is that all numbers have to be greater than
or less than
, we can write two inequalities. Either
, or
(using the inequalities given by
and
, and since
and
are squeezed in between
and
, we only have to consider these two inequalities).
This gives us either or
, and using simple counting, there are
values for
in the first case and
values for
in the second case, and hence our answer is
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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