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Revision as of 09:48, 4 December 2023
Contents
Problem
An arithmetic sequence of positive integers has terms, initial term
, and common difference
. Carl wrote down all the terms in this sequence correctly except for one term, which was off by
. The sum of the terms he wrote was
. What is
?
Solution 1
Since one of the terms was either more or
less than it should have been, the sum should have been
or
The formula for an arithmetic series is This can quickly be rederived by noticing that the sequence goes
, and grouping terms.
We know that or
. Let us now show that
is not possible.
If , we can simplify this to be
Since every expression in here should be an integer, we know that either
and
or
and
The latter is not possible, since
and
The former is also impossible, as
Thus,
.
We can factor as
. Using similar reasoning, we see that
can not be paired as
and
, but rather must be paired as
and
with a factor of
somewhere.
Let us first try Our equation simplifies to
We know that
so we try the smallest possible value:
This would give us
(Indeed, this is the only possible
.)
There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer
For the sake of completeness, we can explore It turns out that we reach a contradiction in this case, so we are done.
~Technodoggo
Solution 2
There are terms, the
th term is
, summation is
.
The summation of the set is . First,
: its only possible factors are
, and as said by the problem,
, so
must be
or
. Let's start with
. Then,
, and this means
,
. Summing gives
. We don't need to test any more cases, since the problem writes that all
are the same, so we don't need to test other cases.
-HIA2020
Solution 3
We must have the sum of terms of the arithmetic sequence is , which is
or
.
Since we have is prime, it cannot be the sum of the arithmetic sequence.
We see that is just
.
We can write any arithmetic sequence with an odd amount of terms like this:
where b is the middle term and d is the common difference.
By the sum of an arithmetic sequence, we have and
and therefore
or
.
Then .
We must have that m is either or
, so m is either
or
.
So or
.
Taking, we have no answer choices that give
, and then taking
gives the only answer that works is
.
Therefore we have
~ESAOPS
Solution 4
The formula for the sum of an arithmetic sequence is , where
is the first term,
is the last term, and
is the number of terms. Let
be the first term,
be the common difference, and
be the number of terms of Carl's sequence. Since the sum the sequence is
less or
more than
, we have
The right-hand side is either
or
. We know that it has to be divisible by
so we can find the factors of
and
. Checking all the primes less than
, we find that
is prime and
.
Because , the sum must be
and the only possible values of
are
and
. We can test both cases.
Case 1:
Substituting for
gives us
. Since the sequence consists of only positive integers,
is an integer. We know that
but if
, then
. Hence, this case is not possible.
Case 2:
Substituting for
gives us
. Using the same logic from case 1, we get
, so
. Solving for
, we get
. Therefore,
, so the answer is
.
~azc1027
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=F30LJeoaNWo
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.