Difference between revisions of "2016 AMC 10B Problems/Problem 18"
Coltsfan10 (talk | contribs) |
Coltsfan10 (talk | contribs) (→Solution 2.1) |
||
| Line 28: | Line 28: | ||
==Solution 2.1== | ==Solution 2.1== | ||
| − | At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to <math>30</math>, and that is because <math>\frac{345}{30}=11.5</math>, which means <math>11</math> and <math>12</math> must be the middle 2 numbers; however, a sequence of length <math>30</math> with middle numbers <math>11</math> and <math>12</math> would go into the negatives, so any number from 30 onwards wouldn't work, and since <math>1</math> is a trivial, non-counted solution, we get <math>\boxed{\textbf{(E) }7}</math> | + | At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to <math>30</math>, and that is because <math>\frac{345}{30}=11.5</math>, which means <math>11</math> and <math>12</math> must be the middle 2 numbers; however, a sequence of length <math>30</math> with middle numbers <math>11</math> and <math>12</math> would go into the negatives, so any number from 30 onwards wouldn't work, and since <math>1</math> is a trivial, non-counted solution, we get <math>\boxed{\textbf{(E) }7}</math> -ColtsFan10 |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}} | {{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:40, 23 June 2020
Problem
In how many ways can 
be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution 1
Factor 
.
Suppose we take an odd number 
of consecutive integers, with the median as 
. Then 
with 
.
Looking at the factors of , the possible values of are 
with medians as 
respectively.
Suppose instead we take an even number 
of consecutive integers, with median being the average of and 
. Then 
with 
.
Looking again at the factors of , the possible values of are 
with medians 
respectively.
Thus the answer is 
.
Solution 2
We need to find consecutive numbers (an arithmetic sequence that increases by 
) that sums to . This calls for the sum of an arithmetic sequence given that the first term is , the last term is 
and with 
elements, which is: 
.
So, since it is a sequence of consecutive numbers starting at and ending at 
. We can now substitute with . Now we substiute our new value of into to get that the sum is 
.
This simplifies to 
. This gives a nice equation. We multiply out the 2 to get that 
. This leaves us with 2 integers that multiplies to 
which leads us to think of factors of . We know the factors of are: 
. So through inspection (checking), we see that only 
and 
work. This gives us the answer of ways.
~~jk23541
Solution 2.1
At the very end of Solution 2, where we find the factors of 690, instead of inspection, notice that all numbers will work until you get to 
, and that is because 
, which means 
and 
must be the middle 2 numbers; however, a sequence of length with middle numbers and would go into the negatives, so any number from 30 onwards wouldn't work, and since is a trivial, non-counted solution, we get -ColtsFan10
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
