Difference between revisions of "2007 AMC 10B Problems/Problem 8"
(→Problem 8) |
|||
Line 16: | Line 16: | ||
It's pretty clear that there's a pattern: <math>8</math> sets, <math>6</math> sets, <math>4</math> sets. The amount of sets per case decreases by <math>2</math>, so it's obvious Case <math>4</math> has <math>2</math> sets. The total amount of possible five-digit numbers is <math>8+6+4+2=\boxed{\textbf{(D)}\ 20}</math>. | It's pretty clear that there's a pattern: <math>8</math> sets, <math>6</math> sets, <math>4</math> sets. The amount of sets per case decreases by <math>2</math>, so it's obvious Case <math>4</math> has <math>2</math> sets. The total amount of possible five-digit numbers is <math>8+6+4+2=\boxed{\textbf{(D)}\ 20}</math>. | ||
− | |||
− | |||
− | |||
− | |||
− |
Revision as of 12:06, 5 November 2023
CHIKEN NUGGIEs
Solution 2
Case : The numbers are separated by
.
We this case with and
. Following this logic, the last set we can get is
and
. We have
sets of numbers in this case.
Case : The numbers are separated by
.
This case starts with and
. It ends with
and
. There are
sets of numbers in this case.
Case : The numbers start with
and
. It ends with
and
. This case has
sets of numbers.
It's pretty clear that there's a pattern: sets,
sets,
sets. The amount of sets per case decreases by
, so it's obvious Case
has
sets. The total amount of possible five-digit numbers is
.