Difference between revisions of "2016 AMC 10B Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | How many four-digit integers <math>abcd</math>, with <math>a \ | + | How many four-digit integers <math>abcd</math>, with <math>a \neq 0</math>, have the property that the three two-digit integers <math>ab<bc<cd</math> form an increasing arithmetic sequence? One such number is <math>4692</math>, where <math>a=4</math>, <math>b=6</math>, <math>c=9</math>, and <math>d=2</math>. |
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math> | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20</math> | ||
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==Solution== | ==Solution== |
Revision as of 11:26, 29 June 2016
Contents
[hide]Problem
How many four-digit integers , with
, have the property that the three two-digit integers
form an increasing arithmetic sequence? One such number is
, where
,
,
, and
.
Solution
The numbers are and
. Note that only
can be zero, and that
.
To form the sequence, we need . This can be rearranged as
. Notice that since the left-hand side is a multiple of
, the right-hand side can only be
or
. (A value of
would contradict
.) Therefore we have two cases:
and
.
Case 1
If , then
, so
. This gives
.
If
, then
, so
. This gives
.
If
, then
, so
, giving
.
There is no solution for
.
Case 2
This means that the digits themselves are in arithmetic sequence. This gives .
Counting the solutions, the answer is .
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.