Difference between revisions of "2000 AMC 8 Problems/Problem 11"
Mrdavid445 (talk | contribs) |
Talkinaway (talk | contribs) (Solution) |
||
Line 1: | Line 1: | ||
− | The number <math>64</math> has the property that it is divisible by | + | ==Problem== |
+ | |||
+ | The number <math>64</math> has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property? | ||
<math>\textbf{(A)}\ 15 \qquad | <math>\textbf{(A)}\ 15 \qquad | ||
Line 6: | Line 8: | ||
\textbf{(D)}\ 18 \qquad | \textbf{(D)}\ 18 \qquad | ||
\textbf{(E)}\ 20</math> | \textbf{(E)}\ 20</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Casework by the units digit <math>u</math> will help organise the answer. | ||
+ | |||
+ | <math>u=0</math> gives no solutions, since no real numbers are divisible by <math>0</math> | ||
+ | |||
+ | <math>u=1</math> has <math>4</math> solutions, since all numbers are divisible by <math>1</math>. | ||
+ | |||
+ | <math>u=2</math> has <math>4</math> solutions, since every number ending in <math>2</math> is even (ie divisible by <math>2</math>). | ||
+ | |||
+ | <math>u=3</math> has <math>1</math> solution: <math>33</math>. <math>\pm 10</math> or <math>\pm 20</math> will retain the units digit, but will stop the number from being divisible by <math>3</math>. <math>\pm 30</math> is the smallest multiple of <math>10</math> that will keep the number divisible by <math>3</math>, but those numbers are <math>3</math> and <math>63</math>, which are out of the range of the problem. | ||
+ | |||
+ | <math>u=4</math> has <math>2</math> solutions: <math>24</math> and <math>44</math>. Adding or subtracting <math>10</math> will kill divisiblty by <math>4</math>, since <math>10</math> is not divisible by <math>4</math>. | ||
+ | |||
+ | <math>u=5</math> has <math>4</math> solutions: every number ending in <math>5</math> is divisible by <math>5</math>. | ||
+ | |||
+ | <math>u=6</math> has <math>1</math> solution: <math>36</math>. <math>\pm 10</math> or <math>\pm 20</math> will kill divisibility by <math>3</math>, and thus kill divisiblilty by <math>6</math>. | ||
+ | |||
+ | <math>u=7</math> has no solutions. The first multiples of <math>7</math> that end in <math>7</math> are <math>7</math> and <math>77</math>, but both are outside of the range of this problem. | ||
+ | |||
+ | <math>u=8</math> has <math>1</math> solution: <math>48</math>. <math>\pm 10, \pm 20, \pm 30</math> will all kill divisibility by <math>8</math> since <math>10, 20, </math> and <math>30</math> are not divisibile by <math>8</math>. | ||
+ | |||
+ | <math>u=9</math> has no solutions. <math>9</math> and <math>99</math> are the smallest multiples of <math>9</math> that end in <math>9</math>. | ||
+ | |||
+ | Totalling the solutions, we have <math>0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17</math> solutions, giving the answer <math>\boxed{C}</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC8 box|year=2000|num-b=10|num-a=12}} |
Revision as of 19:52, 30 July 2011
Problem
The number has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property?
Solution
Casework by the units digit will help organise the answer.
gives no solutions, since no real numbers are divisible by
has
solutions, since all numbers are divisible by
.
has
solutions, since every number ending in
is even (ie divisible by
).
has
solution:
.
or
will retain the units digit, but will stop the number from being divisible by
.
is the smallest multiple of
that will keep the number divisible by
, but those numbers are
and
, which are out of the range of the problem.
has
solutions:
and
. Adding or subtracting
will kill divisiblty by
, since
is not divisible by
.
has
solutions: every number ending in
is divisible by
.
has
solution:
.
or
will kill divisibility by
, and thus kill divisiblilty by
.
has no solutions. The first multiples of
that end in
are
and
, but both are outside of the range of this problem.
has
solution:
.
will all kill divisibility by
since
and
are not divisibile by
.
has no solutions.
and
are the smallest multiples of
that end in
.
Totalling the solutions, we have solutions, giving the answer
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |