Difference between revisions of "2000 AMC 8 Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | Casework by the units digit <math>u</math> will help | + | Casework by the units digit <math>u</math> will help organize the answer. |
<math>u=0</math> gives no solutions, since no real numbers are divisible by <math>0</math> | <math>u=0</math> gives no solutions, since no real numbers are divisible by <math>0</math> | ||
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<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=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 | + | <math>u=4</math> has <math>2</math> solutions: <math>24</math> and <math>44</math>. Adding or subtracting <math>10</math> will kill divisibility 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=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 | + | <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 divisibility 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=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 | + | <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 divisible 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>. | <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> | + | 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>, which is 17. |
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/m5D5-2YB0tI Soo, DRMS, NM | ||
==See Also== | ==See Also== |
Latest revision as of 15:59, 1 January 2024
Contents
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 organize 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 divisibility 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 divisibility 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 divisible by
.
has no solutions.
and
are the smallest multiples of
that end in
.
Totalling the solutions, we have solutions, giving the answer
, which is 17.
Video Solution
https://youtu.be/m5D5-2YB0tI Soo, DRMS, NM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.