Difference between revisions of "2017 AMC 10B Problems/Problem 1"
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<math>\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15</math> | <math>\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15</math> | ||
− | + | ==Solution 1== | |
− | + | Let her <math>2</math>-digit number be <math>x</math>. Multiplying by <math>3</math> makes it a multiple of <math>3</math>, meaning that the sum of its digits is divisible by <math>3</math>. Adding on <math>11</math> increases the sum of the digits by <math>1+1 = 2,</math> (we can ignore numbers such as <math>39+11=50</math>) and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be <math>2</math> more than a multiple of <math>3</math>. There are two such numbers between <math>71</math> and <math>75</math>: <math>71</math> and <math>74.</math> Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed: | |
− | |||
− | Let her <math>2</math>-digit number be <math>x</math>. Multiplying by <math>3</math> makes it a multiple of <math>3</math>, meaning that the sum of its digits is divisible by <math>3</math>. Adding on <math>11</math> increases the sum of the digits by <math>1+1 = 2,</math> and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be <math>2</math> more than a multiple of <math>3</math>. There are two such numbers between <math>71</math> and <math>75</math>: <math>71</math> and <math>74.</math> Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed: | ||
<cmath></cmath> | <cmath></cmath> | ||
For <math>71,</math> we reverse the digits, resulting in <math>17.</math> Subtracting <math>11</math>, we get <math>6.</math> We can already see that dividing this by <math>3</math> will not be a two-digit number, so <math>71</math> does not meet our requirements. | For <math>71,</math> we reverse the digits, resulting in <math>17.</math> Subtracting <math>11</math>, we get <math>6.</math> We can already see that dividing this by <math>3</math> will not be a two-digit number, so <math>71</math> does not meet our requirements. | ||
Line 18: | Line 16: | ||
Therefore, our answer is <math>\boxed{\bold{(B)} 12}</math>. | Therefore, our answer is <math>\boxed{\bold{(B)} 12}</math>. | ||
− | + | ==Solution 2== | |
Working backwards, we reverse the digits of each number from <math>71</math>~<math>75</math> and subtract <math>11</math> from each, so we have | Working backwards, we reverse the digits of each number from <math>71</math>~<math>75</math> and subtract <math>11</math> from each, so we have | ||
<cmath>6, 16, 26, 36, 46</cmath> | <cmath>6, 16, 26, 36, 46</cmath> | ||
The only numbers from this list that are divisible by <math>3</math> are <math>6</math> and <math>36</math>. We divide both by <math>3</math>, yielding <math>2</math> and <math>12</math>. Since <math>2</math> is not a two-digit number, the answer is <math>\boxed{\textbf{(B)}\ 12}</math>. | The only numbers from this list that are divisible by <math>3</math> are <math>6</math> and <math>36</math>. We divide both by <math>3</math>, yielding <math>2</math> and <math>12</math>. Since <math>2</math> is not a two-digit number, the answer is <math>\boxed{\textbf{(B)}\ 12}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | You can just plug in the numbers to see which one works. When you get to <math>12</math>, you multiply by <math>3</math> and add <math>11</math> to get <math>47</math>. When you reverse the digits of <math>47</math>, you get <math>74</math>, which is within the given range. Thus, the answer is <math>\boxed{\textbf{(B)}\ 12}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 4 (Fastest Way)== | ||
+ | Let x be the original number. The last digit of <math>3x+11</math> must be <math>7</math> so the last digit of <math>3x</math> must be <math>6</math>. The only answer choice that satisfies this is <math>\boxed{\textbf{(B)}\ 12}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Subtract <math>11</math> from the numbers <math>71</math> through <math>75</math>. This yields <math>71-11 = 60</math>, <math>72-11 = 61</math>, <math>73-11 = 62</math>, <math>74-11 = 63</math>, and <math>75-11 = 64</math>. Of these, the only ones divisible by <math>3</math> are <math>60</math> and <math>63</math>. Therefore, the only possible values are <math>71</math> and <math>74</math>. Switching the digits of each, we get <math>17</math> and <math>47</math>. Subtracting <math>11</math> from each, we get the numbers <math>6</math> and <math>36</math>. Dividing each by <math>3</math>, we get <math>2</math> and <math>12</math>. The only two-digit number is <math>12</math>, so the answer is <math>\boxed{\textbf{(B)}\ 12}</math>. | ||
+ | |||
+ | ~TheGoldenRetriever | ||
+ | |||
+ | ==Video Solution (HOW TO CRITICALLY THINK!!!)== | ||
+ | https://youtu.be/EbFw47vZASs | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/PQnA07go4GM | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/zTGuz6EoBWY | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|before=First Problem|num-a=2}} | {{AMC10 box|year=2017|ab=B|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:51, 20 August 2023
Contents
Problem
Mary thought of a positive two-digit number. She multiplied it by and added
. Then she switched the digits of the result, obtaining a number between
and
, inclusive. What was Mary's number?
Solution 1
Let her -digit number be
. Multiplying by
makes it a multiple of
, meaning that the sum of its digits is divisible by
. Adding on
increases the sum of the digits by
(we can ignore numbers such as
) and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be
more than a multiple of
. There are two such numbers between
and
:
and
Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed:
For
we reverse the digits, resulting in
Subtracting
, we get
We can already see that dividing this by
will not be a two-digit number, so
does not meet our requirements.
Therefore, the answer must be the reversed steps applied to
We have the following:
Therefore, our answer is
.
Solution 2
Working backwards, we reverse the digits of each number from ~
and subtract
from each, so we have
The only numbers from this list that are divisible by
are
and
. We divide both by
, yielding
and
. Since
is not a two-digit number, the answer is
.
Solution 3
You can just plug in the numbers to see which one works. When you get to , you multiply by
and add
to get
. When you reverse the digits of
, you get
, which is within the given range. Thus, the answer is
.
Solution 4 (Fastest Way)
Let x be the original number. The last digit of must be
so the last digit of
must be
. The only answer choice that satisfies this is
.
Solution 5
Subtract from the numbers
through
. This yields
,
,
,
, and
. Of these, the only ones divisible by
are
and
. Therefore, the only possible values are
and
. Switching the digits of each, we get
and
. Subtracting
from each, we get the numbers
and
. Dividing each by
, we get
and
. The only two-digit number is
, so the answer is
.
~TheGoldenRetriever
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
Video Solution
Video Solution
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.