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: | |
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− | 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. | ||
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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>. | 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/PQnA07go4GM | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/zTGuz6EoBWY | ||
+ | |||
+ | ~IceMatrix | ||
==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 07:23, 4 November 2021
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 .
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
~IceMatrix
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 |
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