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Difference between revisions of "2017 AMC 8 Problems/Problem 8"

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Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the <math>2</math>-digit number is even, and thus, the digit in the tens place must be <math>9</math>. The only even <math>2</math>-digit number starting with <math>9</math> and divisible by <math>7</math> is <math>98</math>, which has a units digit of <math>\boxed{\textbf{(D)}\ 8}.</math>
 
Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the <math>2</math>-digit number is even, and thus, the digit in the tens place must be <math>9</math>. The only even <math>2</math>-digit number starting with <math>9</math> and divisible by <math>7</math> is <math>98</math>, which has a units digit of <math>\boxed{\textbf{(D)}\ 8}.</math>
  
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==Solution 2==
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/VFJxCDC1YUg
  
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(Statement 1) Cannot be true, because only one of these four statements is true, and (Statement 1) states that the number is prime, which would make (Statement 2) and (Statement 3) false, which is not possible. And since the number being described is even, it must end with an even number (0,2,4,6,8). And since the number being described is a two-digit number, the first digit must be 9 (according to statement 4) because we are looking for the units digit (the digit in the one's place of a number). And so if we plug in the number 9 to all of the even answers, we will get three possible outcomes. <math>94</math>, <math>96</math>, and <math>98</math>. Because the number described is divisible by 7 (according to statement 3), the only possible answer for the number being described would be <math>98</math>. So, the answer is <math>\boxed{\textbf{(D)}\ 8}.</math>
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~Education, the Study of Everything
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==Solution 3==
 
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Like solutions 1 and 2, Statement 1 can't be true because it would contradict both Statements 2 and 3. Therefore, the other three must be true. We know the following:
 
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*If the number is even and has 9 as a digit, 9 must be the tens digit.
 
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*Since it's divisible by both 2 and 7, the number is a multiple of 14.
 
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The only multiple of 14 that has a tens digit of 9 is <math>98</math>. Thus, our answer is <math>\boxed{\textbf{(D)}\ 8}.</math>
 
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~Ligonmathkid2
 
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:31, 26 May 2024

Problem

Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."

(1) It is prime.

(2) It is even.

(3) It is divisible by 7.

(4) One of its digits is 9.

This information allows Malcolm to determine Isabella's house number. What is its units digit?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

Notice that (1) cannot be true. Otherwise, the number would have to be prime and be either even or divisible by 7. This only happens if the number is 2 or 7, neither of which are two-digit numbers, so we run into a contradiction. Thus, we must have (2), (3), and (4) be true. By (2), the $2$-digit number is even, and thus, the digit in the tens place must be $9$. The only even $2$-digit number starting with $9$ and divisible by $7$ is $98$, which has a units digit of $\boxed{\textbf{(D)}\ 8}.$

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/VFJxCDC1YUg

~Education, the Study of Everything

Video Solution

https://youtu.be/3Q4FTeSCuwo

~savannahsolver

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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. AMC logo.png

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