2017 AMC 8 Problems/Problem 8

Revision as of 13:00, 18 January 2021 by Hashtagmath (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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?

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


Notice that (1) cannot be true. Otherwise, the number would have to prime and either be 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) 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 $\boxed{\textbf{(D)}\ 8}.$

Solution 2

(Statement 1) Cannot be true, because only one of these four statements are 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. $94$, $96$, and $98$. Because the number described is divisible by 7 (according to statement 3), the only possible answer for the number being described would be $98$. So the answer is $\boxed{\textbf{(D)}\ 8}.$

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS