Difference between revisions of "2017 AMC 8 Problems/Problem 8"
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Notice that (1) cannot be true, as we quickly see that we cannot have <math>2</math> of the <math>3</math> remaining conditions be true without running into a contradiction. Thus, we must have (2), (3), and (4) 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, as we quickly see that we cannot have <math>2</math> of the <math>3</math> remaining conditions be true without running into a contradiction. Thus, we must have (2), (3), and (4) 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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==See Also== | ==See Also== |
Revision as of 14:50, 22 November 2017
Problem 8
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?
Solution
Notice that (1) cannot be true, as we quickly see that we cannot have of the remaining conditions be true without running into a contradiction. Thus, we must have (2), (3), and (4) true. By (2), the -digit number is even, and thus the digit in the tens place must be . The only even -digit number starting with and divisible by is , which has a units digit of
~nukelauncher
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.