Difference between revisions of "2017 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | Notice that (1) cannot be true. Otherwise, the number would have to prime and either | + | 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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− | ( | + | ==Video Solution (CREATIVE THINKING!!!)== |
+ | https://youtu.be/VFJxCDC1YUg | ||
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+ | ~Education, the Study of Everything | ||
==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?
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 -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
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~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.