Difference between revisions of "2016 AMC 8 Problems/Problem 5"
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The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>. | The number <math>72+1=73</math> satisfies both conditions. We subtract the biggest multiple of <math>11</math> less than <math>73</math> to get the remainder. Thus, <math>73-11(6)=73-66=\boxed{\textbf{(E) }7}</math>. | ||
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+ | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5) | ||
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{{AMC8 box|year=2016|num-b=4|num-a=6}} | {{AMC8 box|year=2016|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:06, 5 July 2020
The number is a two-digit number.
• When is divided by
, the remainder is
.
• When is divided by
, the remainder is
.
What is the remainder when is divided by
?
Solution
From the second bullet point, we know that the second digit must be . Because there is a remainder of
when it is divided by
, the multiple of
must end in a
. We now look for this one:
The number satisfies both conditions. We subtract the biggest multiple of
less than
to get the remainder. Thus,
.
Video Solution
https://www.youtube.com/watch?v=LqnQQcUVJmA (has questions 1-5)
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.