Difference between revisions of "2014 AMC 8 Problems/Problem 21"
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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math> | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8</math> | ||
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+ | ==Solution 1== | ||
+ | The number<math>\mod{3}</math> is congruent to sum of a number's digits<math>\mod{3}</math> is congruent to the number<math>\mod{3}</math>. <math>74A52B1 \pmod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \pmod{3}</math> is also congruent to 0. <math>7+4+5+2+1 \equiv 1 \pmod{3}</math>, so <math>A+B\equiv 2 \pmod{3}</math>. As we know, <math>326AB4C\equiv 0 \pmod{3}</math>, so <math>3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}</math>, and therefore <math>A+B+C\equiv 0 \pmod{3}</math>. We can substitute 2 for <math>A+B</math>, so <math>2+C\equiv 0 \pmod{3}</math>, and therefore <math>C\equiv 1\pmod{3}</math>. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is <math>\boxed{\textbf{(A) }1}</math>. | ||
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+ | ==Solution 2== | ||
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+ | Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. <math>7 + 4 + 5 + 2 + 1 = 19</math>. To be a multiple of <math>3</math>, <math>A + B</math> has to be either <math>2</math> or <math>5</math> or <math>8</math>... and so on. We add up the numerical digits in the second number; <math>3 + 2 + 6 + 4 = 15</math>. We then add two of the selected values, <math>5</math> to <math>15</math>, to get <math>20</math>. We then see that C = <math>1, 4</math> or <math>7, 10</math>... and so on, otherwise the number will not be divisible by three. We then add <math>8</math> to <math>15</math>, to get <math>23</math>, which shows us that C = <math>1</math> or <math>4</math> or <math>7</math>... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be <math>1, 4,</math> and <math>7</math>. However, in the answer choices, there is no <math>7</math> or <math>4</math> or anything greater than <math>7</math>, but there is a <math>1</math>, so <math>\boxed{\textbf{(A) }1}</math> is our answer. | ||
==Video Solution for Problems 21-25== | ==Video Solution for Problems 21-25== | ||
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https://youtu.be/7TOtBiod55Q ~savannahsolver | https://youtu.be/7TOtBiod55Q ~savannahsolver | ||
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=20|num-a=22}} | {{AMC8 box|year=2014|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:10, 31 January 2023
Contents
Problem 21
The -digit numbers
and
are each multiples of
. Which of the following could be the value of
?
Solution 1
The number is congruent to sum of a number's digits
is congruent to the number
.
must be congruent to 0, since it is divisible by 3. Therefore,
is also congruent to 0.
, so
. As we know,
, so
, and therefore
. We can substitute 2 for
, so
, and therefore
. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is
.
Solution 2
Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. . To be a multiple of
,
has to be either
or
or
... and so on. We add up the numerical digits in the second number;
. We then add two of the selected values,
to
, to get
. We then see that C =
or
... and so on, otherwise the number will not be divisible by three. We then add
to
, to get
, which shows us that C =
or
or
... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be
and
. However, in the answer choices, there is no
or
or anything greater than
, but there is a
, so
is our answer.
Video Solution for Problems 21-25
https://www.youtube.com/watch?v=6S0u_fDjSxc
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=2593
Video Solution
https://youtu.be/7TOtBiod55Q ~savannahsolver
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.