Difference between revisions of "2014 AMC 8 Problems/Problem 21"
(→Solution) |
(→1) |
||
Line 6: | Line 6: | ||
The sum of a number's digits <math>\mod{3}</math> is congruent to the number <math>\pmod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{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>. | The sum of a number's digits <math>\mod{3}</math> is congruent to the number <math>\pmod{3}</math>. <math>74A52B1 \mod{3}</math> must be congruent to 0, since it is divisible by 3. Therefore, <math>7+4+A+5+2+B+1 \mod{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>. | ||
− | ==Solution== | + | ==Solution 2== |
+ | |||
+ | Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 5 or 8. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add 5 to 15, to get 20. We then see that C = 1 or 7, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7 in order to be a multiple of three. We take the common numbers we got from both these equations, which are 1 and 7. However, in the answer choices, there is no 7, but there is a 1, so [(A) 1] is your answer. :) | ||
==See Also== | ==See Also== |
Revision as of 19:39, 29 July 2019
Contents
Problem
The -digit numbers
and
are each multiples of
. Which of the following could be the value of
?
Solution
The 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. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 5 or 8. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add 5 to 15, to get 20. We then see that C = 1 or 7, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7 in order to be a multiple of three. We take the common numbers we got from both these equations, which are 1 and 7. However, in the answer choices, there is no 7, but there is a 1, so [(A) 1] is your answer. :)
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
Another solution is using the divisbility rule of 3. Just add the digits of each number together.