Difference between revisions of "2017 AMC 10A Problems/Problem 20"
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<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Note that <math>n \equiv S(n) \pmod{9}</math>. This can be seen from the fact that <math>\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}</math>. Thus, if <math>S(n) = 1274</math>, then <math>n \equiv 5 \pmod{9}</math>, and thus <math>n+1 \equiv S(n+1) \equiv 6 \pmod{9}</math>. The only answer choice that is <math>6 \pmod{9}</math> is <math>\boxed{\textbf{(D)}\ 1239}</math>. | Note that <math>n \equiv S(n) \pmod{9}</math>. This can be seen from the fact that <math>\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}</math>. Thus, if <math>S(n) = 1274</math>, then <math>n \equiv 5 \pmod{9}</math>, and thus <math>n+1 \equiv S(n+1) \equiv 6 \pmod{9}</math>. The only answer choice that is <math>6 \pmod{9}</math> is <math>\boxed{\textbf{(D)}\ 1239}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>. | ||
| + | By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>. | ||
| + | By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. | ||
| + | We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards. | ||
| + | After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9. | ||
| + | So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | ||
| + | ~ProGameXD | ||
==See Also== | ==See Also== | ||
Revision as of 18:07, 9 February 2017
Contents
Problem
Let 
equal the sum of the digits of positive integer 
. For example, 
. For a particular positive integer , 
. Which of the following could be the value of 
?
Solution 1
Note that 
. This can be seen from the fact that 
. Thus, if , then 
, and thus 
. The only answer choice that is 
is 
.
Solution 2
We can find out that the least number of digits the number 
is 
, with 
's and one 
.
By randomly mixing the digits up, we are likely to get: 
...
....
By adding 1 to this number, we get: ...
...
.
We can subtract 6 from every available choice, and see if the number is divisible by 9 afterwards.
After subtracting 6 from every number, we can conclude that 
(originally 
) is the only number divisible by 9.
So our answer is .
~ProGameXD
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AMC 10 Problems and Solutions | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
