Difference between revisions of "2017 AMC 10A Problems/Problem 20"
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==Solution 1== | ==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>\equiv 6 \pmod{9}</math> is <math>\boxed{\textbf{(D)}\ 1239}</math>. |
==Solution 2== | ==Solution 2== | ||
− | We can find out that the least number of digits the number <math> | + | One divisibility rule that we can use for this problem is that a multiple of <math>9</math> will always have its digits sum to a multiple of <math>9</math>. We can find out that the least number of digits the number <math>n</math> has is <math>142</math>, with <math>141</math> <math>9</math>'s and <math>1</math> <math>5</math>, assuming the rule above. No matter what arrangement or different digits we use, the divisibility rule stays the same. To make the problem simpler, we can just use the <math>141</math> <math>9</math>'s and <math>1</math> <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 <math>1</math> to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. |
− | By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>. | + | Knowing that <math>n+1</math> is divisible by <math>9</math> when <math>6</math>, we can subtract <math>6</math> from every available choice, and see if the number is divisible by <math>9</math> afterwards. |
− | By adding <math>1</math> to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. | ||
− | Knowing that | ||
After subtracting <math>6</math> from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by <math>9</math>. | After subtracting <math>6</math> from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by <math>9</math>. | ||
So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | ||
− | |||
− | |||
==Solution 3== | ==Solution 3== | ||
The number <math>n</math> can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice. | The number <math>n</math> can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice. | ||
− | If <math> | + | If <math>A</math> is correct, then <math>n</math> must be some number <math>99999999...9</math>, because when we add one to <math>99999999...9</math> we get <math>10000000...00</math>. Thus, if <math>1</math> is the correct answer, then the equation <math>9x=1274</math> must have an integer solution (i.e. <math>1274</math> must be divisible by <math>9</math>). But since it does not, <math>1</math> is not the correct answer. |
− | If <math> | + | If <math>B</math> is correct, then <math>n</math> must be some number <math>29999999...9</math>, because when we add one to <math>29999999...9</math>, we get <math>30000000...00</math>. Thus, if <math>3</math> is the correct answer, then the equation <math>2+9x=1274</math> must have an integer solution. But since it does not, <math>3</math> is not the correct answer. |
Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if <math>S(n+1)=N</math>, then <math>n</math> must be a number whose initial digits sum to <math>N-1</math>, and whose other, terminating digits, are all <math>9</math>. Thus, we can evaluate the three final possibilities by seeing if the equation <math>(N-1)+9x=1274</math> has an integer solution. | Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if <math>S(n+1)=N</math>, then <math>n</math> must be a number whose initial digits sum to <math>N-1</math>, and whose other, terminating digits, are all <math>9</math>. Thus, we can evaluate the three final possibilities by seeing if the equation <math>(N-1)+9x=1274</math> has an integer solution. | ||
− | The equation does not have an integer solution for <math>N=12</math>, so <math> | + | The equation does not have an integer solution for <math>N=12</math>, so <math>C</math> is not correct. However, the equation does have an integer solution for <math>N=1239</math> (<math>x=4</math>), so <math>\boxed{D}</math> is the answer. |
+ | ==Solution 4 (Intuition)== | ||
+ | If adding <math>1</math> to <math>n</math> does not carry any of its digits, then <math>S(n+1)=S(n)+1</math> (ex: <math>25+1=26</math>. Sum of digits <math>7 \rightarrow 8</math>). But since no answer choice is <math>1275</math>, that means <math>n</math> has some amount of <math>9</math>'s from right to left. | ||
+ | |||
+ | When <math>n+1</math>, some <math>9</math>'s will bump to 0, not affected its<math>\pmod 9</math>. But the first non-9 digit (from right to left) will be bumped up by 1. So <math>S(n) + 1 \pmod {9} \equiv S(n+1) \pmod{9}</math>. For example, <math>34999+1=35000</math>, and the sum of digits <math>7+27 \rightarrow 8+0</math>. | ||
+ | |||
+ | Since <math>S(n) \equiv 5 \pmod{9}</math>, that means <math>S(n+1) \equiv 6 \pmod{9}</math>. The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) } 1239}.</math> | ||
+ | ~BakedPotato66 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zfChnbMGLVQ?t=3996 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/pbt6p5s8J50 | ||
==See Also== | ==See Also== | ||
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{{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2017|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 02:16, 24 July 2021
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
One divisibility rule that we can use for this problem is that a multiple of will always have its digits sum to a multiple of . We can find out that the least number of digits the number has is , with 's and , assuming the rule above. No matter what arrangement or different digits we use, the divisibility rule stays the same. To make the problem simpler, we can just use the 's and . By randomly mixing the digits up, we are likely to get: ....... By adding to this number, we get: ....... Knowing that is divisible by when , we can subtract from every available choice, and see if the number is divisible by afterwards. After subtracting from every number, we can conclude that (originally ) is the only number divisible by . So our answer is .
Solution 3
The number can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.
If is correct, then must be some number , because when we add one to we get . Thus, if is the correct answer, then the equation must have an integer solution (i.e. must be divisible by ). But since it does not, is not the correct answer.
If is correct, then must be some number , because when we add one to , we get . Thus, if is the correct answer, then the equation must have an integer solution. But since it does not, is not the correct answer.
Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if , then must be a number whose initial digits sum to , and whose other, terminating digits, are all . Thus, we can evaluate the three final possibilities by seeing if the equation has an integer solution.
The equation does not have an integer solution for , so is not correct. However, the equation does have an integer solution for (), so is the answer.
Solution 4 (Intuition)
If adding to does not carry any of its digits, then (ex: . Sum of digits ). But since no answer choice is , that means has some amount of 's from right to left.
When , some 's will bump to 0, not affected its. But the first non-9 digit (from right to left) will be bumped up by 1. So . For example, , and the sum of digits .
Since , that means . The only answer choice that meets this requirement is
~BakedPotato66
Video Solution
https://youtu.be/zfChnbMGLVQ?t=3996
~ pi_is_3.14
Video Solution
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.