Difference between revisions of "2017 AMC 10A Problems/Problem 20"

Revision as of 13:39, 8 June 2021

Problem

Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ , $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$ ?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution 1

Note that $n \equiv S(n) \pmod{9}$ . This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$ . Thus, if $S(n) = 1274$ , then $n \equiv 5 \pmod{9}$ , and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that is $6 \pmod{9}$ is $\boxed{\textbf{(D)}\ 1239}$ .

Solution 2

One divisibility rule for division that we can use in the problem is that a multiple of $9$ has its digit always add up to a multiple of $9$ . We can find out that the least number of digits the number $N$ is $142$ , with $141$ $9$ 's and $1$ $5$ , assuming the rule above. No matter what arrangement or different digits we use, the divisor rule stays the same. To make the problem simpler, we can just use the $141$ $9$ 's and $1$ $5$ . By randomly mixing the digits up, we are likely to get: $9999$ ... $9995999$ ... $9999$ . By adding $1$ to this number, we get: $9999$ ... $9996000$ ... $0000$ . Knowing that this number is ONLY divisible by $9$ when $6$ is subtracted, 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 $1233$ (originally $1239$ ) is the only number divisible by $9$ . So our answer is $\boxed{\textbf{(D)}\ 1239}$ .

Solution 3

The number $n$ 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 $A$ is correct, then $n$ must be some number $99999999...9$ , because when we add one to $99999999...9$ we get $10000000...00$ . Thus, if $1$ is the correct answer, then the equation $9x=1274$ must have an integer solution (i.e. $1274$ must be divisible by $9$ ). But since it does not, $1$ is not the correct answer.

If $B$ is correct, then $n$ must be some number $29999999...9$ , because when we add one to $29999999...9$ , we get $30000000...00$ . Thus, if $3$ is the correct answer, then the equation $2+9x=1274$ must have an integer solution. But since it does not, $3$ 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 $S(n+1)=N$ , then $n$ must be a number whose initial digits sum to $N-1$ , and whose other, terminating digits, are all $9$ . Thus, we can evaluate the three final possibilities by seeing if the equation $(N-1)+9x=1274$ has an integer solution.

The equation does not have an integer solution for $N=12$ , so $C$ is not correct. However, the equation does have an integer solution for $N=1239$ ( $x=4$ ), so $\boxed{D}$ is the answer.

Solution 4 (Intuition)

If adding $1$ to $n$ does not carry any of its digits, then $S(n+1)=S(n)+1$ (ex: $25+1=26$ . Sum of digits $7 \rightarrow 8$ ). But since no answer choice is $1275$ , that means $n$ has some amount of $9$ 's from right to left.

When $n+1$ , some $9$ 's will bump to 0, not affected its $\pmod 9$ . But the first non-9 digit (from right to left) will be bumped up by 1. So $S(n) + 1 \pmod {9} \equiv S(n+1) \pmod{9}$ (ex: $34999+1=3500$ . Sum of digits $7+27 \rightarrow 8+0$ ).

Since $S(n) \equiv 5 \pmod{9}$ , that means $S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that meets this requirement is $\boxed{\textbf{(D) } 1239}.$

~BakedPotato66

Video Solution

https://youtu.be/zfChnbMGLVQ?t=3996

~ pi_is_3.14

Video Solution

https://youtu.be/pbt6p5s8J50

@@ Line 26: / Line 26: @@
 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} = S(n+1) \pmod{9}</math>. Since <math>S(n) \cong 5 \pmod{9}</math>, <math>S(n+1) \cong 6 \pmod{9}</math>.
+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> (ex: <math>34999+1=3500</math>. Sum of digits <math>7+27 \rightarrow 8+0</math>).
-The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) } 1239}.</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

2017 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AMC 10A Problems/Problem 20"

Revision as of 13:39, 8 June 2021

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4 (Intuition)

Video Solution

Video Solution

See Also