AIME 2016 I · 第 8 题

Solution 1

To minimize $s(p)$, the numbers $1$, $2$, and $3$ (which sum to $6$) must be in the hundreds places. For the units digit of $s(p)$ to be $0$, the numbers in the ones places must have a sum of either $10$ or $20$. However, since the tens digit contributes more to the final sum $s(p)$ than the ones digit, and we are looking for the minimum value of $s(p)$, we take the sum's units digit to be $20$. We know that the sum of the numbers in the tens digits is $\sum\limits_{i=1}^9 (i) -6-20=45-6-20=19$. Therefore, $m=100\times6+10\times19+20=810$.

To find $n$, realize that there are $3!=6$ ways of ordering the numbers in each of the places. Additionally, there are three possibilities for the numbers in the ones place: $(4,7,9)$, $(5,7,8)$, and $(5,6,9)$. Therefore there are $6^3\times3=648$ ways in total. $|m-n|=|810-648|=\fbox{162}$.

Solution 2

Like solution 1, to minimize $s(p)$, we must have the hundreds digits to be 1, 2, and 3.

Then, notice that the units digits must sum to 20, as it is impossible to get them to sum to 10 without 1, 2, or 3. Then, we choose the largest numbers to be in the units places to minimize $s(p)$, resulting in us using 9, 7, and 4.

Then, 5, 6, and 8 are in the tens slots, giving us three placeholder numbers, which are 159, 267, and 384, in which $159+267+384 = 810$.

To show 810 is the minimum possible notice that we must have the units digits to sum to 20, meaning we carry over a 2. Then, if we assume the middle is less than 21, say 20, then we arrive at a contradiction as we cannot make 20 using any of the others, and thus less than 20 as well is not possible, forcing us to carry over another 2. Then, if we carry over the 2, the smallest possible hundred digits are $1+2+3+2 = 8$, and the smallest tens digit is 1 from 21. Thus, our smallest number must be 810.

To find all $s(p) = 810$, we notice that the 100ds digits must be 1, 2, and 3 (from our proof earlier). In addition, we are to carry over a 2 from the tens places. (one can also see that if we have our 100ds digits $a, b,$ and $c$, then $a+b+c = 6$ where $a \neq b \neq c$ and $a,b,c > 0$, meaning that we have only $1+2+3$ that satisfies the condition). Then, say the middle digits are $d, e,$ and $f$. Then, since we are forced to carry over a 2 from the units digits, we have that $d+e+f = 21-2 = 19$. Then, we have $d,e,f > 1$ and $d \neq e \neq f \neq 1,2,3$. Then, we can do simple casework to obtain only three sets of numbers that work, those being, (4,6,9), (4,7,8), (5,6,8). We then double check the units digits to see if they truly do sum to 20, giving us (5,7,8), (5,6,9), and (4,7,9) respectively, which indeed sum to 20. Thus, we have three possible sums. Each of these three sums has $3! \times 3! \times 3! = 6^3 = 216$ total permutations within digits, and thus there are $216 \times 3 = 648$ total permutations (which is our value of $n$).

Thus, $|m-n| = |810 - 648| = \boxed{162}$.

~Pinotation

Video Solutions

https://www.youtube.com/watch?v=WBtMUzgqfwI

https://www.youtube.com/watch?v=QBHakfd2gnQ