Mock AIME 5 Pre 2005 Problems/Problem 2

Problem

Two 5-digit numbers are called "responsible" if they are: $\begin{align*} &\text {i. In form of abcde and fghij such that fghij = 2(abcde)}\\ &\text {ii. all ten digits, a through j are all distinct.}\\ &\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}$

If two "responsible" numbers are small as possible, what is the sum of the three middle digits of $\text {abcde}$ and last two digits on the $\text {fghij}$ ? That is, $b + c + d + i + j$ .

Solution

We have $a\geq1$ . If $a=1$ , then $f=0$ , but $f\geq1$ , a contradiction. If $a=2$ and $b=0$ , then $g=0=b$ , a contradiction. If $a=2$ and $b=1$ , then $f=1=b$ , a contradiction. If $a=2$ and $b=2$ , then $a=2=b$ , a contradiction. If $a=2$ and $b=3$ , then $f=1=g$ , a contradiction. If $a=2$ and $b=4$ , then $g=2=b$ , a contradiction. If $a=2$ and $b=5$ , then $g=2=b$ , a contradiction. If $a=2$ and $b=6$ , then $f=1$ and $g=3$ . Thus, we must have cde=2(hij), where $c, d, e, h, i, j$ are distinct digits from the list $0, 4, 5, 7, 8, 9$ . If $h\geq5$ , then we have $c\geq10$ , a contradiction. Thus, we must have $h=4$ , and therefore $c=8, 9$ . If $c=8$ , then we have $i\leq4$ , so $i=0, 4$ . If we have $i=4$ then $h=i=4$ , a contradiction. If we have $i=0$ then $d=0$ (as $1$ is not in the list of permitted digits). Thus, we must have $c=9$ . If we have $j=7$ , then $e=4=c$ , a contradiction. If we have $j=8$ , then $e=6$ , which is not in the list of permitted digits, a contradiction. If we have $j=0$ , then $e=0=j$ , a contradiction. Thus, we must have $j=5$ , and therefore $e=0$ . But now we must have $d$ odd as $j=5$ . Thus, we have $d=7$ and $i=8$ . Thus, our minimal responsible pair of two 5-digit numbers is abcde=26970, fghij=13485. So, we have b+c+d+i+j=6+9+7+8+5=35.

~ AbbyWong

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Mock AIME 5 Pre 2005 Problems/Problem 2

Problem

Solution