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Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 2"
 
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==Problem==
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Two 5-digit numbers are called "responsible" if they are:
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<cmath>\begin{align*}
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&\text {i. In form of abcde and fghij such that fghij = 2(abcde)}\\
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&\text {ii. all ten digits, a through j are all distinct.}\\
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&\text {iii.} a + b + c + d + e + f + g + h + i + j = 45\end{align*}</cmath>
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If two "responsible" numbers are small as possible, what is the sum of the three middle digits of <math>\text {abcde}</math> and last two digits on the <math>\text {fghij}</math>? That is, <math>b + c + d + i + j</math>.
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==Solution==
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We have <math>a\geq1</math>.
 
We have <math>a\geq1</math>.
 
If <math>a=1</math>, then <math>f=0</math>, but <math>f\geq1</math>, a contradiction.
 
If <math>a=1</math>, then <math>f=0</math>, but <math>f\geq1</math>, a contradiction.
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fghij=13485.
 
fghij=13485.
 
So, we have b+c+d+i+j=6+9+7+8+5=35.
 
So, we have b+c+d+i+j=6+9+7+8+5=35.
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~AbbyWong
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~ AbbyWong

Latest revision as of 16:27, 23 February 2025

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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