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Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 2"

(Created page with "We have <math>a\geq1</math>. If <math>a=1</math>, then <math>f=0</math>, but <math>f\geq1</math>. If <math>a=2</math> and <math>b=0</math>, then <math>g=0=b</math>, a contradi...")
 
Line 8: Line 8:
 
If <math>a=2</math> and <math>b=5</math>, then <math>g=2=b</math>, a contradiction.
 
If <math>a=2</math> and <math>b=5</math>, then <math>g=2=b</math>, a contradiction.
 
If <math>a=2</math> and <math>b=6</math>, then <math>f=1</math> and <math>g=3</math>.
 
If <math>a=2</math> and <math>b=6</math>, then <math>f=1</math> and <math>g=3</math>.
−
Thus, we must have
+
Thus, we must have cde=2(hij), where <math>c, d, e, h, i, j</math> are distinct digits from the list <math>0, 4, 5, 7, 8, 9</math>.
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cde=2(hij),
 
−
where <math>c, d, e, h, i, j</math> are distinct digits from the list <math>0, 4, 5, 7, 8, 9</math>.
 
 
If <math>h\geq5</math>, then we have <math>c\geq10</math>, a contradiction. Thus, we must have <math>h=4</math>, and therefore <math>c=8, 9</math>.
 
If <math>h\geq5</math>, then we have <math>c\geq10</math>, a contradiction. Thus, we must have <math>h=4</math>, and therefore <math>c=8, 9</math>.
 
If <math>c=8</math>, then we have <math>i\leq4</math>, so <math>i=0, 4</math>.
 
If <math>c=8</math>, then we have <math>i\leq4</math>, so <math>i=0, 4</math>.
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abcde=26970,
 
abcde=26970,
 
fghij=13485.
 
fghij=13485.
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So, we have
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So, we have b+c+d+i+j=6+9+7+8+5=<math>[b]35[/b]</math>.
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$b+c+d+i+j=6+9+7+8+5=[b]35[/b].
 

Revision as of 05:37, 5 September 2023

We have $a\geq1$. If $a=1$, then $f=0$, but $f\geq1$. 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, 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=$[b]35[/b]$.

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