Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 2"
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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>. | + | If <math>a=1</math>, then <math>f=0</math>, but <math>f\geq1</math>, a contradiction. |
If <math>a=2</math> and <math>b=0</math>, then <math>g=0=b</math>, a contradiction. | If <math>a=2</math> and <math>b=0</math>, then <math>g=0=b</math>, a contradiction. | ||
If <math>a=2</math> and <math>b=1</math>, then <math>f=1=b</math>, a contradiction. | If <math>a=2</math> and <math>b=1</math>, then <math>f=1=b</math>, a contradiction. | ||
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If <math>a=2</math> and <math>b=4</math>, then <math>g=2=b</math>, a contradiction. | If <math>a=2</math> and <math>b=4</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=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 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>. |
| − | 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>. | ||
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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If we have <math>j=8</math>, then <math>e=6</math>, which is not in the list, a contradiction. | If we have <math>j=8</math>, then <math>e=6</math>, which is not in the list, a contradiction. | ||
If we have <math>j=0</math>, then <math>e=0=j</math>, a contradiction. Thus, we must have <math>j=5</math>, and therefore <math>e=0</math>. | If we have <math>j=0</math>, then <math>e=0=j</math>, a contradiction. Thus, we must have <math>j=5</math>, and therefore <math>e=0</math>. | ||
| − | But now we must have <math>d</math> odd as <math>j=5</math>. Thus, we have <math>d=7</math> and <math>i=8</math>. | + | But now we must have <math>d</math> odd as <math>j=5</math>. Thus, we have <math>d=7</math> and <math>i=8</math>. Thus, our minimal responsible pair of two 5-digit numbers is |
| − | Thus, our minimal responsible pair of two 5-digit numbers is | ||
abcde=26970, | abcde=26970, | ||
fghij=13485. | fghij=13485. | ||
| − | So, we have b+c+d+i+j=6+9+7+8+5= | + | So, we have b+c+d+i+j=6+9+7+8+5=35. |
Revision as of 05:38, 5 September 2023
We have 
.
If 
, then 
, but 
, a contradiction.
If 
and 
, then 
, a contradiction.
If and 
, then 
, a contradiction.
If and 
, then 
, a contradiction.
If and 
, then 
, a contradiction.
If and 
, then 
, a contradiction.
If and 
, then , a contradiction.
If and 
, then 
and 
. Thus, we must have cde=2(hij), where 
are distinct digits from the list 
.
If 
, then we have 
, a contradiction. Thus, we must have 
, and therefore 
.
If 
, then we have 
, so 
.
If we have 
then 
, a contradiction.
If we have 
then 
(as 
is not in the list of permitted digits). Thus, we must have 
.
If we have 
, then 
, a contradiction.
If we have 
, then 
, which is not in the list, a contradiction.
If we have 
, then 
, a contradiction. Thus, we must have 
, and therefore 
.
But now we must have 
odd as . Thus, we have 
and 
. 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.
