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...") |
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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>. |
− | 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. | ||
− | So, we have | + | So, we have b+c+d+i+j=6+9+7+8+5=<math>[b]35[/b]</math>. |
− |
Revision as of 06:37, 5 September 2023
We have .
If
, then
, but
.
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=
.