Difference between revisions of "Mock AIME 5 Pre 2005 Problems/Problem 2"
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If we have <math>i=0</math> then <math>d=0</math> (as <math>1</math> is not in the list of permitted digits). Thus, we must have <math>c=9</math>. | If we have <math>i=0</math> then <math>d=0</math> (as <math>1</math> is not in the list of permitted digits). Thus, we must have <math>c=9</math>. | ||
If we have <math>j=7</math>, then <math>e=4=c</math>, a contradiction. | If we have <math>j=7</math>, then <math>e=4=c</math>, 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=8</math>, then <math>e=6</math>, which is not in the list of permitted digits, 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. |
− | 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, 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>. 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=35. | So, we have b+c+d+i+j=6+9+7+8+5=35. | ||
+ | ~AbbyWong |
Latest revision as of 06:41, 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 of permitted digits, 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.
~AbbyWong