Difference between revisions of "1992 AIME Problems/Problem 6"
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We now have to consider the cases where <math>b,c = 0</math>, specifically when <math>1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}</math>. We can see that <math>1100, 1200, 1300, 1400, 1500</math>, and <math>2000</math> all work, giving a grand total of <math>150 + 6 = \boxed{156}</math> ordered pairs. | We now have to consider the cases where <math>b,c = 0</math>, specifically when <math>1abc \in \{1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000\}</math>. We can see that <math>1100, 1200, 1300, 1400, 1500</math>, and <math>2000</math> all work, giving a grand total of <math>150 + 6 = \boxed{156}</math> ordered pairs. | ||
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=== Solution 3 === | === Solution 3 === | ||
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Thus, since there should be no carrying, in <math>O</math> only integers <math>0</math> to <math>4</math> is possible | Thus, since there should be no carrying, in <math>O</math> only integers <math>0</math> to <math>4</math> is possible | ||
Therefore, the answer is <math>{156}</math> | Therefore, the answer is <math>{156}</math> | ||
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+ | {{AIME box|year=1992|num-b=5|num-a=7}} | ||
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+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 20:00, 14 August 2016
Problem
For how many pairs of consecutive integers in is no carrying required when the two integers are added?
Solution
Solution 1
Consider what carrying means: If carrying is needed to add two numbers with digits and
, then
or
or
. 6. Consider
.
has no carry if
. This gives
possible solutions.
With , there obviously must be a carry. Consider
.
have no carry. This gives
possible solutions. Considering
,
have no carry. Thus, the solution is
.
Solution 2
Consider the ordered pair where
and
are digits. We are trying to find all ordered pairs where
does not require carrying. For the addition to require no carrying,
, so
unless
ends in
, which we will address later. Clearly, if
, then adding
will require no carrying. We have
possibilities for the value of
,
for
, and
for
, giving a total of
, but we are not done yet.
We now have to consider the cases where , specifically when
. We can see that
, and
all work, giving a grand total of
ordered pairs.
Solution 3
There are 3 forms possible.
:
,
:
,
:
,
:
---
Thus, since there should be no carrying, in
only integers
to
is possible
Therefore, the answer is
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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