Difference between revisions of "1992 AIME Problems/Problem 6"
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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| + | === Solution 3 === | ||
| + | There are 3 forms possible. | ||
| + | <math>1OOO</math> : <math>5^3</math> , | ||
| + | <math>1OO9</math> : <math>5^2</math> , | ||
| + | <math>1O99</math> : <math>5^1</math> , | ||
| + | <math>1999</math> : <math>5^0 or 1</math> --- | ||
| + | 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> | ||
Revision as of 11:36, 14 February 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.
| 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 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
Solution 3
There are 3 forms possible.
: 
,
: 
,
: 
,
: 
---
Thus, since there should be no carrying, in 
only integers 
to 
is possible
Therefore, the answer is 
