Difference between revisions of "1992 AIME Problems/Problem 6"
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
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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 | ||
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Solution 3
There are 3 forms possible. : , : , : , : --- Thus, since there should be no carrying, in only integers to is possible Therefore, the answer is