Difference between revisions of "1987 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
Since no carrying over is allowed, the range of possible values of any digit of <math>m</math> is from <math>0</math> to the respective [[digit]] in <math>1492</math> (the values of <math>n</math> are then fixed). Thus, the number of [[ordered pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>. | Since no carrying over is allowed, the range of possible values of any digit of <math>m</math> is from <math>0</math> to the respective [[digit]] in <math>1492</math> (the values of <math>n</math> are then fixed). Thus, the number of [[ordered pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>. | ||
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| + | If you do not understand the above solution, consider this. For every [[positive integer]] <math>m</math>, there is only one [[whole number]] <math>n</math> that you can add to it to obtain the required sum. Also, the total number of [[non-negative integers]] that are smaller than or equal to an integer <math>a</math> is <math>(a + 1)</math> because there are <math>(a - 1)</math> positive integers that are less than it, and <math>0</math>, in addition to itself. | ||
== See also == | == See also == | ||
{{AIME box|year=1987|before=First Question|num-a=2}} | {{AIME box|year=1987|before=First Question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:04, 14 February 2014
Problem
An ordered pair 
of non-negative integers is called "simple" if the addition 
in base 
requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to 
.
Solution
Since no carrying over is allowed, the range of possible values of any digit of 
is from 
to the respective digit in (the values of 
are then fixed). Thus, the number of ordered pairs will be 
.
If you do not understand the above solution, consider this. For every positive integer , there is only one whole number that you can add to it to obtain the required sum. Also, the total number of non-negative integers that are smaller than or equal to an integer 
is 
because there are 
positive integers that are less than it, and , in addition to itself.
See also
| 1987 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
