Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1987 AIME Problems/Problem 1"

(→‎Solution)
m (→‎Problem)
 
(2 intermediate revisions by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
An [[ordered pair]] <math>(m,n)</math> of [[non-negative]] [[integer]]s is called "simple" if the [[addition]] <math>m+n</math> in base <math>10</math> requires no [[carrying]]. Find the number of simple ordered pairs of non-negative integers that sum to <math>1492</math>.  
+
An [[ordered pair]] <math>(m,n)</math> of [[non-negative]] [[integer]]s is called "simple" if the [[addition]] <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to <math>1492</math>.
  
 
== Solution ==
 
== Solution ==
Line 9: Line 9:
  
  
−
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.
+
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, in addition to <math>0</math> and 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}}

Latest revision as of 13:42, 17 September 2016

Problem

An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$.

Solution

Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}$.



If you do not understand the above solution, consider this. For every positive integer $m$, there is only one whole number $n$ 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 $a$ is $(a + 1)$ because there are $(a - 1)$ positive integers that are less than it, in addition to $0$ and 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_1&oldid=80292"