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 "2022 AIME I Problems/Problem 2"

(→‎Video Solution (Mathematical Dexterity)
Line 16: Line 16:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2 (Simple) ==
 +
 +
As in the previous solution, we get <math>99a = 71b+8c</math>. 99 and 71 are big numbers comparatively to 8, so we hypothesize that <math>a</math> and <math>b</math> are equal and <math>8c</math> fills the gap between them. The difference between 99 and 71 is 28, which is a multiple of 4, so if we multiply this by 2, it will be a multiple of 8 and thus the gap can be filled. Therefore, a viable solution is <math>a = 2, b = 2, c = 7</math> and the answer is <math>\boxed{227}</math>
 +
 +
~KingRavi
  
 
==Video Solution (Mathematical Dexterity)==
 
==Video Solution (Mathematical Dexterity)==

Revision as of 20:01, 17 February 2022

Problem

Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution

We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\] Taking both sides modulo $71,$ we have \begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$

Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

~MRENTHUSIASM

Solution 2 (Simple)

As in the previous solution, we get $99a = 71b+8c$. 99 and 71 are big numbers comparatively to 8, so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between 99 and 71 is 28, which is a multiple of 4, so if we multiply this by 2, it will be a multiple of 8 and thus the gap can be filled. Therefore, a viable solution is $a = 2, b = 2, c = 7$ and the answer is $\boxed{227}$

~KingRavi

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=z5Y4bT5rL-s

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_I_Problems/Problem_2&oldid=171155"