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 "2002 AMC 12B Problems/Problem 13"
(→‎Solution 1.1)
Line 24: Line 24:
  
  
−
the normal sequence can be described as N^2+N divided by 2.
+
the normal sequence can be described as <math>N^2+N</math> divided by 2.
  
 
Since have 18 terms adding 18n will increase the consective sequence startering number by 1  
 
Since have 18 terms adding 18n will increase the consective sequence startering number by 1  
  
−
(N^2+N)/2 +18n
+
<math>\frac{(N^2+N)}{2} +18n</math>
  
 
now subsitute 18 as N
 
now subsitute 18 as N
  
−
we get (18^2+18)/2 = 171
+
we get <math>\frac{(18^2+18)}{2} = 171</math>
  
  
−
put I^2 which is integer square and plug in all our results  
+
put <math>I^2</math> which is integer square and plug in all our results  
−
171 + 18n = I^2
+
<math>171 + 18n = I^2</math>
  
−
18n = I^2-171
+
<math>18n = I^2-171
−
I^2-171 = 0 mod(18)
+
I^2-171 = 0 mod(18)</math>
  
 
subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number
 
subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number
  
−
225-177 = mod(18)
+
<math>225-177 = mod(18)</math>
−
54 = mod(18)
+
<math>54 = mod(18)</math>
  
−
54 is dividsble by 18 and is therefore the smallest number possible.
+
54 is divisble by 18 and is therefore the smallest number possible.
  
 
== See also ==
 
== See also ==

Revision as of 10:59, 19 October 2025

Problem

The sum of $18$ consecutive positive integers is a perfect square. The smallest possible value of this sum is

$\mathrm{(A)}\ 169 \qquad\mathrm{(B)}\ 225 \qquad\mathrm{(C)}\ 289 \qquad\mathrm{(D)}\ 361 \qquad\mathrm{(E)}\ 441$

Solution 1

Let $a, a+1, \ldots, a + 17$ be the consecutive positive integers. Their sum, $18a + \frac{17(18)}{2} = 9(2a+17)$, is a perfect square. Since $9$ is a perfect square, it follows that $2a + 17$ is a perfect square. The smallest possible such perfect square is $25$ when $a = 4$, and the sum is $225 \Rightarrow \mathrm{(B)}$.

Solution 2

Notice that all five choices given are perfect squares.

Let $a$ be the smallest number, we have \[a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153\]

Subtract $153$ from each of the choices and then check its divisibility by $18$, we have $225$ as the smallest possible sum. $\mathrm {(B)}$

~ Nafer


Solution 1.1

the normal sequence can be described as $N^2+N$ divided by 2.

Since have 18 terms adding 18n will increase the consective sequence startering number by 1

$\frac{(N^2+N)}{2} +18n$

now subsitute 18 as N

we get $\frac{(18^2+18)}{2} = 171$


put $I^2$ which is integer square and plug in all our results $171 + 18n = I^2$

$18n = I^2-171 I^2-171 = 0 mod(18)$

subsitute the answer choices starting with B because 169 is less than 171 and results in a neagtive number

$225-177 = mod(18)$ $54 = mod(18)$

54 is divisble by 18 and is therefore the smallest number possible.

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 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=2002_AMC_12B_Problems/Problem_13&oldid=263541"