testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
Username:
Password:
Save Password
Forgot your Password?

 All Forums
 Homework Help Forums
 Miscellaneous Math Topics
 An ageless problem.
 New Topic  Reply to Topic
 Printer Friendly
Author Previous Topic Topic Next Topic  

TchrWill
Advanced Member

USA
79 Posts

Posted - 07/08/2013 :  15:54:04  Show Profile  Reply with Quote
The Krane children were born 2 years apart. When Micheal, the youngest of the Six children was 1 year old, the sum of their ages was 1+3+5+7+9+11, or 36, a perfect square. How many years old was Micheal the next time the sum of their ages was a perfect square?
Go to Top of Page

Subhotosh Khan
Advanced Member

USA
9116 Posts

Posted - 07/16/2013 :  18:35:01  Show Profile  Reply with Quote
quote:
Originally posted by TchrWill

The Krane children were born 2 years apart. When Micheal, the youngest of the Six children was 1 year old, the sum of their ages was 1+3+5+7+9+11, or 36, a perfect square. How many years old was Micheal the next time the sum of their ages was a perfect square?



After 'n' years the sum of their ages would be

S = 36 + 6*n............The S must be divisible by 6

and that number would be 144 .... 18 years later.
Go to Top of Page

TchrWill
Advanced Member

USA
79 Posts

Posted - 07/22/2013 :  11:21:48  Show Profile  Reply with Quote
Right on Subhotosh Khan.

Letting M be Michael's age at any time, the sum of the six ages will next be a square when 36 + 6(M - 1) = N^2. The sum of the ages increases in increments of 6 making the square we seek an even square. The sum of the ages progresses as

M's age....1....2....3..4...5....6...7.....8...9....10
Sum........36..42..48.54...60...66....72...78...84...90

M's age. ..11..12.13.14...15...16....17...18...19...20
Sum........96-102--108-114-120--126---132---138--144

Looks like Michael was 19 when the sum of the ages equaled 144 = 12^2, the next square after 36.

Another way of zeroing in on the answer is to simply find what value of M will create an even square. Those above 36 proceed as 64, 100, 144, 196, 256, etc. A few punches on the old calculator shows that M = 19 satisfies N^2 = 144.


Another way of attacking this one is as follows:

1--The first sum of ages is 36 at Michael's age 1.
2--The subsequent sums increase by 6 each year.
3--The sum sought can be defined as N^2 = 36 + 6(M - 1), M being Michael's attained age.
4--Written another way, N^2 - 36 = N^2 - n^2 = 6(M - 1)
5--N^2 - n^2 can be written as (N + n)(N - n)
6--Therefore, (N + n)(N - n) = 6(M - 1)
7--Then, (N + n) = (M - 1) and (N - n) = 6
8--Adding, 2N = M + 5 or N = (M + 5)/2
9--We can now write N^2 = (M - 1)^2/4 = 36 + 6(M - 1)
10--Expanding and simplifying, M^2 - 14M - 95 = 0
11--From the quadratic formula, M = [14+/-sqrt(196 + 380)]/2
12--Solving, M = 19.
Go to Top of Page
  Previous Topic Topic Next Topic  
 New Topic  Reply to Topic
 Printer Friendly
Jump To:
Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.1 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 14 Apr 2014