My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum

LinkBack Thread Tools Display Modes
October 3rd, 2015, 02:46 PM   #1
Joined: Oct 2015
From: milton keynes

Posts: 2
Thanks: 0

Impossible proof by induction

Prove (3^k ) + (7 ^ (k-1)) + 8 is divisible by 12.

Last edited by skipjack; October 3rd, 2015 at 03:08 PM.
Johntee is offline  
October 3rd, 2015, 02:50 PM   #2
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,675
Thanks: 2655

Math Focus: Mainly analysis and algebra
Why is this impossible?
v8archie is offline  
October 3rd, 2015, 04:26 PM   #3
Global Moderator
Joined: May 2007

Posts: 6,807
Thanks: 717

True for k=1.
Assume true for k, show true for k+1
(for k+1) $\displaystyle 3(3^k)+7(7^{k-1})+8$ divisible by 12?
Subtract assumption for k and get $\displaystyle 2(3^k)+6(7^{k-1})$ divisible by 12?
Expression is $\displaystyle 6(3^{k-1}+7^{k-1})$
Term inside parentheses is even! Therefore 6x term is divisible by 12!
mathman is offline  

  My Math Forum > High School Math Forum > Algebra

impossible, induction, proof

« Radicals | slope »

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Proof by induction Tiome_nguyen Abstract Algebra 3 June 2nd, 2012 09:14 AM
Proof by induction A u B = A + B - A n B Spaghett Number Theory 1 October 19th, 2011 07:19 PM
proof by induction Airmax Applied Math 9 May 8th, 2009 12:02 PM
Proof. Show 2^n=1 mod n impossible for n>1 cos5000 Number Theory 4 April 28th, 2008 10:36 AM
proof by induction MaD_GirL Number Theory 5 November 14th, 2007 06:34 PM

Copyright © 2019 My Math Forum. All rights reserved.