
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 27th, 2016, 11:10 AM  #1 
Newbie Joined: Jun 2016 From: baku Posts: 1 Thanks: 0  Problem involving cubed numbers
ABCD  DCBA = A^3 + B^3 + C^3 + D^3 ABCD and DCBA are four digit numbers. A, B, C, D are different numbers between 0 and 9. 
June 28th, 2016, 04:50 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479 
You really did not ask a question. Do you want the answer (which we tend not to give for homework) or help finding the answer. If the latter, this kind of problem teaches you how to use information that is general knowledge but not explicit in the statement of the problem or that is implied but not obvious from the problem itself. Where I would start is with this: a, b, c, and d are nonnegative. And we are dealing with FOUR digit numbers. So a and d are greater than 0. Do you see why? So $a^3 + b^3 + c^3 + d^3 > 0 \implies d \not > a.$ Why? But the problem says a and d are not equal so $1 \le d < a \le 9.$ What thoughts do you have about what to try next? 
June 28th, 2016, 06:13 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,131 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Here's another hint for problems like this: if ABCD is a fourdigit number where the digits have been replaced with letters, then an equivalent description of the number is $\displaystyle 1000A + 100B + 10C + D$ where A, B, C and D are now algebraic variables (and also integers). 
June 28th, 2016, 11:24 AM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
                                                                  As another true example, there was a sign on a wall in a Kroger supermarket deli that read: "No sugar cookies are here!" I asked a worker, "When will the sugar cookies be here?" The hyphen makes all the difference.  
June 28th, 2016, 12:42 PM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479  Quote:
 
June 28th, 2016, 07:56 PM  #6 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
June 29th, 2016, 01:23 AM  #7  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,131 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
Hint... look up the meaning of the word "pedantry"  

Tags 
cubed, involving, numbers, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Probability Questions Involving 64 Numbers  EvanJ  Advanced Statistics  0  December 5th, 2013 05:08 PM 
Equation Involving Complex Numbers  bilano99  Algebra  2  July 25th, 2013 05:42 AM 
Simultaneous equations involving complex numbers?! STRESSS!!  queenie_n  Complex Analysis  5  October 17th, 2012 05:20 PM 
Proof involving a field and prime numbers  Dudealadude  Abstract Algebra  1  January 28th, 2012 02:03 AM 
Series Involving Prime Numbers  everk  Real Analysis  2  September 5th, 2011 01:06 PM 