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DEFABC = 6(ABCDEF)


Date: 12/10/2001 at 06:39:38
From: Vincent Tan
Subject: Real Numbers

Let abcdef be a 6-digit integer such that defabc is 6 times the value 
of abcdef. Find the value of  a + b + c + d + e + f.


Date: 12/10/2001 at 13:16:18
From: Doctor Greenie
Subject: Re: Real Numbers

Hello, Vincent -

This is a very interesting problem - one that lends itself to two 
totally different methods of solution. Thanks for sending it to us.


First approach - as a word arithmetic problem.

One approach to this problem is as a word arithmetic problem, where 
the possible values for each digit are determined using logic and our 
knowledge of the process of multiplication.

We have

   ABCDEF
   x    6
   ------
   DEFABC

The first thing I notice here is that a 6-digit number multiplied by 6 
yields a 6-digit number; this means A must be 1. So we have

   1BCDEF
   x    6
   ------
   DEF1BC

In the product, digit D must be 6, 7, 8, or 9. Looking at the 
multiplication process in the 100s column elminates two of these 
possibilities: if D is 7, then the "carry" from the 10s column to the 
100s column would have to be 9, which is not possible; and if D is 9, 
then that same carry would have to be 7, which is also not possible.  
So D is either 6 or 8, and we have

   1BC6EF        1BC8EF
   x    6   or   x    6
   ------        ------
   6EF1BC        8EF1BC

In the first possibility, B must be 0, so we would have

   10C6EF
   x    6
   ------
   6EF10C

The E in the product must be 2, 3, 4, or 5 (digits 0 and 1 are already 
used); looking at the multiplication process in the 1s and 10s column 
for each of these possible values for digit E shows that the only 
possible digits for E and " are E=3 and F=2; but that make C=2 also.  
The assumption that D=6 has led to a dead end, so we now know D is 8.  
So we have

   1BC8EF
   x    6
   ------
   8EF1BC

Examining the "carry" from the 10000s column to the 100000s column, we 
see that B must be either 3 or 4.  So we have

   13C8EF       14C8EF
   x    6   or  x    6
   ------       ------
   8EF13C       8EF14C

If B is 3, then, looking at the multiplication process in the 10000s 
column, we see that E must be either 0 or 2; but looking at the 
multiplication process in the 1s and 10s columns, we see that neither 
of these values is possible.  So B is 4, and we have

   14C8EF
   x    6
   ------
   8EF14C

Looking at the multiplication process in the 10000s column in this 
case, we see that E can be 5, 7, or 9. But trying each of these 
possible values for E in the multiplication process in the 1s and 10s 
column, we find that the only possible values for E and F are E=5 and 
F=7; these values make C=2.

So we have our unique solution:

   142857
   x    6
   ------
   857142


Second approach - as an algebra problem.

The given problem says

   ABCDEF
   x    6
   ------
   DEFABC

If we think of ABC and DEF as the 3-digits integers x and y, 
respectively, then ABCDEF is 1000x+y and DEFABC is 1000y+x; so we have

    6(1000x+y) = 1000y+x

    6000x+6y = 1000y+x

    5999x = 994y

        5999x        35x         5x
    y = ----- = 6x + --- = 6x + ---
         994         994        142

Since x and y are integers, the fraction 5x/142 must be an integer; 
the smallest integer x for which this fraction is an integer is x=142, 
and we then have

    y = 6x + 5x/142 = 852+5 = 857

The solution to the algebraic problem is x=142 and y=857; so the 
solution to the original problem is (as before)

    142857
    x    6
    ------
    857142

Thanks for a fun problem.  Write back if you have any further 
questions about this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Puzzles

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