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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

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

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