Math Is Fun Forum

XxMathCuriousXx

Member

Registered: 2009-02-05

Posts: 2

ABCDEF × 3 help please!!!!

I need help knowing how numbers can be in place of these letter twice...
Numbers being down  as 285714 and 142857
There are two different ways  to solve it and does A get counted as 1 or 0?
i'm lost
please help!!!

ABCDEF × 3 = BCDEFA

The Solution . . .
285714 × 3=857142
or
142857 × 3=428571

Last edited by XxMathCuriousXx (2009-02-05 14:25:13)

XxMathCuriousXx

Member

Registered: 2009-02-05

Posts: 2

Re: ABCDEF × 3 help please!!!!

Like why those numbers like how you get it... I thought again A is 0 or 1...

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: ABCDEF × 3 help please!!!!

Since they're arranged in the same order both times, you can think of BCDEF as a single variable.
So now you have AZ x 3 = ZA

Translating this to a 'standard' equation:
(100000A + Z) x 3 = 10Z + A

Rearrange for Z:

300000A + 3Z = 10Z +A
299999A = 7Z
Z = (299999/7)A = 42857A

Now, A is a single-digit number, so that means there are only a few possibilities that could work.

A=1 --> Z = 42857, which gives 142857 as a solution.
A=2 --> Z = 85714, which gives 285714
A=3 --> Z = 128571. However, Z is a 5-digit number, so this is too high.
The same goes for any value of A higher than this.

(Technically, you could also put A=0 --> Z=0, giving 000000 as a working answer. Not sure if you want to allow that)

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Why did the vector cross the road?
It wanted to be normal.