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Finding Units

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Date: 4/14/96 at 23:18:30
From: Anonymous
Subject: Math puzzle

Could you please show me how to do this problem.

The two digits in the numerator of a fraction are reversed in its
denominator.  If 1 is subtracted from both the numerator and the
denominator, the value of the resulting fraction is 1/2. The
fraction whose numerator is the difference and whose denominator
is the sum of the units and tens digits equals 2/5. Find the
original fraction.

I tried using 10t + U divided by 10u + t to represent the reversed
numbers, and (10t+u)-1 divided by (10u+t)-1 = 1/2 to represent the
second clue, and t-u divided by t+u = 2/5 to represent the third
clue,but it doesn't work.
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Date: 4/26/96 at 7:50:24
From: Doctor Steven
Subject: Re: Math puzzle

Okay we'll search for this answer but we'll do it systematically.

First let's look to see if we can find some conditions we want.

Say a and b are the digits.  We want (ab - 1)/(ba-1) = 1/2.
So (ba-1) must be even.  This says a is odd.

We also want ba > ab, so ab/ba is less than one, so b > a.

Now we want our denominators ba and (ba-1) to be two digit
numbers.

If a >= 5, then our numerator is > 50, so the denominator would be
three digits.  So a < 5.

This says a is odd, a < 5, and b > a.  We also want b almost 2*a
but this is inexact.

So a can be 1, or a can be 3.

If a = 1 then b had better be 2, or else equation 1 wouldn't hold,
so check this: (ab - 1) = 11 and (ba - 1) = 20, almost 1/2 but not
quite, so a does not equal 1.

So a = 3. Since the numerator is in the thirties we know b = 6, or
b = 7. If b = 6 our denominator would have to be in the 70's, so b
cannot equal 6, so b = 7.

So a = 3, and b = 7 if they equal anything at all.  Check this.

(ab - 1) = 36 and (ba - 1) = 72, so (ab - 1)/(ba - 1) = 1/2.

Check to see if (their difference/their sum) = 2/5.

(7 - 3)/(7 + 3) = 4/10 = 2/5

And the original fraction is:

37/73

Sort of a neat puzzle. The trick was you didn't want to use
systems of equations to solve it, but rather think about the
restrictions on a and b.

-Doctor Steven,  The Math Forum

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