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Finding Howlers


Date: 10/25/1999 at 00:13:03
From: Ruby Nagpal
Subject: Howlers: an easy way to reduce fractions

An example of a howler is 16/64. You can cross out the 6 on the top 
and the bottom, and then you are left with 1/4, which is actually 
correct. I am to find 2-digit, 3-digit, and 4-digit howlers also and a 
way to prove that I have found all of them (this is probably only for 
the 2-digit howlers)

The way that I have worked on this is to try every combination of 
fractions, but I need some guidance as to a formula or a method to do 
this. Any help would be greatly appreciated.

Ruby


Date: 10/25/1999 at 13:01:51
From: Doctor Peterson
Subject: Re: Howlers: an easy way to reduce fractions

Hi, Ruby.

I suppose you could try every case, but algebra will work better. 
Let's think about the two-digit case. We can write any two-digit 
number "AB" this way:

     10A + B

So a fraction "AB/BC" (with the ones digit of the numerator the same 
as the tens digit of the denominator) is

     10A + B
     -------
     10B + C

Set this equal to A/C, and you have an equation with three variables. 
You won't get a single solution using algebra, but you can use the 
result to help in your search, so there are fewer cases to try. For 
instance, you can solve for C in terms of A and B, and then think 
about which values of A and B will give a single digit for C.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/25/1999 at 14:17:59
From: Doctor Rob
Subject: Re: Howlers: an easy way to reduce fractions

Thanks for writing to Ask Dr. Math, Ruby.

For two-digit numbers, the approach is this:

     (10*a+b)/(10*b+c) = a/c     (canceling the "b" digits)

            (10*a+b)*c = a*(10*b+c)

          10*a*c + b*c = 10*a*b + a*c

                 9*a*c = b*(10*a-c)

Here a, b, and c are digits, and since the numbers are two-digit ones, 
neither a nor b can be 0. Since division by 0 is forbidden, c cannot 
be 0 either, so 1 <= a <= 9, 1 <= b <= 9, 1 <= c <= 9.

Now you can observe that there is a solution with a = b = c. This 
accounts for things like 11/11 = 1/1, which isn't really a howler, 
since it is obviously correct (you can't tell that a bogus digit 
cancellation occurred). If any two of a, b, and c are equal, the third 
is equal to the same value, as you can readily verify by solving the 
above equation for it. So we can assume that a, b, and c are all 
different.

Now there are three cases to consider:

CASE 1: 9 is a divisor of b.

So b = 9, and

     a*c = 10*a - c,

       0 = 10*a - c - a*c,

      10 = (10-c)*(a+1).

The divisors of 10 are 1, 2, 5 and 10, but the factorization above 
cannot be 10*1, because neither a nor c is 0. This gives the solution 
a = b = c = 9, and two others.


CASE 2: 3 is a divisor of b, but 9 is not.

Then b = 3 or b = 6. For b = 3 we get the equation

     10 = (10-3*c)*(3*a+1).

The factorization must be 1*10 for this to work (the other possible 
factorizations lead to fractional or 0 values for a and c), so
a = b = c = 3.

For b = 6 we get the equation

     40 = (20-3*c)*(3*a+2).

The factorization can be 2*20, 5*8 or 8*5 (the other possible 
factorizations lead to fractional or 0 values for a and c. The first 
leads to a = b = c = 6, and the other 2 give 2 more solutions.


CASE 3: 3 is not a divisor of b.

Then 9 is a divisor of 10*a - b, and so 9 is a divisor of a - b. By 
the inequalities satisfied by a and b, we have

     -8 = 1 - 9 <= a - b <= 9 - 1 = 8

The only multiple of 9 in this range is 0, so a - b = 0, a = b = c.  
This produces 6 more of this kind of solution.

This produces a total of 4 howlers, and 9 non-howlers:

     11/11, 22/22, ..., 99/99.

For multiple-digit cancellations, you can adopt a similar approach.
The 18 possible equations would be, for two-digit cancellations of
three-digit numbers:

     (100*a+10*b+c)/(100*b+10*c+d) = a/d
     (100*a+10*b+c)/(100*b+10*d+c) = a/d
     (100*a+10*b+c)/(100*c+10*b+d) = a/d
     (100*a+10*b+c)/(100*c+10*d+b) = a/d
     (100*a+10*b+c)/(100*d+10*b+c) = a/d
     (100*a+10*b+c)/(100*d+10*c+b) = a/d
     (100*a+10*b+c)/(100*a+10*c+d) = a/d
     (100*a+10*b+c)/(100*a+10*d+c) = a/d
     (100*a+10*b+c)/(100*c+10*a+d) = a/d
     (100*a+10*b+c)/(100*c+10*d+a) = a/d
     (100*a+10*b+c)/(100*d+10*a+c) = a/d
     (100*a+10*b+c)/(100*d+10*c+a) = a/d
     (100*a+10*b+c)/(100*a+10*b+d) = a/d
     (100*a+10*b+c)/(100*a+10*d+b) = a/d
     (100*a+10*b+c)/(100*b+10*a+d) = a/d
     (100*a+10*b+c)/(100*b+10*d+a) = a/d
     (100*a+10*b+c)/(100*d+10*a+b) = a/d
     (100*a+10*b+c)/(100*d+10*b+a) = a/d

For one-digit cancellations of three-digit numbers, you could have

     (100*a+10*b+c)/(100*c+10*d+e) = (10*a+b)/(10*d+e)

and 8 other similar equations.

Naturally, with more digits, the equations become more and more 
complicated, and the analysis more and more involved. The basic 
techniques, however, don't change: use the equation, use the 
inequalities the digits have to satisfy, use the divisibility 
properties of powers of 10 and powers of 10 less 1, and split into 
cases when needed.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Discrete Mathematics
High School Number Theory

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