Finding HowlersDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/