Cancellation is the process of dividing the numerator and the denominator of a fraction by a common factor to produce a simpler fraction
E.g. 30/18 = (5 * 6)/(3 * 6) = 5/3
30/18 can be simplified to 5/3 by canceling “6” from the fraction.
An amusing activity in mathematics searches for fractions that remain unchanged by the action of “Anomalous Cancellation”.
We write a solution in the form ab/bc = a/c. And, we ignore the (trivial) solutions having a = b or b = 0 and assume that a, b, c are one-digit numbers in the range 1 ≤ a, b, c ≤ 9
For example, deleting the digit 6 from each of the numerator and denominator of the fraction
16 1
—- = —-
64 4
Other examples,
49 4 1
—- = — = —-
98 8 2
19 1
—- = —-
95 5
26 2
—- = —
65 5
Of course deleting digits this way is an invalid mathematical operation, and the equalities obtained here are purely coincidental.
We could find these with a simple computer program which checks all 9^3 possibilities.
But where’s the fun in that?
Consider these four examples:
16/64 = 1/4, 19/95 = 1/5, 26/65 = 2/5, 49/98 = 4/8
Let’s rather do a little bit of algebra, and see what we get.
We write the basic equation as (10a + b)/(10b + c) = a/c
(10a + b)c = (10b + c)a
10ac + bc = 10ba + ca
10ac – 10ba = ca – bc
10a(c – b) = c(a – b) [Eq #1]
Suppose that a > b. Then a/c = ab/bc > 1 and a > c.
Since the right-hand side of [Eq #1] is then positive,
it must be the case that c > b,
But then c(a – b) = 10a(c – b) > 10(c – b) > 0,
and a – b > 10(c – b) > 0.
This forces 8 ≥ a – b ≥ 10(c – b) ≥ 10, which is impossible.
So a < b and c < b as well.
Since 2 and 5 divide the left-hand side of [Eq #1],
it follows that either
5 divides c and 2 divides a – b, or 2 divides c and 5 divides a – b.
Suppose the first case holds. Since c is a one-digit number,
it follows that c = 5 and then 2a(5 – b) = a – b.
This, in turn, gives that a divides a – b, and so a divides b.
These conditions on a, b, c imply that a is one of 1, 2, 3, 4,
which allow us to solve for b in [Eq #1],
The second and third examples appear here.
Suppose the second case holds.
Then 5 = a – b which leads again to a being one of 1, 2, 3, 4
from which we can solve for b, c.
The first and fourth examples appear here.
Amazing infinite chain of coincidences:
1/4 = 16/64 = 166/664 = 1666/6664 = …
1/5 = 19/95 = 199/995 = 1999/9995 = …
2/5 = 26/65 = 266/665 = 2666/6665 = …
4/8 = 49/98 = 499/998 = 4999/9998 = …
Additional larger examples abound:
484/847 = 4/7 654/545 = 6/5 4324/3243 = 4/3
4582/1264 = 58/16 3346/1673 = 34/17 4455/3564 = 45/36.
Other examples,
187 / 880 = 17/80 199/995 = 1/5 742/424 = 7/4
654/545 = 6/5 484/847 = 4/7 499/998 = 4/8
166/664 = 1/4 266/665 = 2/5 138/1840 = 3/40
4277/2730 = 47/30 4784/7475 = 48/75 6188/1820 = 68/20
And from Wolfram Math World the following fractions:
13/325 = 1/25 124/217 = 4/7 127/762 = 1/6
138/184 = 3/4 139/973 = 1/7 145/435 = 1/3
148/185 = 4/5 154/253 = 14/23 161/644 = 11/44 = 1/4
163/326 = 1/2 176/275 = 16/25 182/819 = 2/9
187/286 = 17/26 187/385 = 17/35 187/748 = 1/4
218/981 = 2/9 273/728 = 3/8 275/374 = 25/34
286/385 = 26/35 316/632 = 1/2 327/872 = 3/8
364/637 = 4/7 412/721 = 4/7 436/763 = 4/7
Fun With Num3ers 