Each of the small letters in bold represents a different base 36 digit from 0 to Z to satisfy this cryptarithmetic equation. None of the numbers can contain any leading zero.
ab.c + b.cd = (ab.c)*(b.cd)
Prove that abc is always divisible by bcd.
Note: Adjacent numerals are multi-digit base 36 numbers, and not the product. For example, if a=1, b=2, c=3 and d=4, then b.cd represents the base-36 number 2.34 and, a.bc represents the base 36 number 1.23.
The equation, ab.c + b.cd = (ab.c)*(b.cd), is true when
b = 1, c = 0, and the product of a and d equals the value of the base.
For base-36 the following are valid values for a and d (in base-36, I=18 and C=12):a d
I 2 : I1.0 + 1.02 = I1.0 * 1.02
C 3 : C1.0 + 1.03 = C1.0 * 1.03
9 4 : 91.0 + 1.04 = 91.0 * 1.04
6 6 : 61.0 + 1.06 = 61.0 * 1.06
4 9 : 41.0 + 1.09 = 41.0 * 1.09
3 C : 31.0 + 1.0C = 31.0 * 1.0C
2 I : 21.0 + 1.0I = 21.0 * 1.0I
is the base, abc
and, as b=1
, may be written as a*B2 + 1*B1+0*B0
and simplified to a*B2 + Bbcd
and, as b=1, c=0
and, as B=a*d
, i.e., d=B/a
, may be written as 1*B2 + 0*B1 + (B/a)*B0
and simplified to B2 + B/a
Multiplying the equation for bcd
gives the equation for abc
, therefore abc
-times, hence a multiple of, bcd
Posted by Dej Mar
on 2011-04-02 11:31:23