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 multidigit base 36 numbers, and not the product. For example, if a=1, b=2, c=3 and d=4, then b.cd represents the base36 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 base36 the following are valid values for a and d (in base36, 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
Where
B is the base,
abc is
B*ab.c and, as
b=1 and
c=0, may be written as
a*B^{2} + 1*B^{1}+0*B^{0 }and simplified to
a*B^{2} + Bbcd is
B^{2}*b.cd and, as
b=1, c=0 and, as
B=a*d, i.e.,
d=B/a, may be written as
1*B^{2} + 0*B^{1} + (B/a)*B^{0 }and simplified to
B^{2} + B/aMultiplying the equation for
bcd by
a gives the equation for
abc, therefore
abc is
atimes, hence a multiple of,
bcd.

Posted by Dej Mar
on 20110402 11:31:23 