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Sum, Times, I Wonder..(2) (Posted on 2011-04-02) Difficulty: 2 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 1 of 1
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

Where B is the base, abc is B*ab.c and, as b=1 and c=0, may be written as a*B2 + 1*B1+0*B0 and simplified to a*B2 + B

bcd is B2*b.cd 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 by a gives the equation for abc, therefore abc is a-times, hence a multiple of, bcd.
  Posted by Dej Mar on 2011-04-02 11:31:23
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