If a, b, and c are different numerals between 1 and 9 (inclusive), and
aaaaaa = b * c * bc * (c+c-b) * (c+b+b) * (c-b),
what are a, b, and c?
Note: Adjacent numerals are multi-digit numbers made of those numerals, not the product; i.e., if m=2 and n=3, then mn=23.
(In reply to
Answer by K Sengupta)
We know that:
aaaaaa= 111,111*a
=(1001)*(111)*a
= 7*11*13*3*37*a
Hence, bc=13 or 37
CASE 1: bc=13
Then, lhs= 111,111*a
rhs= 1*3*13*5*5*2
= 3*13*50
This is possible if a=0 in the lhs, giving the lhs as 0, which is a cryptarithmetic contradiction.
CASE 2: bc=37
Then, lhs= 111111*a
rhs= 3*7*37*11*13*4
= (7*11*13)*(3*37)*4
=(1001)*(111)*4
= (111,111)*4
It is now evident that lhs=rhs only when:
a=4
Consequently, (a,b,c) =(4,3,7) is the required solution to the given cryptarithmetic puzzle.
Edited on April 5, 2022, 10:41 pm
Puzzle Solution
