A certain set of six consecutive non-zero digits can be rearranged to form three two-digit numbers whose product consists of the same six digits. What are the three two-digit numbers?
Possible selection 6 consecutive non-zero digits.
1,2,3,4,5,6
2,3,4,5,6,7
3,4,5,6,7,8
4,5,6,7,8,9
So, 4,5 is common in every sequence
Case(i):
If 5 divides one of the 3 2-digit numbers, then all of
them must be odd to satisfy the given condition.
(If one of them is even then last digit is 0, which contradicts)
Case(ii):
Sum of digits of the sequences=6/2*(1st+last)=3*k
which is 21,27,33,39
The sequences 1,3,4 are multiples of 3 but not 9.
So, the 6-digit number in these cases is divisible by 3 but not 9.
one of 2-digit no.s is divisible by 3, whereas other 2 are not.
In other case, it is divisible by 27.
Observation: These 2 digit no.s are not divisible by 11.
So, if abcdef is the 6-digit number then
a+c+e is not equal to b+d+f (or) difference between them is not 11.
Are any of these useful in getting to the solution?
Edited on July 19, 2007, 10:18 am
Some thoughts!
