A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,

code is of the form abcde
let's fill the constraints first:
a can be filled in 2 ways (2 or 4)
c can be filled in 4 ways (0,3,6, or 9)
b can be filled in 8 ways (can't be same as a or c)
d can be filled in 9 ways (can't be same as c)
e can be filled in 9 ways (can't be same as d)
total = 2*4*8*9*9 = 5184.

IMO it should be 2*8*3*9*9= 3888 (B)
I do not understand how 0 can be considered a multiple of 3. In that way 0 is multiple of each and every number.

Hey raks85,

Yes, 0 can be written as a multiple of any integer.
Regards,
Ashutosh
e-GMAT

Hello everyone,

Thanks for participation.

We have posted the official answer to the question.

Regards,
Ashutosh
e-GMAT

egmat

I have a doubt -in digit abcde d can also take the value of e right? d is sandwiched between c and e ( as b is sandwiched between a and c) . de is also a consecutive pair- so shouldn't the no of ways to fill d be 8? and that of e be 9

I'm also not clear as why d can be 9 ways since it cannot be the same as c or e and there are no restrictions on d. Are c-d and d-e not Consecutive pairs as well? Bunuel, please help me understand my mistake. Thanks.

Posted from my mobile device

I had the exact same doubt. One possible explanation is :

Going as per solution, If the code is abcde, then,

i) First we choose, "a" - this has 2 ways of being chosen (2,4).
ii) then we choose "c" - this has 4 ways of being chosen (yes, 0 is a multiple- 0,3,6,9).
iii) next, we lock on "b" - this has 8 ways of being chosen (SINCE "b" cannot be same as "a" or "c");
moving forward,
iv) to choose "d" - we have 9 possible ways here because "d" cannot be same as "c", AND HERE we have TO realize that "e" has not yet been chosen...
hence, now that "d" is locked,
v) we move forward to choose "e" - 9 ways to choose "e" because "e" cannot be same as "d".

with this condition, both "d" and "e" will be unique.

=> Final answer is 2x8x4x9x9 = 5184

I hope my reasoning is clear and unambiguous.

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