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

Question

A five-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible.

A. 2,016
B. 3,888
C. 4,608
D. 5,184
E. 6,236

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EgmatQuantExpert · Answer

Solution

Given:

• A five-digit code uses the digits 0,1,2,3,4,5,6,7,8,9.
• The first digit must be 2 or 4
• 3rd digit must be a multiple of 3
• No two consecutive digits should be same

To find:  
 • The number of different codes possible.

Approach and Working:

Let the 5-digit number is abcde.

• Ways to fill ‘a’ = 2(2 or 4)

• Ways to fill ‘c’ = 4 (0,3,6,9)

• Hence, we filled two different digits in ‘a’ and ‘c’. 
 o So, out of 10 digits, b cannot take 2 digits. 
o Thus, ‘b’ can be filled in 8 ways.

• Now, ‘d’ can be filled in 9 ways as similar digit cannot be in ‘c’ and ‘d’.

• Similarly, ‘e’ can be filled in 9 ways.

Thus, total ways to fill= 2*8*4*9*9= 5184

Hence, the correct answer is option d.

Answer: d

Hero8888 · Answer

A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible.

1st cell - 2 options (2,4) - the first digit must be 2 or 4
3rd cell - 4 options (0,3,6,9) - multiples of 3
2nd cell - 8 options - can not be surrounded by the same digits that are located in cells 1 and 3, so we have to pick from 10-2=8
4th cell - 9 options
5th cell - 9 options

2*8*4*9*9 = 5,184

Answer D

thefibonacci · Answer

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.

raks85 · Answer

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.

EgmatQuantExpert · Answer

Hey  raks85 ,

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

Bubbasch · Answer

What if the question states, "A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be next to each other, and repetitions are not allowed. How many different codes are possible?"

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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