XAT 2011 Question 89

Question 89

Ln a bank the account numbers are all 8 digit numbers, and they all start with the digit 2. So, an account number can be represented as $$2x_1x_2x_3x_4x_5x_6x_7$$. An account number is considered to be a ‘magic’ number if $$x_{1}x_{2}x_{3}$$ is exactly the same as $$x_{4}x_{5}x_{6}$$ or $$x_{5}x_{6}x_{7}$$ or both. $$X_{i}$$ can take values from 0 to 9, but 2 followed by seven $$0_{s}$$ is not a valid account number. What is the maximum possible number of customers having a ‘magic’ account number?

Solution

Account number = $$2 x_1x_2x_3x_4x_5x_6x_7$$

Case 1 : $$x_1x_2x_3$$ is exactly same as $$x_4x_5x_6$$

=> $$x_1x_2x_3 = x_4x_5x_6 = 000$$ and $$x_7$$ = 1 to 9   (as '20000000' is not valid)

=> 9 possibilities

Now, $$x_1x_2x_3 = x_4x_5x_6$$ = 001 to 999 and $$x_7$$ = 0 to 9

=> $$999 \times 10 = 9990$$ possibilities.

Case 2 : $$x_1x_2x_3$$ is exactly same as $$x_5x_6x_7$$

=> $$x_1x_2x_3 = x_5x_6x_7 = 000$$ and $$x_4$$ = 1 to 9   (as '20000000' is not valid)

=> 9 possibilities

Now, $$x_1x_2x_3 = x_5x_6x_7$$ = 001 to 999 and $$x_4$$ = 0 to 9

=> $$999 \times 10 = 9990$$ possibilities.

Subtracting common possibilities in above cases.

=> $$x_4x_5x_6 = x_5x_6x_7$$

=> $$x_4 = x_5 = x_6 = x_7$$

Except '0', the possibilities are 1111,2222,....,9999 => 9 possibilities.

$$\therefore$$ Maximum possible number of customers having a ‘magic’ account number

= $$(9 + 9990) + (9 + 9990) - (9)$$

= $$19989$$



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