Original post: thunderdogg
Posted: Mon Nov 26, 2007 10:14 pm
Question 6 from this problem set:
6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?
a) 27.
b) 36.
c) 72.
d) 112.
e) 422.
---
So um... how does a 5 digit credit card have a 6th digit that is used to determine the 5th digit. This problem set might be a waste of time. ONly on problem 6 and the questions are clearly not very good.
This problem actually makes sense. We have 5 slots which we need to fill with different numbers. And we are given that the first and the second are 1 & 2 respectively.
_1_ _2_ __ __ __
The third digit is bigger than 6, therefore it could be either 7 or 8 or 9, since the range is [1:9].
The fourth digit is divisible by 3, therefore it could be either 3 or 6 or 9.
The fifth digit is 3 times the sixth, therefore it's multiple of 3, so in a given range it could be either 3 or 6 or 9. By the way, the sixth digit could be either 1 or 2 or 3, since the product of 5th and 6th can't be more than 9. So the reference to the 6th digit in the question is absolutely sensible.
From this point, it looks like a simple combinatorics problem. We have 3 options in each of three decisions. So to calculate possible numbers we need to multiply 3 by 3 by 3. So the answer is 27 (A).
