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Difficulty:
75%
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Question Stats:
55% (02:28) correct
45%
(02:33)
wrong
based on 152
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A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?
(A) 375
(B) 625
(C) 1,875
(D) 3,750
(E) 5,625
(A) 375
(B) 625
(C) 1,875
(D) 3,750
(E) 5,625
08 Oct 2019, 12:34
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rohan2345
I wouldn't call this a sub 600 level problem. It's a combinatorics (counting) problem with a couple of different constraints you need to take into account. If you aren't confident about these, it's fine to skip them.
You solve these by breaking them down into smaller, easier problems. One thing you don't know here is which digit is the number. You do know that there's only one number, and you know it isn't the first or last. So, there are three possibilities:
letter, number, letter, letter, letter
letter, letter, number, letter, letter
letter, letter, letter, number, letter
Work on the first one first. Letters can be repeated, so there are 5 possibilities for each letter. There are also 3 possibilities for the number. So, there are 5*3*5*5*5 codes of that form. Similarly, there are 5*3*5*5*5 codes of each of the other two forms. In total, that gives us 5*3*5*5*5*3 codes.
What I noticed next is that this number will be odd (since it's only divisible by 5s and 3s) and divisible by 9 (since it's divisible by 3 twice.) I'm really trying to avoid having to do the arithmetic! Eliminate (D), because it's even. You can check whether the other answer choices are divisible by 9 by summing together their digits. Only (E) has a digit sum that's divisible by 9, so only (E) can be the right answer.
16 Sep 2024, 05:32
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