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I need help on the below problems. Thank you!!

a. Intel will issue IDs for up to 5000 employees. ID Intel uses IDs beginning with a single capital letter followed by a string of numbers (if needed), what is the minimum number of digits necessary for the ID?

b. Intel also issues vehicle gate passes for up to 5000 employees. If Intel uses gate passes starting with three capital letters followed by a string of numbers (if need), what is the minimum number of digits necessary for the gate pass?

c. Some areas of the Intel plant, such as the clean room, require special combination codes to enter in an electronic lock using a numerical keypad. The codes consist of a series of digits. Intel secutiry has determined as a matter of policy that the probability of breaking a combination code at random should be less than one in a million. If 50 unique codes have been issued to 50 employees, what is the minimum number of digits required?

d. Intel cafeteria management plans a menu of available dishes every quarter. Dishes consists of a soup, an entree, anda dessert, menus always consist of four soups and five desserts. Assuming that more than 100 menu combinations of soup-entree-dessert are required by the cafeteria management for sufficient variety, at least how many entrees are required?

user59117
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1 Answers1

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I’ll work (a); (b) and (d) are very similar, and I’ll add a hint for (c), where it may be a little harder to see what’s wanted.

In (a) suppose that $n$ digits are used after the capital letter. Then there are $26$ possible capital letters, and $10$ possible digits in each of the next $n$ positions, so there are $26\cdot10^n$ possible gate passes. You want to find the smallest non-negative integer $n$ that gives you at least $5000$ passes. Clearly $n=2$ is too small, since it allows only $2600$ passes, but $n=3$ does the job: it allows $26000$ passes, over $5$ times as many as you actually need.

In (c) you should again let $n$ be the number of digits needed. This gives you $10^n$ combination codes. You’ve issued $50$ of these codes, and you want to be sure that if someone picks an $n$-digit code at random, the probability of getting one of the $50$ that you’ve issued is less than $\frac1{1,000,000}$. That probability is $\frac{50}{10^n}$, the number of codes issued divided by the total number of possible codes.

Brian M. Scott
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  • Thanks Brian! I am confused as to how to get the n for letter c. – user59117 Feb 14 '13 at 22:50
  • @user59117: Clearly $n$ must be at least $2$ just in order to have at least $50$ combinations. Does $n=2$ work? No: in that case you’ve used $\frac{50}{100}=\frac12$ of the combinations, and someone choosing a combination at random has a $50$% chance of picking one that works. How about $n=3$? That’s better: the intruder has only a $\frac{50}{1000}=\frac1{20}$ chance, but that’s still way more than $\frac1{1,000,000}$. If you don’t see a shortcut, keep nudging $n$ up one unit at a time until the probability is less than $\frac1{1,000,000}$. – Brian M. Scott Feb 14 '13 at 22:53
  • I used n=8, then 50 / 10 raised to 8 gave me 0.0000005. Thus, the answer is 8 digits, correct? – user59117 Feb 14 '13 at 23:15
  • @user59117: Yes, that’s right. – Brian M. Scott Feb 14 '13 at 23:18
  • Thanks!! Can you give me a hint for letters b & d? – user59117 Feb 14 '13 at 23:22
  • @user59117: (b) is completely trivial: how many passes can you make just using three letters? (d) follows the same pattern as (a) and (b): if $n$ is the number of entrées, how many combinations can you make? How big must $n$ be in order for that to be at least $100$? – Brian M. Scott Feb 14 '13 at 23:28
  • Ok so for letter b, 26 raised to 3, therefore 0 digit? And for d, 6 entrees, as 4x5x6=120 and if I only have 3 entrees it will give me less than 100 combinations. ?? – user59117 Feb 14 '13 at 23:33
  • @user59117: Right: the three letters are already enough, and $5$ entrées gives you exactly $100$ combinations, which is one short of what you need, so a sixth entrée is required. – Brian M. Scott Feb 14 '13 at 23:44
  • Thank you so much... appreciate your help! :) – user59117 Feb 14 '13 at 23:49
  • @user59117: You’re very welcome. – Brian M. Scott Feb 14 '13 at 23:53