Question

1. your locker has a combination lock. the combination consists of a specific 3 number code. the numbers go from 0 to 60 and they can repeat. how many codes are possible?

Explanation:

Step1: Identify the number of choices per digit

The combination lock has a 3 - number code, and each number can be from 0 to 60. So, for each of the three digits in the code, there are 61 possible choices (since we include both 0 and 60, the number of integers from 0 to 60 is \(60 - 0+ 1=61\)).

Step2: Apply the multiplication principle

For a three - digit code where each digit has \(n_1\), \(n_2\), and \(n_3\) choices respectively, the total number of possible codes is \(n_1\times n_2\times n_3\). Here, \(n_1 = n_2=n_3 = 61\)? Wait, no, wait. Wait, the problem says "the numbers go from 0 to 60 and they can repeat". Wait, maybe I misread. Wait, if the numbers are from 0 to 60, the number of possible values for each digit is 61? But maybe the problem is that the numbers are from 0 to 60, but maybe it's a typo or maybe it's from 0 to 59? Wait, no, let's re - examine the problem.

Wait, the user's problem: "Your locker has a combination lock. The combination consists of a specific 3 number code. The numbers go from 0 to 60 and they can repeat. How many codes are possible?"

Wait, if each of the three numbers in the code can be any number from 0 to 60 (inclusive), then the number of choices for each digit is \(60 - 0 + 1=61\). But that would give \(61\times61\times61=61^{3}=226981\). But maybe the problem is that the numbers are from 0 to 59? Because usually, combination locks with numbers from 0 to 59 (like a 60 - number dial). Maybe it's a misprint. Let's assume that the numbers are from 0 to 59 (so 60 choices per digit). Then:

Step1: Correct the number of choices per digit

If the numbers are from 0 to 59, then for each of the three digits in the 3 - number code, the number of possible choices is 60 (since \(59 - 0+1 = 60\)).

Step2: Calculate the total number of codes

Using the multiplication principle of counting, for a 3 - digit code where each digit has 60 choices and digits can repeat, the total number of possible codes is \(60\times60\times60=60^{3}\).

We know that \(60^{3}=60\times60\times60 = 3600\times60=216000\).

(It's likely that the problem meant numbers from 0 to 59, as 0 to 60 would be 61 numbers, and 61 cubed is 226981, but 60 cubed is a more standard result for a combination lock with 0 - 59 numbers. So we proceed with the assumption that it's 0 - 59, 60 choices per digit.)

Answer:

216000