We are asked to find the greatest 3-digit number which, when divided by 75, 45, and 60, leaves:
(A) No remainder
(B) A remainder of 4 in each case.
Part A: No remainder
The number should be divisible by 75, 45, and 60 with no remainder. This means we need to find the least common multiple (LCM) of 75, 45, and 60, and then find the greatest 3-digit number divisible by this LCM.
Step 1: Find the LCM of 75, 45, and 60
Prime factorization of 75: 75 = 3 × 5²
Prime factorization of 45: 45 = 3² × 5
Prime factorization of 60: 60 = 2² × 3 × 5
To find the LCM, we take the highest powers of all prime factors:
Highest power of 2: 2²
Highest power of 3: 3²
Highest power of 5: 5²
So, the LCM of 75, 45, and 60 is: LCM = 2² × 3² × 5² = 4 × 9 × 25 = 900
Step 2: Find the greatest 3-digit number divisible by 900
The greatest 3-digit number is 999. To check if it's divisible by 900: 999 ÷ 900 ≈ 1.11, which is not an integer.
The largest multiple of 900 within the 3-digit range is: 900 × 1 = 900
So, the greatest 3-digit number divisible by 75, 45, and 60 with no remainder is 900.
Part B: Remainder 4 in each case
Now, we need to find the greatest 3-digit number that, when divided by 75, 45, and 60, leaves a remainder of 4 in each case. This means the number must be 4 more than a multiple of the LCM of 75, 45, and 60.
Step 1: Use the LCM from Part A
From Part A, we found that the LCM of 75, 45, and 60 is 900.
Step 2: Find the number of the form 900k + 4
We need to find the largest 3-digit number of the form 900k + 4, where k is an integer.
The largest 3-digit number is 999, so we solve the inequality: 900k + 4 ≤ 999
900k ≤ 995
k ≤ 995 ÷ 900
k ≤ 1.1
So, the largest integer k is 1.
For k = 1: 900 × 1 + 4 = 904
Therefore, the greatest 3-digit number that leaves a remainder of 4 when divided by 75, 45, and 60 is 904.
Final Answer:
The greatest 3-digit number divisible by 75, 45, and 60 with no remainder is 900.
The greatest 3-digit number that leaves a remainder of 4 when divided by 75, 45, and 60 is 904.