What is the least number that divides 24, 30, 40, and 108, leaving a remainder of 16?
Understand the Problem
The question is asking for the least number that divides the numbers 24, 30, 40, and 108, leaving a remainder of 16. This involves finding a common divisor based on the given conditions.
Answer
The least number is $18$.
Answer for screen readers
The least number that divides 24, 30, 40, and 108 leaving a remainder of 16 is 18.
Steps to Solve
- Adjust each number for the remainder To find the least number that divides each of the given numbers (24, 30, 40, 108) leaving a remainder of 16, we first need to subtract 16 from each of these numbers.
Calculate:
- $24 - 16 = 8$
- $30 - 16 = 14$
- $40 - 16 = 24$
- $108 - 16 = 92$
Now we have the new set of numbers: 8, 14, 24, and 92.
- Find the greatest common divisor (GCD) Next, we need to find the GCD of the adjusted numbers (8, 14, 24, 92). We will start by finding the prime factorization of each number:
- 8: $2^3$
- 14: $2 \times 7$
- 24: $2^3 \times 3$
- 92: $2^2 \times 23$
The common factor in all of these is $2$, so the GCD is $2$.
- Identify the least number To find the least number that divides the original numbers leaving a remainder of 16, we take the GCD we found and add back the remainder.
Thus, the least number is: $$ \text{Least number} = \text{GCD} + \text{remainder} = 2 + 16 = 18 $$
The least number that divides 24, 30, 40, and 108 leaving a remainder of 16 is 18.
More Information
The concept of finding a number that leaves a specific remainder when dividing is often related to modular arithmetic. The process of adjusting the base numbers to eliminate the remainder simplifies the calculation of common divisors.
Tips
- Forgetting to adjust the original numbers: Remember to subtract the remainder from all numbers before finding the GCD.
- Incorrect factorization: Ensure you are correctly identifying prime factors without errors.
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