Let's solve this problem step-by-step:
1. Understanding the Problem:
We need to find the least number that, when divided by 15, 45, and 60, leaves a remainder of 6 in each case.
2. Expressing the Condition Mathematically:
Let's denote the unknown number by [tex]\( N \)[/tex]. According to the problem, [tex]\( N \)[/tex] satisfies the following conditions:
[tex]\[
N \equiv 6 \pmod{15}
\][/tex]
[tex]\[
N \equiv 6 \pmod{45}
\][/tex]
[tex]\[
N \equiv 6 \pmod{60}
\][/tex]
3. Rewriting the Expressions:
We can rewrite [tex]\( N \)[/tex] as [tex]\( N = 15k + 6 \)[/tex], [tex]\( N = 45m + 6 \)[/tex], and [tex]\( N = 60n + 6\)[/tex] for some integers [tex]\( k \)[/tex], [tex]\( m \)[/tex], and [tex]\( n \)[/tex].
4. Finding the Common Value:
Since [tex]\(N\)[/tex] leaves a remainder of 6 when divided by 15, 45, and 60, let's consider the number [tex]\( N - 6 \)[/tex]. This number must be divisible by all three divisors:
[tex]\[
N - 6 = k \cdot 15 = m \cdot 45 = n \cdot 60
\][/tex]
5. Finding the Least Common Multiple (LCM):
To find a number that is divisible by 15, 45, and 60, we need to calculate the Least Common Multiple (LCM) of these numbers.
6. Calculating the LCM of 15, 45, and 60:
- The prime factorization of the numbers are:
[tex]\[
15 = 3 \times 5
\][/tex]
[tex]\[
45 = 3^2 \times 5
\][/tex]
[tex]\[
60 = 2^2 \times 3 \times 5
\][/tex]
- The LCM takes the highest power of each prime that appears:
[tex]\[
\text{LCM}(15, 45, 60) = 2^2 \times 3^2 \times 5 = 180
\][/tex]
7. Finding the Least Number [tex]\( N \)[/tex]:
Since [tex]\( N - 6 \)[/tex] must be a multiple of 180:
[tex]\[
N - 6 = 180
\][/tex]
[tex]\[
N = 180 + 6 = 186
\][/tex]
8. Conclusion:
The least number which, when divided by 15, 45, and 60, leaves a remainder of 6 in each case is:
[tex]\[
\boxed{186}
\][/tex]
