To solve these problems, let's break down the steps one by one.
### Part 6: Find the least number which, when increased by 3, is divisible by 36, 40, and 64.
First, we need to determine the least common multiple (LCM) of the three numbers provided (36, 40, and 64). The LCM is the smallest number that is a multiple of all three numbers.
To find the LCM, we can break down each number into its prime factors:
- \(36 = 2^2 \times 3^2\)
- \(40 = 2^3 \times 5\)
- \(64 = 2^6\)
Now take the highest powers of each prime number found in the factorization:
- For \(2\), the highest power is \(2^6\) (from 64).
- For \(3\), the highest power is \(3^2\) (from 36).
- For \(5\), the highest power is \(5\) (from 40).
Multiplying these together gives us the LCM:
\(LCM = 2^6 \times 3^2 \times 5 = 64 \times 9 \times 5 = 576 \times 5 = 2880\).
Therefore, the least number which, when increased by 3, should be divisible by the LCM 2880. In other words, we are looking for a number such that \(number + 3 = k \times 2880\) for some integer \(k\).
Since \(2880\) is divisible by \(2880\), if we subtract \(3\) from \(2880\), we get \(2877\), which is not divisible by \(2880\). Therefore, the next number to check is \(2880 \times 2 = 5760\). Subtracting \(3\) gives us \(5757\), which is also not divisible by \(2880\). The pattern here is that adding multiples of the LCM to our initial choice won't work because \(n + 3 = k \times LCM\) implies \(n\) already being divisible by the LCM, and since \(3\) is not a factor of \(2880\), we can't achieve divisibility this way.
However, since we know that \(2877\) is three less than \(2880\) and adding multiples of \(2880\) to \(2877\) will always result in a number three less than a multiple of \(2880\), the least number we are looking for is actually \(2877\), since it's the smallest number that satisfies \(number + 3\) being divisible by the LCM we found.
### Part 7: The question is incomplete.
Unfortunately, the second part of your question (part 7) is incomplete, so I'm not able to provide an answer or guide you through the solution. If you can provide the complete question, I'd be more than happy to help you with it.
