As a mathematical expert, I am well-versed in solving problems involving the least common multiple (LCM) and finding numbers that satisfy specific divisibility conditions. The task at hand is to find the smallest number that is divisible by 35, 56, and 91, and additionally leaves a remainder of 7 when divided by these numbers.
To begin with, let's understand what the least common multiple is. The LCM of a set of numbers is the smallest number that is a multiple of each of the numbers in the set. It is a fundamental concept in number theory and is used to solve problems related to divisibility and common multiples.
In the case of the numbers 35, 56, and 91, we first need to find their LCM. The prime factorization of these numbers is as follows:
- 35 = 5 × 7
- 56 = 2^3 × 7
- 91 = 7 × 13
The LCM is found by taking the highest power of each prime number that appears in the factorization of these numbers. In this case, the LCM would be:
LCM = 2^3 × 5 × 7 × 13 = 8 × 5 × 7 × 13 = 3640
Now, the problem specifies that the number we are looking for must leave a remainder of 7 when divided by 35, 56, and 91. This is a condition that modifies the LCM we have found. The number we are looking for is not just a multiple of the LCM, but a multiple of the LCM plus 7.
So, the smallest number that satisfies the condition is:
Number = LCM + 7 = 3640 + 7 = 3647
This number, 3647, when divided by 35, 56, and 91, will leave a remainder of 7 in each case. It is the smallest such number because any smaller number would not be divisible by all three numbers and would not satisfy the remainder condition.
To summarize, the process involves finding the LCM of the given numbers and then adjusting it to meet the specific remainder condition. The result is a unique number that satisfies all the criteria of the problem.
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