When Mike sorts his marbles, he notices that if he put them in groups of 5, he has one left over. When he sorts them in groups of 7, he also has one left over, but in groups of 6, he has none left over. What is the smallest number of marbles that he could have?

Look at the multiples of 6.

6, 12, 18, 24, 30, 36, 42, 48

Which of these numbers also fits the description in the problem?

Ah, the curious case of Mike and his mischievous marbles! Well, it seems like he's in a bit of a numerical conundrum here. Let's put on our detective hats and solve this puzzle.

To find the smallest number of marbles that satisfies these conditions, we need to find the smallest multiple that leaves a remainder of 1 when divided by both 5 and 7, but leaves no remainder when divided by 6.

Let's start with multiples of 5 and add 1 to see if it leaves a remainder of 1 when divided by 7. We've got 6, 11, 16... and so on. After some analysis, we find that the first multiple that satisfies both conditions is 16.

But hold your chuckles, we're not done yet! For our final test, we need to check if this number leaves no remainder when divided by 6. *drumroll, please*

Ta-da! 16 is indeed divisible by 6 with no remainder. So the smallest number of marbles Mike could have is 16.

Now, excuse me while I juggle some calculations of my own.

To find the smallest number of marbles that satisfy these conditions, we need to find the smallest common multiple of 5, 7, and 6.

Let's start by finding the common multiple of 5 and 7 that leaves a remainder of 1 for both. We can use trial and error:

5 = 1 (mod 7) means that 5 is congruent to 1 modulo 7.
Thus, we can write 5 as 7n + 1, where n is an integer.

Substituting this into the equation 5k + 1 = 7m + 1, we get:
5k = 7m + 1 - 1
5k = 7m

We want to find the lowest value of k that satisfies this equation.

Taking multiples of 7 (since 7m is divisible by 7):
7, 14, 21, 28, 35, ...

Taking multiples of 5 (5k), we observe that the first multiple that has a common multiple with 7 is 35.

Hence, the smallest common multiple of 5 and 7 (leaving a remainder of 1 for both) is 35.

Now, let's find the smallest common multiple of 35 and 6 that leaves no remainder:

To find the lowest common multiple (LCM) of two numbers (35 and 6), we can use the formula:

LCM(a, b) = (a * b) / GCD(a, b)

where GCD(a, b) represents the greatest common divisor of a and b.

The GCD(35, 6) can be found using the Euclidean Algorithm:
35 = 5 * 6 + 5
6 = 1 * 5 + 1
5 = 5 * 1 + 0

The GCD(35, 6) is 1.

Using the formula, we find the LCM(35, 6):
LCM(35, 6) = (35 * 6) / 1 = 210.

Therefore, the smallest number of marbles that Mike could have is 210.

To find the smallest number of marbles that satisfy these conditions, we need to find the least common multiple (LCM) of 5, 7, and 6.

To start, we can find the LCM of 5 and 7. One way to do this is to list the multiples of both numbers until we find the smallest common multiple.

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, ...

Multiples of 7: 7, 14, 21, 28, 35, 42, 49, ...

The smallest common multiple of 5 and 7 is 35.

Now, we need to find the LCM of 35 and 6. Again, we can list the multiples of both numbers.

Multiples of 35: 35, 70, 105, 140, 175, 210, ...

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, ...

The smallest common multiple of 35 and 6 is 210.

Therefore, the smallest number of marbles that Mike could have is 210.