ok i got the multiples of 7 so what do i do next ...Our class planned a holiday party for disadvantaged kids.Some of us baked cookies for the party.On the day of the party,we found we could divide the cookies into packets of two, three, four, five, or six and have just one cookie left over in each case.If we divided them into packets of seven, there would be no cookies left over. What is the least number of cookies the class could have baked?__________

with your multiples of 7, figure out which multple of 7 will be multiples of the other+1

for instance, take the multiple of 7 : 49

is 48 divisible by 2,3,4,5,6? Yes, on 2,3,4,6, but not five. So try another multiple of 7

Since divided into bags of 2,3,4,5,6, all left one cookie, we have also another way.

Find the LCM (lowest common multiple) of 2,3,4,5,6. Call this L.

Check if L+1 is divisible by 7. If it is, L is your answer. If it is not, try 2L+1, 3L+1, ... until it is divisible by 7. You will not have to try more than 7 times.

If you are not sure how to find the LCM of 2,3,4,5,6, post again.

To find the least number of cookies the class could have baked, we need to find the least common multiple (LCM) of 2, 3, 4, 5, and 6, and then add 1 to it. Since dividing the cookies by 7 leaves no cookies remaining, the LCM of 2, 3, 4, 5, and 6 must also be divisible by 7.

The multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...

Now, let's list the multiples of the other numbers and see where they intersect with the multiples of 7:

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ...
Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, ...
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...

From the lists above, we can see that the multiples of 7 where we have exactly 1 less cookie in each case are: 21, 42, 63, and so on.

The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60 (since it is the smallest number that appears in the multiples of all those numbers).

Adding 1 to the LCM, we get the possible number of cookies that the class could have baked: 60 + 1 = 61.

Therefore, the least number of cookies the class could have baked is 61.

To find the least number of cookies the class could have baked, we need to find the smallest number that is divisible by each of the numbers given (2, 3, 4, 5, and 6) and also one more than a multiple of 7.

First, let's find the multiples of 7. The multiples of 7 are:

7, 14, 21, 28, 35, ...

We are looking for a number that has a remainder of 1 when divided by 2, 3, 4, 5, and 6. This means we need to find a number that is one more than a multiple of 7 and also satisfies these divisibility conditions.

Starting with the smallest multiple of 7 (7), we add 1 to each multiple to see if it satisfies the divisibility conditions:

7 + 1 = 8 (not divisible by 2, move on)
14 + 1 = 15 (not divisible by 2, move on)
21 + 1 = 22 (not divisible by 2, move on)
28 + 1 = 29 (not divisible by 2, move on)
35 + 1 = 36 (divisible by 2)

Now we have found a number (36) that is divisible by 2 and also one more than a multiple of 7. However, we need to check if it is divisible by the other numbers (3, 4, and 5).

36 is divisible by 3 since the sum of its digits (3 + 6 = 9) is divisible by 3.

Next, let's check if 36 is divisible by 4. Since the last two digits of 36 are 36, which is divisible by 4, 36 is divisible by 4.

Lastly, let's check if 36 is divisible by 5. Since the last digit of 36 is 6, which is divisible by 5, 36 is divisible by 5.

Therefore, the least number of cookies the class could have baked is 36.

To summarize, we found the least number of cookies the class could have baked by finding a number that is one more than a multiple of 7, divisible by 2, 3, 4, and 5. Starting with the smallest multiple of 7, we added 1 until we found a number (36) that satisfies all the conditions.