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?__________
To find the least number of cookies the class could have baked, we need to look for a number that satisfies the given conditions: dividing the cookies into packets of 2, 3, 4, 5, or 6 leaves 1 cookie left over in each case, while dividing them into packets of 7 leaves no cookies left over.
To approach this problem, we can use a technique known as the Chinese remainder theorem. Let's consider the remainders when dividing the number of cookies by each of the numbers 2, 3, 4, 5, 6. We'll start by finding the least common multiple (LCM) of these numbers.
The LCM of 2, 3, 4, 5, 6 is 60. This means that any number of cookies that leaves remainders of 1 when divided by 2, 3, 4, 5, and 6 will also leave a remainder of 1 when divided by 60.
Now, let's consider the remainder when dividing 60 by 7. The remainder is 4. If we subtract 4 from 60, we get 56, which is divisible by 7 without any remainder. Therefore, any number of cookies that leaves remainders of 1 when divided by 2, 3, 4, 5, 6, and 0 (or no remainder) when divided by 7 will also leave a remainder of 1 when divided by 60.
The least number of cookies meeting these conditions is 60 + 1 = 61. Hence, the class could have baked a minimum of 61 cookies.
To find the least number of cookies that the class could have baked, we need to find the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, or 6, and leaves no remainder when divided by 7.
The least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60. So, the number of cookies must be a multiple of 60.
Let's find the smallest multiple of 60 that satisfies the conditions. Starting from 60, we can add 60 each time until we find a number that satisfies all the conditions.
60 + 60 = 120 (remainder 0 for all cases except 6)
120 + 60 = 180 (remainder 0 for all cases except 6)
180 + 60 = 240 (remainder 0 for all cases except 6)
240 + 60 = 300 (remainder 0 for all cases except 6)
300 + 60 = 360 (remainder 0 for all cases except 6)
360 + 60 = 420 (remainder 0 for all cases except 5)
420 + 60 = 480 (remainder 0 for all cases except 6)
480 + 60 = 540 (remainder 0 for all cases except 6)
540 + 60 = 600 (remainder 0 for all cases except 6)
600 + 60 = 660 (remainder 0 for all cases except 5)
660 + 60 = 720 (remainder 0 for all cases except 4)
720 + 60 = 780 (remainder 0 for all cases except 6)
780 + 60 = 840 (remainder 0 for all cases)
Therefore, the least number of cookies the class could have baked is 840.