What is the least number of cookies we could have baked? Please show all work.?

Question

Our class planned a 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.

Answer 1

n is the number of students.

n mod 2 = 1, so it's odd.
n mod 3 = 1
n mod 4 = 1,
n mod 5 = 1, so it ends in 1 (or 6 but 6 is even)
n mod 6 = 1
n mod 7 = 0

So really maybe the best thing to do is to find common multiples of 2,3,4,5,and 6, add 1, and see if they're divisible by 7.

The least common multiple would be 4*3*5 = 60, but 61 isn't divisible by 7.

Next is 120, but 121 doesn't work.

181, nope.

241, nope.

301...works!

Answer 2

Let the number of cookies = 7x (since it is a multiple of 7)

7x has to be one more than a multiple of the Least common multiple of 2,3,4,5,6
LCM of 2,3,4,5,6 = (5 * (LCM of 2,3,4,6))
= 5* 12
= 60

so,
7x = 60n + 1

So,
x = (60n + 1) / 7
where x and n are both Positive Integers
(n is the number of packets we would get if we packed 60 in a packet)

Now, to solve this equation,
METHOD 1:
Try various values of n to fit this equation:
n --> 60n+1 --> x
1 --> 61 --> 8.something
2 --> 121 --> 17.something
3 --> 181 --> 25.something
4 --> 241 --> 34.something
5 --> 301 --> 43 <----- This is the number

So, minimum number of cookies is 301

METHOD 2:
Plot on a single graph, the equations:
y = 7x
y = 60n+1

you will reach the intersection at 301 (x = 43 or n = 5)

METHOD 3:
To complicate the solution instead of following the substitution method, you can use Linear programming:
x = (60n + 1) / 7 <--- Basis equation
n > 0 <--- To ensure positivity of n
minimise n <-- target