# Answer to Question #5031 in Algebra for zachary

Question #5031

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 divid the cookies into packets of two, three, four, five, six and just have one cookie left over in each case. If we divided them in to packets of seven, there would be no cookies left over. What is the least number of cookies the class could have baked?

Expert's answer

Let's denote the number of cookis by N. Now we can formalize the given conditions:

(N-1)|2& (2 divides N-1)

(N-1)|3& (3 divides N-1)

(N-1)|4& (4 divides N-1)

(N-1)|5& (5 divides N-1)

(N-1)|6& (6 divides N-1)

N|7 (7 divides N& ) ==> N=7K

We see that if (N-1)|2 and (N-1)|3 then (N-1)|6 because 2*3=6.

There exists such integer M that N-1 = 2*3*4*5*M ==> N = 2*3*4*5*M+1.

Also we know that N = 7K

So, we get an Diophantine equation:

2*3*4*5*M+1 = 7K, or

120M+1 = 7K.

Let's solve it by enumeration method:

M=1: 121 = 7K ==> K isn't integer;

M=2: 241 = 7K ==> K isn't integer;

...

M=6: 721 = 7K ==> K = 103.

So, N = 721.

(N-1)|2& (2 divides N-1)

(N-1)|3& (3 divides N-1)

(N-1)|4& (4 divides N-1)

(N-1)|5& (5 divides N-1)

(N-1)|6& (6 divides N-1)

N|7 (7 divides N& ) ==> N=7K

We see that if (N-1)|2 and (N-1)|3 then (N-1)|6 because 2*3=6.

There exists such integer M that N-1 = 2*3*4*5*M ==> N = 2*3*4*5*M+1.

Also we know that N = 7K

So, we get an Diophantine equation:

2*3*4*5*M+1 = 7K, or

120M+1 = 7K.

Let's solve it by enumeration method:

M=1: 121 = 7K ==> K isn't integer;

M=2: 241 = 7K ==> K isn't integer;

...

M=6: 721 = 7K ==> K = 103.

So, N = 721.

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