SOLUTION: if the eggs in a basket are removed 2 at a time, one eggs will remain. If they are removed 3 at a time, 2 eggs remain. If the eggs are removed 4,5 or 6 at a time, then 3,4 and 5

Question 836189: if the eggs in a basket are removed 2 at a time, one eggs will remain. If they are removed 3 at a time, 2 eggs remain. If the eggs are removed 4,5 or 6 at a time, then 3,4 and 5 eggs remian, respectively. But if they are taken out 7 at a time, no eggs will be left over. Find the least numbe of eggs that could be in the basket.
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!

Let the number be N

if the eggs in a basket are removed 2 at a time, one eggs will remain.

So N is 1 less than a multiple of 2, so N=2A-1 for some positive integer A.

If they are removed 3 at a time, 2 eggs remain.

So N is 2 more, and therefore 1 less than, a multiple of 3, so N=3B-1 for some positive integer B.

If the eggs are removed 4...at a time, then 3...eggs remain,

So N is 3 more, and therefore 1 less than, a multiple of 4, so N=4C-1 for some positive integer C

If the eggs are removed ...5... at a time, then ...4... eggs remain,

So N is 4 more, and therefore 1 less than, a multiple of 5, so N=5D-1 for some positive integer D.

If the eggs are removed ...6 at a time, then ...5 eggs remain,

So N is 5 more, and therefore 1 less than, a multiple of 6, so N=6E-1 for some positive integer E

But if they are taken out 7 at a time, no eggs will be left over.

So N is a multiple of 7, so N=7F for some positive integer F

So

N = 2A-1 = 3B-1 = 4C-1 = 5D-1 = 6E-1 = 7F

Add 1 to all those:

N+1 = 2A = 3B = 4C = 5D = 6E = 7F+1

N+1=7F+1 has to be a common multiple of 2,3,4,5,6.  The least
common multiple of 2,3,4,5,6 is 60.  So N+1 is a multiple of
60.  So N is 1 less than a multiple of 60.

So we find the least multiple of 60 that is 1 more than a multiple
of 60.

The first multiple of 60 is 60 itself.  1 less than 60 is 59, but
59 is not a multiple of 7.

The next multiple of 60 is 120.  1 less than 120 is 119, and sure
enough, 119 is a multiple of 7. since 119 = 17*7.

Answer: 119 

Edwin

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