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
