The least number of eggs that could be in the basket is 119.
Multiple of the number:
The multiple is the numbers you get when you multiply a certain number by an integer.
Given,
If the eggs in a basket are removed 2 at a time, 1 egg will remain.
If the eggs are removed 3 at a time, 2 eggs will remain.
If the eggs are removed 4, 5, or 6 at a time, then 3, 4, and 5 eggs will remain, respectfully.
But if they are taken out 7 at a time, no eggs will be left over.
Here we need to find the least number of eggs in the basket.
Let the number of eggs in the basket be N.
We know that,
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
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
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
Similarly,
For 5,
=> N = 5D - 1
For 6,
=> N = 6E - 1
So, if they are taken out 7 at a time, no eggs will be left over.
So,
N => 7F
Where A,B,C,D,E, and F are any positive integer.
Therefore,
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
Through this,
N + 1 = 7F + 1
has to be a common multiple of 2,3,4,5,6.
So, the LCM of 2,3,4,5,6 is 60.
So N+1 is a multiple of 60 which means 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 multiples are, 60, 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.
So,
=> 119 = 17*7.
Therefore, the least number of eggs that could be in the basket is 119.
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