When eggs in a basket are removed two, three, four, five or six at a time, there remain, respectively, one, two, three, four, or five eggs. When they are taken out seven at a time, none are left over. Find the smallest number of eggs that could have been contained in the basket.

59

Step-by-step explanation:

If we remove 2 eggs at a time, it will remain 1 eggs, so the number of eggs is an odd number.

If we remove 3 eggs at a time, it will remain 2 eggs, so possible numbers of eggs (just odd numbers) are:

5, 11, 17, 23, 29, 35, 41, 47, 53, 59,  ...

If we remove 4 eggs at a time, it will remain 3 eggs, so possible numbers of eggs are:

7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, ...

Comparing with the values from removing 3 eggs, the equal values are:

11, 23, 35, 47, 59, ...

If we remove 5 eggs at a time, it will remain 4 eggs, so possible numbers of eggs are:

9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, ...

Comparing with the possible values previously calculated, we have just the value 59 (and the bigger values that weren't calculated)

If we remove 6 eggs at a time, it will remain 5 eggs, so possible numbers of eggs are:

11, 17, 23, 29, 35, 41, 47, 53, 59, ...

In this group we have the value 59. So the smallest number of eggs that could have been contained in the basket is 59.

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.

4 puzzles in 3 hours

step-by-step explanation: