The key to solving this is to design a pattern after each bunny has gone.
After the first bunny has gone: every egg would be out of their corresponding basket.
After the second bunny has gone: every second egg would be placed back into their basket, because they were out of their basket. This means basket 2, 4, 6, and so on, would have their eggs placed back in.
After the third bunny has gone: every third egg would be affected. That is, the third egg would be placed back in, but the sixth egg would be placed back out because it was previously placed back in.
We can generalise this to determine which eggs are out, and which are in, by their factors, because the factors are the causes of each egg's status. For example, egg number 2 has a factor of 1 and itself. Thus, we know that the egg's status remains unchanged. In fact, every number with an even number of factors would have their basket status unchanged.
Thus, the only group of numbers that will never satisfy this are perfect squares. Perfect squares have an odd number of factors, because they will have a factor that, when multiplied by itself, produce the desired number.
Let's take 4, as an example. Since its factors are 1, 2, and 4, then bunny number 1 will take the egg out, bunny 2 will place it back in, and bunny 4 will take the egg out.
Thus, we know that 4 is a solution. Using this, we can further generalise a statement. "Every factor will pair up with another factor, unless the desired number is produced as a perfect square" and we know this is true.
Thus, we can conclude that the only solutions to this question are the perfect squares from 1 to 100: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.
2.
Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.
3.
Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following (). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,
4.
The same method used to solve problem (3) can be applied to this problem.