Each time Shannon pushes the button on a machine, a bell rings 7 times. Each

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kendra

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time she turns the switch on...? machine, the bell rings 3 times. During one hour, Shannon caused the bell on the machine to ring 23 times. How many times did she push the button?

2. How many positive three-digit integers are there such that each digit is an odd number less than 6?
(a) 3
(b) 6
(c) 9
(d) 18
(e) 27

3. x^2 + 16x +a = (x+b)^2
In the equation above, a and b are constants. If the equation is true for all values of x, what is the value of a?
(a) 4
(b) 8
(c) 16
(d) 64
(e) 256

PLEASE EXPLAIN YOUR ANSWERS =]
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
1) There are a few ways to solve this. Suppose that you have absolutely no mathematical machinery under your belt and are reduced to guessing. You can see that 23 is not divisible by either 3 or 7, so you can check: can he flip the switch once? That leaves 20 bell rings. 20 is not divisible by 7, so you try again. Can he flip the switch twice? That leaves 17 bell rings. Still no good. What about three times? That leaves 14 bell rings, which is divisible by 7! So, one possible setup can be two button pushes and three switch flips. (If you keep going, you will note that there are no more solutions).

2) You can use the digits 1, 3, 5 for each digit in your three-digit number. So you have three decimal places and three choices for each place, so that gives you 3 * 3 * 3 = 3^3 = 27 numbers satisfying the criterion.

3) Let's expand out the right hand side. (x + b)^2 = x^2 + 2bx + b^2. All we have to do is compare the coefficients of the left side to the right side. We see that 2b = 16 and b^2 = a. From the first relation, we know that b = 8. Therefore, a = (8)^2 = 64.

That should do it.
 
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