#### Why did Snap, Crackle and Pop get scared?

**Answer: ** They heard there was a cereal killer on the loose.

**Solution: **

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#### I am a two-digit number. All my digits are even. No two digits are the same. None of my digits are prime numbers. I am not a multiple of ten. My tens digit is bigger than my other numbers. If you followed all the previous steps, there should be three options remaining the number is the option where if you add all the digits it’s exactly in the middle (in how big the number is) of all the other options with their digits added together. What number am I?

**Answer: ** If you followed all the steps apart from the last one there will be three options remaining: 64, 84, and 86. You then had to add up the digits, 64=6+4=10, 84=8+4=12, and 86=8+6=14. Finally, you then had to take up the middle biggest number (12) and put it back as it was before the digits were added together and your answer should be 84.

**Solution: **

## Show Answer

If you followed all the steps apart from the last one there will be three options remaining: 64, 84, and 86. You then had to add up the digits, 64=6+4=10, 84=8+4=12, and 86=8+6=14. Finally, you then had to take up the middle biggest number (12) and put it back as it was before the digits were added together and your answer should be 84.

## Show Answer

If you followed all the steps apart from the last one there will be three options remaining: 64, 84, and 86. You then had to add up the digits, 64=6+4=10, 84=8+4=12, and 86=8+6=14. Finally, you then had to take up the middle biggest number (12) and put it back as it was before the digits were added together and your answer should be 84.

#### Lighter than air, I float away, with one mans touch I shall decay. What am I?

**Answer: ** Bubble.

**Solution: **

#### There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, …). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, …). This continues until all 100 people have passed through the room. What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

**Answer: ** First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

**Solution: **

## Show Answer

First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 …….. That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it’s original state. So why aren’t all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it’s self. Clearly the sixth person will only flick the bulb once and so the pairs don’t cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

## Show Answer

First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 …….. That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it’s original state. So why aren’t all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it’s self. Clearly the sixth person will only flick the bulb once and so the pairs don’t cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

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