**5. Math Riddle 5**

There are two ducks in front of two other ducks. There are two ducks behind two other ducks. There are two ducks beside two other ducks. How many ducks are there?

**Answer: **Four ducks. They are in square formation.

**4. Math Riddle 4**

Harry says, “The day before yesterday I was 25 and the next year I will be 28. This is only true on one day in a year.” What day is Harry’s birthday?

**Answer: **Harry’s birthday is on the 31^{st} December. He must have made the statement on the 1^{st} January as the day before yesterday (30^{th} December), he was 25. He turned 26 on the 31^{st} December (his birthday). He made the statement in a new year so in that year he would be 27 and “the next year”, he would be 28.

**3. Math Riddle 3**

Today is Connor’s birthday. A year ago on his birthday he had five candles and he lit them all except for the last one. Today he is going to light all the candles. How old is Connor today? *(He is not turning five, by the way.)*

**Answer: **Connor is 31 years old today as he is lighting the candles in binary.

If you convert the number 30 into binary you will get 11110 which is why Connor lit all the candles except for the last one a year ago. However, the number 31 in binary is 11111 hence Connor lights all the candles this time.

**2. Math Riddle 2**

If,

29-1=30

9-1=10

14-1=15

Based on the above logic, can you prove that

11-1=10?

**Answer: **The trick to solving this math riddle is to think of the numbers in their Roman numerical form. The Roman equivalent for the number 1 is I so if you remove from the left side, you get the number on the right side. For example:

29-1=30

XXIX-I=XXX

14-1=15

XIV-I=XV

9-1=10

IX-I=X

Therefore…

11-1=10

XI-I=X.

**1. Math Riddle 1**

There are 100 people standing in a circle all numbered from 1 to 100. Person 1 has a sword and he uses it to kill person 2 and then gives the sword to person 3. Person 3 kills person 4 with the sword and then gives it to person 5. This is continued until only one person is left standing. Which number survives?

**Answer: **Person 73 will be the last man standing. This math riddle can be solved using the Josephus problem approach. If you notice the pattern of eliminating every 2^{nd} person, all the even numbered people will get killed the first time you go around the circle. Whenever the number of people are in powers of 2, person 1 will always survive, so the solution requires getting the nearest smallest number that is the power of 2 or alternatively the greatest power of 2 that is less than the number of people in the circle.

For example, there are 100 people in the circle, so if we write out the powers of 2, the number 100 lies between 2^{6 }[two to the power of six] (64) and 2^{7 }[two to the power of seven] (128). In this case, 2^{6 }or 64 is the greatest power of 2 that is less than the total number of people in the circle. Therefore if you construct the following equation:

n=2x(X-Y)+1, where n=survivor, X=number of people in the circle and Y=greatest power of 2 that is less than or equal to the total number of people in the circle, you can input the necessary values to work out the survivor:

n=2x(100-64)+1

n=73.

You can, of course, solve this the long way by writing down every single number and working your way through it. Whatever works I guess.