1 S(6)

2 S(7)

3 C(1,3)

4 C(1,4)

5 J(2,7,19)

6 J(4,6,14)

7 J(3,5,11)

8 S(1)

9 S(5)

10 J(1,1,7)

11 S(6)

12 Z(5)

13 J(1,1,6)

14 C(1,3)

15 S(7)

16 Z(6)

17 S(6)

18 J(1,1,5)

n | m | 0 | 0 | 0 | 0 | 0 |

n | m | n | n | 0 | 1 | 1 |

n+1 | m | n | n | 1 | 1 | 1 |

n+2 | m | n | n | 2 | 1 | 1 |

n+3 | m | n | n | 3 | 1 | 1 |

… | … | … | … | … | … | … |

n+n=2n | m | n | n | n | 1 | 1 |

2n | m | n | n | 0 | 2 | 1 |

2n+1 | m | n | n | 1 | 2 | 1 |

2n+2 | m | n | n | 2 | 2 | 1 |

… | … | … | … | … | … | … |

2n+n=3n | m | n | n | n | 2 | 1 |

3n | m | n | n | 0 | 3 | 1 |

… | … | … | … | … | … | … |

… | … | … | … | … | … | … |

n(n)=n^{2} |
m | n | n | n | n-1 | 1 |

n^{2} |
m | n | n | 0 | n | 1 |

n^{2} |
m | n^{2} |
n | 0 | 1 | 2 |

n^{2}+1 |
m | n^{2} |
n | 1 | 1 | 2 |

n^{2}+2 |
m | n^{2} |
n | 2 | 1 | 2 |

… | … | … | … | … | … | … |

n^{2}+n^{2}=2n^{2} |
m | n^{2} |
n | n^{2} |
1 | 2 |

2n^{2} |
m | n^{2} |
n | 0 | 2 | 2 |

… | … | … | … | … | … | … |

… | … | … | … | … | … | … |

n(n^{2})=n^{3} |
m | n^{2} |
n | n^{2} |
n | 2 |

n^{3} |
m | n^{2} |
n | 0 | n | 2 |

n^{3} |
m | n^{3} |
n | 0 | 1 | 3 |

… | … | … | … | … | … | … |

… | … | … | … | … | … | … |

… | … | … | … | … | … | … |

n(n^{m-1})=n^{m} |
m | n^{m-1} |
n | n^{m-1} |
n | m-1 |

n^{m} |
m | n^{m} |
n | 0 | 1 | m |

Example 1.

2 | 4 | 0 | 0 | 0 | 0 | 0 |

2 | 4 | 2 | 2 | 0 | 1 | 1 |

3 | 4 | 2 | 2 | 1 | 1 | 1 |

4 | 4 | 2 | 2 | 2 | 1 | 1 |

4 | 4 | 2 | 2 | 0 | 2 | 1 |

4 | 4 | 4 | 2 | 0 | 1 | 2 |

5 | 4 | 4 | 2 | 1 | 1 | 2 |

6 | 4 | 4 | 2 | 2 | 1 | 2 |

7 | 4 | 4 | 2 | 3 | 1 | 2 |

8 | 4 | 4 | 2 | 4 | 1 | 2 |

8 | 4 | 4 | 2 | 0 | 2 | 2 |

8 | 4 | 8 | 2 | 0 | 1 | 3 |

9 | 4 | 8 | 2 | 1 | 1 | 3 |

10 | 4 | 8 | 2 | 2 | 1 | 3 |

11 | 4 | 8 | 2 | 3 | 1 | 3 |

12 | 4 | 8 | 2 | 4 | 1 | 3 |

13 | 4 | 8 | 2 | 5 | 1 | 3 |

14 | 4 | 8 | 2 | 6 | 1 | 3 |

15 | 4 | 8 | 2 | 7 | 1 | 3 |

16 | 4 | 8 | 2 | 8 | 1 | 3 |

16 | 4 | 8 | 2 | 0 | 2 | 3 |

16 | 4 | 16 | 2 | 0 | 1 | 4 |

Example 2.

3 | 3 | 0 | 0 | 0 | 0 | 0 |

3 | 3 | 3 | 3 | 0 | 1 | 1 |

4 | 3 | 3 | 3 | 1 | 1 | 1 |

5 | 3 | 3 | 3 | 2 | 1 | 1 |

6 | 3 | 3 | 3 | 3 | 1 | 1 |

6 | 3 | 3 | 3 | 3 | 2 | 1 |

6 | 3 | 3 | 3 | 0 | 2 | 1 |

7 | 3 | 3 | 3 | 1 | 2 | 1 |

8 | 3 | 3 | 3 | 2 | 2 | 1 |

9 | 3 | 3 | 3 | 3 | 2 | 1 |

9 | 3 | 3 | 3 | 3 | 2 | 1 |

9 | 3 | 3 | 3 | 0 | 3 | 1 |

9 | 3 | 9 | 3 | 0 | 1 | 2 |

10 | 3 | 9 | 3 | 1 | 1 | 2 |

11 | 3 | 9 | 3 | 2 | 1 | 2 |

12 | 3 | 9 | 3 | 3 | 1 | 2 |

13 | 3 | 9 | 3 | 4 | 1 | 2 |

14 | 3 | 9 | 3 | 5 | 1 | 2 |

15 | 3 | 9 | 3 | 6 | 1 | 2 |

16 | 3 | 9 | 3 | 7 | 1 | 2 |

17 | 3 | 9 | 3 | 8 | 1 | 2 |

18 | 3 | 9 | 3 | 0 | 2 | 2 |

19 | 3 | 9 | 3 | 1 | 2 | 2 |

20 | 3 | 9 | 3 | 2 | 2 | 2 |

21 | 3 | 9 | 3 | 3 | 2 | 2 |

22 | 3 | 9 | 3 | 4 | 2 | 2 |

23 | 3 | 9 | 3 | 5 | 2 | 2 |

24 | 3 | 9 | 3 | 6 | 2 | 2 |

25 | 3 | 9 | 3 | 7 | 2 | 2 |

26 | 3 | 9 | 3 | 8 | 2 | 2 |

27 | 3 | 9 | 3 | 9 | 2 | 2 |

27 | 3 | 9 | 3 | 0 | 3 | 2 |

27 | 3 | 27 | 3 | 0 | 1 | 3 |

For any positive integers n and m,

1 J(1,3,3)

2 J(2,3,23)

3 J(2,3,22)

4 S(6)

5 S(7)

6 C(1,3)

7 C(1,4)

8 J(2,7,25)

9 J(4,6,17)

10 J(3,5,14)

11 S(1)

12 S(5)

13 J(1,1,10)

14 S(6)

15 Z(5)

16 J(1,1,9)

17 C(1,3)

18 S(7)

19 Z(6)

20 S(6)

21 J(1,1,8)

22 J(1,1,22)

23 Z(1)

24 S(1)

Note: if n=m=0, the answer is undetermined and the program never stops.

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2 S(1)

3 S(1)

4 S(1)

5 S(3)

6 J(3,4,1)

7 S(2,3,9)

8 S(1,1,2)

0 | 0 | 0 | 0 |

3 | 1 | 0 | 0 |

3 | 1 | 3 | 0 |

6 | 2 | 3 | 0 |

9 | 3 | 3 | 0 |

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A bank teller made a mistake today. The teller switched the dollars and cents when they cashed a check for Mrs. Jones, giving her dollars instead of cents and cents instead of dollars.

After buying a newspaper for 5 cents, Mrs. Jones realized that she had remaining exactly twice as much as the original check.

**
**What was the amount of the original check?

**Answer**

Let original check be x dollars and y cents. So the money given is y dollars and x cents. Therefore:

100y + x – 5 = 2(100x + y)

So,

199x – 98y = -5

We require integer values of x and y that solve this equation (a Diophantine equation). Following the algorithm to solve a Diophantine equation:

199=2*98+3

98=32*3+2

3=1*2+1

2=2*1

1=3-1*2=3-1(98-32*3)=33*3-1*98=33(199-2*98)-1*98=33*199-66*98-1*98=33*199-67*98

(* denotes multiplication).

So, a=33 and b=67 solves the equation 199a-98b = 1. Multiplying both sides by -5, we get:

199(-5a)-98(-5b) = -5.

Hence, x = -5×33 = -165 and y = -5×67 = -335, solves the equation 199x – 98y = -5. However, these are not the only solutions. The general solution is given by:

x = -165+98n and y = -335+199n, where n is an integer.

Check:

199(-165+98n)-98(-335+199n) = -199*165+199x98n+98×335-98x199n = -199×165+98×335 = -5

We require a solution where both x and y are positive. The first solution of this kind occurs when n=2, giving:

x = -165+98*2 = 31 and y = -335+199*2 = 63.

Hence, the original check was 31$63c.

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If we toss a coin 10 times, are we more likely to get heads ten times or heads 7 times and tails 3 times?

**Answer
**

Heads 7 times and tails 3 times.There is only one way of getting heads 10 times (HHHHHHHHHH), but there are many ways of getting heads 7 times and tails 3 times (for example: HHHHHHHTTT, TTTHHHHHHH, HHHTHTTHHH, HTHTHHTHHH).

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If we toss a coin 10 times, which of the two following sequences is more likely to occur?

HHHHHHHHHH

HHTTHTHTTT

(H stands for ‘heads’ and T stands for ‘tails’)

**Answer
**

The two sequences are equally likely. The probability of either one of them occurring is 1/2 to the power of 10

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Event A – possible equally probable outcomes: 0 and 1

Event B – possible equally probable outcomes: 0 and 1

Event A and event B each occurred once. The outcome of one of the two events was a 0 and occurred during a time interval t within a time interval T. The other event occurred during the time interval T. What is the probability that the outcome of the other event was also a 0?

Use the result to confirm the solution to problem 2.

What happens as t tends to 0? What if t = T? What is the range of this probability?

**Answer 4**

Independently of whether event A occurs before, after or simultaneously to event B, the only relevant combinations of these outcomes are:

- Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T)
**AND**Event B – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T)

- Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T)
**AND**Event B – outcome = 0, during time interval T but not during time interval t (probability = (1/2)x(T-t)/T) = (T-t)/(2T)

- Event A – outcome = 0, during time interval t (probability = (1/2)x(t/T) = t/2T)
**AND**Event B – outcome = 1, during time interval T (probability = 1/2)

- Event A – outcome = 0, during time interval T but not during time interval t (probability = (1/2)x(T-t/T) = (T-t)/2T)–
**AND**Event B – outcome = 0 during time interval t (probability = (1/2)x(t/T) = t/2T)

- Event A – outcome = 1 at any time in time interval T (probability = 1/2)
**AND**Event B – outcome = 0 during time interval t (probability = (1/2)x(t/T) = t/(2T))

The probabilities of each of these combinations are:

Probability of combination 1) = (t/2T)(t/2T) = t^{2}/4T^{2}

Probability of combination 2) = (t/2T)((T-t)/2T) = t(T-t)/4T^{2}

Probability of combination 3) = (t/2T)(1/2) = t/4T

Probability of combination 4) = ((T-t)/(2T))(t/2T) = t(T-t)/4T^{2}

Probability of combination 5) = (1/2)( t/2T) = t/4T

Combinations 1), 2) and 4) are formed by two 0s, hence the probability of the second outcome being a 0 is:

(t^{2}/4T^{2}+t(T-t)/4T^{2}+t(T-t)/4T^{2}) / (t^{2}/4T^{2}+t(T-t)/4T^{2}+t/4T+t(T-t)/4T^{2}+t/4T) = ((2tT-t^{2})/4T^{2}) / ((4tT-t^{2})/ 4T^{2}) =

(2T-t) / (4T-t)

In problem 2 the outcomes were: boy or girl. So, 0 corresponds to a boy and 1 to a girl. One of the children was a boy born on a Tuesday, hence t = 1 day. All of the different days of the week occur over the course of a 7 day period, and the other child must have been born on one of the 7 days of the week; thus T = 7. Substituting these values we get:

(2×7-1) / (4×7-1) = 13/27.

As t tends to 0, the probability tends to: 2T/4T = ½.

If t = T, the probability becomes: (2T-T)/(4T-T) = T/(3T) = 1/3.

Therefore this probability must always be greater than (or equal to if we allow the events to occur in zero time) ½ and less than or equal to 1/3.

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A person has two children. One is a boy born at an exact specified instant in time (consider a small time interval that tends to zero). What is the probability that the other child is also a boy.

**Answer 3**

Let us assume that the probability of a boy being born is equal to the probability of a girl being born (both 50%) and also assume that the births of the two children were not simultaneous. The only possibilities are:

- Another boy was born before the boy born at the exact specified instant time.

- Another boy was born after the boy born at the exact specified instant time.

- A girl was born before the boy born at the exact specified instant time.

- A girl was born after the boy born at the exact specified instant time.

These four possibilities are all equally probable. In two of them the other child is a boy and in two of them the other child is a girl, hence the probability of the other child being a boy is 50%.

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A person has two children. One is a boy born on a Tuesday. What is the probability that the other child is also a boy?

**Answer
**

Let event A and event B be the births of the two children. Let us assume that the probability of a boy being born is equal to the probability of a girl being born (both 50%). The possible relevant outcomes are:

Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

Event A – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)

Event A – Girl born on any day (probability = 1/2)

and

Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

Event B – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)

Event B – Girl born on any day (probability = 1/2)

(Check: 1/2 + 3/7 + 1/14 = 1)

Independently of whether event A occurs before, after or simultaneously to event B, the only relevant combinations of these outcomes are:

- Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)
**AND**Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

- Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)
**AND**Event B – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)

- Event A – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)
**AND**Event B – Girl born on any day (probability = 1/2)

- Event A – Boy born on any day except Tuesday (probability = (1/2)x(6/7) = 6/14 = 3/7)
**AND**Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14)

- Event A – Girl born on any day (probability = 1/2)
**AND**Event B – Boy born on a Tuesday (probability = (1/2)x(1/7) = 1/14

** **

The probabilities of each of these combinations are:

Probability of combination 1) = (1/14)(1/14) = 1/196

Probability of combination 2) = (1/14)(3/7) = 3/98

Probability of combination 3) = (1/14)(1/2) = 1/28

Probability of combination 4) = (3/7)(1/14) = 3/98

Probability of combination 5) = (1/2)(1/14) = 1/28

Combinations 1), 2) and 4) are formed by two boys, hence the probability of the second child being a boy is:

((1/196)+(3/98)+(3/98)) / ((1/196)+(3/98)+(1/28)+(3/98)+(1/28)) = (13/196) / (27/196) = 13/27

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A person has two children. One is a boy. What is the probability that the other child is also a boy?

**Answer
**

Let event A and event B be the births of the two children. Let us assume that the probability of a boy being born is equal to the probability of a girl being born (both 50%). The possible outcomes are:

Event A – Boy (50% probability)

Event A – Girl (50% probability)

and

Event B – Boy (50% probability)

Event B – Girl (50% probability)

Independently of whether event A occurs before, after or simultaneously to event B, the only possible combinations of these outcomes are:

- Event A – Boy
**AND**Event B – Boy

- Event A – Boy
**AND**Event B – Girl

- Event A – Girl
**AND**Event B – Boy

- Event A – Girl
**AND**Event B – Girl

Therefore, if we are told that one of the children is a boy, we can have any of the first three outcomes, but not the last outcome. In only one of the equally probably first three outcomes we have two boys, hence the probability of the other child being a boy in 1/3.

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