**Puzzle 41**

A man made a journey such that the first twelfth of it was on foot, the next seventh was on a bicycle, and the next sixth was on a boat. He then boarded a train and at a station three miles later, he was joined by a friend who got off at a station six miles before the man got off at the end of his journey, one half the length of which he was on the train with his friend. How long was the man’s journey?

Show Solution*x*= the length of the man;s journey in miles, then

*x*=

*x*/12 +

*x*/7 +

*x*/6 + 3 +

*x*/2 + 6. Solving this linear equation yields

*x*= 84, so the man’s journey is 84 miles long.

**Puzzle 42**

What is the value of ?

Show Solution*x*= . The part of this fraction that begins with the second occurrence of the digit 1 is exactly the same as the original fraction, so we can say

*x*= . Multiplying both sides of this equation by 2 –

*x*produces a quadratic equation whose standard form is

*x*²

*–*2

*x*+ 1 = 0 and whose solution is

*x*= 1, so the value of the fraction is 1.

**Puzzle 43**

A prime number is a positive integer greater than 1 whose only positive factors are 1 and the number itself. An arithmetic sequence is an ordered set of numbers in which there is a common difference between successive terms of the sequence. A) Find the smallest product of three prime numbers that are in arithmetic sequence. B) Find three integers in arithmetic sequence whose product is a prime number.

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**Puzzle 44**

Below is a diagram containing 6 line segments. Add 5 more segments to this diagram to make 9. One solution might be to draw two segments that are extensions of two of the segments that are already there and then draw three other separate and distinct segments, thereby creating a diagram containing 9 segments, two of which are longer than the other 7. Of course, now that I’ve given you that solution, you must come up with a different one.

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**Puzzle 45**

This puzzle is based on a card trick that I have performed on occasion. Bear with me as I describe it without the advantage of demonstrating as I talk. I suggest you follow the instructions with a deck in your hand. I give a standard deck of 52 cards to a member of the audience (you) and then turn around so I can’t see the rest of the proceedings. Shuffle the cards to your satisfaction. Then, holding the deck face down, turn over the top card, *think* the numerical value of the card as you place it face down on the table (aces count as ones and picture cards count as tens) and continue to count from that number up to 12 as you place one card per number from the deck face down on top of the first one. Repeat this process to make a second pile and a third, etc. until there are not enough cards left in the deck to count up to 12 from the value of the top card, in which case, spread the remaining cards face down on the table. Then, I turn around and tell you the sum of the numerical values of the cards on the bottoms of the piles, and you turn over the piles and confirm my assertion. The puzzle has two parts: A) What do I do to determine the sum of those hidden cards? (Hint: I use the number of piles and the number of cards that are spread out. If you can’t figure it out, you can get the answer to this part by clicking on the *Show Solution* *A* button below.) B) More importantly, why does what I do yield the correct sum ?

*n*piles. Then the numbers of cards in the piles would be 13 – , 13 – , 13 – , …, 13 – , respectively. Let

*p*= the number of leftover cards that are spread out on the table. Then (13 – ) + (13 – ) + (13 – ) + … + (13 – ) +

*p*= 52. Since this equation contains

*n*13’s, it is equivalent to 13

*n*– ( + + + … + ) +

*p*= 52 , so 13

*n*– 52 +

*p*= + + + … + , so 13(

*n*– 4) +

*p*= + + + … + . The last equation says that if I do what I say that I do in Part A, I’ll get the sum of the numerical values of all the bottom cards.