Hints and Solutions to 5 Classic Math and Logic Puzzles
The first part of this article contains short hints to help you solve the five brain teasers in the original article 5 Classic Math and Logic Puzzles. The second part contains the full solutions. See how many you can solve with only a hint. Don't scroll past the hints section if you don't want the answer spoiled!
Hints to Solve the 5 Puzzles
Puzzle #1
When Wesley puts an equal number of coins on each side of the balance scale there are two possible outcomes  one side is heavier than the other, or both sides weigh the same. What does this tell Wesley about which pile the fake coin is in?
Puzzle #2
If some people need to cross the bridge more than others, it should be the two faster people. The slower people should cross as few times as possible. Also, each person's crossing speed is limited by the speed of their partner (if they have one).
Puzzle #3
You don't need a calculator to check if the arithmetic is correct or incorrect, nor do you need to work out each side of the equation by hand.
Puzzle #4
Alison doesn't know her uncle's age because there are multiple possibilities for each of the first digits. But some of the last digits are represented only once. If Alison knows that Bethany is equally clueless, what does this tell you about the last digit?
Puzzle #5
There are only finitely many sets of three positive integers whose product is 144, but the sums of these sets of numbers vary widely. If Tom can't figure out what the set of numbers is given both the product and the sum, what does this tell you about the sum? If a 6yearold has two younger siblings, what does this tell you about the age differences?
Full Solutions and Explanations to the 5 Puzzles
Puzzle #1
There are many ways to do it that guarantee Wesley will find the coin using the scale four or fewer times. Here is one possible method. First Wesley puts 9 coins on each side of the scale and leaves 11 remaining on the side. If one side of the scale is heavier than the other, he knows the fake coin is in that pile of 9, otherwise if both sides weigh the same he knows the fake coin is in the pile of 11. Assuming the fake coin is in one of the piles of 9, he splits it into three piles of 3 coins each and compares two of the piles on the scale, knowing the fake coin will either be on the scale or in the remaining pile of 3. After identifying which pile of 3 contains the fake coin, he then balances two of the coins on the scale. The fake coin will either be the one that imbalances the scale or the one left on the side.
Assuming the fake coin is in the pile of 11, he splits it into two piles of 3 coins each one pile of 5 and compares the two piles of 3 with the scale. Either the fake coin imbalances the scale or is the pile of 5. If the fake coin is in one of the piles of 3, he only need to use the scale once more to identify it. If the fake coin is the pile of 5, he puts 2 of the coins on the scale and leaves 3 on the side. If the fake coin is in the pile of 3 on the side, he only needs to use the scale once more to identify it.
This method uses the scale at most four times. In fact, with this method the probability of only needing the scale three times is 89.66%.
In general, if a person has N coins and one of them is fake, the person needs to use the scale K times, where K = ceiling[ Log_3 (N) ].
Puzzle #2
Call the four friends W, X, Y, and Z. First, W and X cross the bridge together  2 minutes. Next, W comes back alone with the flashlight  1 minute. Next, Y and Z cross the bridge together  8 minutes. Next, X comes back alone with the flashlight  2 minutes. Finally, W and X cross the bridge together  2 minutes. Total time is 2 + 1 + 8 + 2 + 2 = 15 minutes.
Puzzle 3:
This is a trick question because if you evaluate the expression 123456^17 + 654321^17 on a pocket calculator it will display
7.386600338206 E 98
and if you evaluate the expression 5213281192281563766701593^4 you will get
7.386600338206 E 98
This might lead you to believe that the two expressions are in fact equal. However, you can use modular arithmetic to see that the last digit of the the first expression is "7" while the last digit of the second expression is "1."
When you take a number that ends in "6" and raise it to a power you get another number that ends in "6." And when you take a number that ends in "1" and raise it to a power you get another number that ends in "1." Therefore, 123456^17 + 654321^17 is a number that ends in "7" since 6 + 1 = 7.
When you take a number that ends in a "3" and raise it to a power, the end digits cycle through the values 3, 9, 7, 1, 3, 9, 7, 1, 3, 9, 7, 1, ... in a predictable pattern. If the exponent is 4, 8, 12, 16, 20, or any number divisible by 4 then the last digit will be a "1." Since the two expressions have different last digits, they are not equal.
Puzzle #4
Francis is 87 years old. A chart of the possible ages by first and last digit is helpful here.
Alison doesn't know her uncle's age based on the first digit because there are multiple choices for each first digit. The only way that she could state with certainty that Bethany is similarly in the dark is if she knows, based on the first digit, that Bethany also has multiple choices for the second digit. This can only occur when the first digit is 9 or 8. (If the first digit were 6, the second digit could be 9, which is unique. Similarly, if the first digit were 7, the second digit could be 8 which is also unique.)
When Alison reveals what she knows, Bethany figures out that the first digit is 8 or 9. If she was unclear at first, it's because the second digit is one that has multiple possibilities  4, 5, 6, or 7. If she is now certain, it's because the second digit is unique among the remaining choices. This means her uncle's age is either 87, 95, or 96.
When Bethany reveals that she knows the age, Alison figures that the only possibilities are 87, 95, and 96. If this is enough information for her to also guess the age, it's because the first digit is unique among the remaining choices. This means Francis is 87 years old.
Puzzle #5
The ages of the children are 8, 6, and 3. To solve this riddle let's examine what each clue tells us about the ages. The first clue tells us the product of the ages is 144. The second clue tells us that among sets of three integers whose product is 144, the sum is not unique. The third clue tells us that there is an oldest child (the two older kids are not twins) and that the age gap between the oldest and youngest is no more than 6. The fourth clue tells us there is a youngest child (the two youngest children are not twins). Together, the third and fourth clues tell us there are no twins at all.
We can examine all the possibilities systematically with a table for all the sets of three integers whose product is 144.
Ages
 Sum
 Gap
 Twins


1, 1, 144
 146
 143
 yes

1, 2, 72
 75
 71
 no

1, 3, 48
 52
 47
 no

1, 4, 36
 41
 35
 no

1, 6, 24
 31
 23
 no

1, 8, 18
 27
 17
 no

1, 9, 16
 26
 15
 no

1, 12, 12
 25
 11
 yes

2, 2, 36
 40
 34
 yes

2, 3, 24
 29
 22
 no

2, 4, 18
 24
 16
 no

2, 6, 12
 20
 10
 no

2, 8, 9
 19
 7
 no

3, 3, 16
 22
 13
 yes

3, 4, 12
 19
 9
 no

3, 6, 8
 17
 5
 no

4, 4, 9
 17
 5
 yes

4, 6, 6
 16
 2
 yes

Of all the possible age sets, the only ones with nonunique sums are {2, 8, 9}, {3, 4, 12}, {3, 6, 8}, and {4, 4, 9} since the first two have a sum of 19 and the second two have a sum of 17.
Of these, the only ones with the correct age gap are {3, 6, 8} and {4, 4, 9}.
Of these two, the only one that doesn't contain twins is {3, 6, 8}.
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