# Brain Teasers & Math Puzzles About Frogs

Humans have always had a complicated relationship with frogs and toads. People used to have an aversion to frogs and toads as much as to spiders and snakes, but nowadays frogs are ubiquitous in kitschy home decor. Frogs appear frequently in folklore and religion, from Grimm's fairy tales to the Bible and Native American traditions. And of course, there are many classic riddles about frogs and toads. Here are several examples of math puzzles and brain teasers with a froggy theme, along with their solutions.

## (1) The Toad That Swallowed a Ring

The queen loves toads and keeps 16 identical toads in a terrarium. One day, the toads got out and one of them swallowed her favorite ring. She knows she will have to kill the toad to get her ring back, but she doesn't know which toad ate it, and she doesn't want to kill more toads than necessary. All she knows is that every toad weighs the same, except for the culprit who now weighs a little more due to having the ring inside. She has a balance scale to compare relative weights. What is the most efficient way to find the toad who swallowed the ring using the balance scale?

Solution: Divide the toads into piles of five, five, and six. Weigh the two piles of five against each other. There are two possibilities

(1) The five-piles weigh the same. In that case, the ring is in the pile of six toads. Divide the six-pile into smaller piles of two, two, and two. Pick two of the two-piles and weigh them against each other. This results in two subcases:

• If the piles weigh the same, the toad is in the two-pile you left aside, in which case you can put one toad from this pile on each side of the scale to find the heavy toad.
• If one pile is heavier than the other, then the heavy side contains one innocent and one guilty toad. Weigh the two toads against each other to find the guilty toad.

(2) If one of the five-piles is heavier than the other, split this group into piles of two, two, and one. Weigh the two two-piles against each other. This results in two subcases:

• If the piles weigh the same, the toad you left aside is the guilty toad.
• If one pile is heavier than the other, it contains one guilty and one innocent toad. Compare their weights using the scale and you'll find the guilty toad.

All in all, the queen can find the toad who swallowed her ring using the scale at most three times.

Sometimes people see the number 16 in this puzzle and immediately think of a halving strategy, since 16 is a power of 2. That is, first you weigh two piles of eight, then two piles of four, then two piles of two, and finally two piles of one. But this strategy uses the scale four times, so it is not as efficient as the solution above.

## (2) Frog Pairing

Lucie is a batrachologist, a scientist who studies frogs. Lucie ask her assistant Lisa to go to the pond to collect three male and three female frogs of a certain species and to put them in three tanks in a certain way: two males in the first tank, two females in the second tank, and one of each sex in the third tank. Lisa collects three frogs of each sex, but forgets Lucie's special instructions; she just puts two frogs in each tank randomly without regard to sex. What is the probability that Lisa did it right anyway by sheer luck?

Solution: For this problem we first need to count the total number of ways to divide six frogs into three pairs. If the female frogs are called A, B, and C and the male frogs are called X, Y, and Z, then there are 15 total ways to split them into three groups of two without following Lucie's special instructions.

(AB, CX, YZ) (AB, CY, XZ) (AB, CZ, XY)
(AC, BX, YZ) (AC, BY, XZ) (AC, BZ, XY)
(AX, BC, YZ) (AX, BY, CZ) (AX, BZ, CY)
(AY, BC, XZ) (AY, BX, CZ) (AY, BZ, CX)
(AZ, BC, XY) (AZ, BX, CY) (AZ, BY, CX)

Of these 15 groupings, there are only nine that have two males paired up, two females paired up, and a pair consisting of a male and female. These are

(AB, CX, YZ) (AB, CY, XZ) (AB, CZ, XY)
(AC, BX, YZ) (AC, BY, XZ) (AC, BZ, XY)
(AX, BC, YZ) (AY, BC, XZ) (AZ, BC, XY)

Therefore, the probability that Lisa does it right by accident is 9/15, or 60%. For some people, it is counterintuitive that Lisa has a greater than 50% chance of doing it right without trying.

## (3) Find the Lying Frog

Theo has four enchanted frogs, three who always tell the truth and one who always lies. He often gets them mixed up, although the frogs themselves know who among them are the truth-tellers and who is the liar. One day the frogs are sitting in a row on his couch. Asking only yes or no questions, how can Theo figure out which one the liar is with only two questions?

Solution: Call the frogs 1, 2, 3, and 4 from left to right. Theo can figure out who the lying frog is with only two questions. First he asks frog number 1, "Is frog number 2 the liar?" This results in two cases:

1. If frog number 1 answers "yes" it means that either frog 1 or 2 is the liar. That means frogs 3 and 4 must be truth tellers. For the second question, Theo asks frog number 4, "Is frog number 1 the liar?" The answer to this question will pinpoint who the lying frog is.
2. If frog number 1 answers "no" it means that frogs 1 and 2 are truth-tellers and either frog 3 or 4 is the liar. For the second question, Theo asks frog number 1 "Is frog number 4 the liar?" The answer to this question identifies the lying frog.

There are many other two-question solutions, this is but one example. Two is the fewest number of questions needed to find the liar.

## (4) Finding the Frog Prince

Lucie the batrachologist has given up trying to find a husband the normal way, so she heads down to the witch's pond where 1/3 of the frogs are actually princes who have been turned into frogs by the witch. They can be released from their curse with a kiss. The pond has 3N frogs, 2N of them are normal and N of them are princes. The witch says Lucie can kiss up to N different frogs to find a prince. If all N of the frogs she kisses are normal frogs, the witch will turn Lucie into a raccoon. What should the value of N be so that Lucie has at least a 95% chance of getting a prince?

Solution: This puzzle is equivalent to finding the value of N so that her chances of being turned into a raccoon are less than 5% or 0.05. If there are 2N normal frogs and N princes, the chances of getting all normal frogs in N tries is the product of the following N terms

[(2N)/(3N)] * [(2N-1)/(3N-1)] * [(2N-2)/(3N-2)] * ... * [(N+1)/(2N+1)]

For example, if N = 1, then there are three frogs in the pond -- one prince and two normal frogs. Her probability of getting a normal frog on one try is 2/3 ≈ 66.7%. Her chance of getting a prince is 33.3%.

If N = 2, then there are six frogs in the pond -- two princes and four normal. Her probability of getting two normal frogs in two tries is (4/6)(3/5) = 40%. Her chance of getting a prince is 60%.

If N = 3, then there are nine frogs in the pond -- three princes and six normal. Her probability of getting three normal frogs in three tries is (6/9)(5/8)(4/7) ≈ 23.8%. Her chance of getting a prince is 77.2%.

As you can see, as N increases, her chances of getting all normal frogs decreases and her chances of getting a prince increases. Just with trial and error you can discover that N = 6 gives her more than 95% chance of getting a prince. With N = 6 the math works out to

1 - (12/18)(11/17)(10/16)(9/15)(8/14)(7/13) ≈ 0.950226

## (5) Leap Frog

Nine frogs are lined up to play a game of leap frog. Instead of playing the usual way where the last frog leaps to the front of the line, they devise a new way to play. In this new version one round consists of the following sequence of leaps:

• The last frog leaps over the frog immediately ahead of her.
• The new last frog now leaps over the two frogs immediately ahead of her.
• The new last frog now leaps over the four frogs immediately ahead of her.
• The new last frog now leas over all eight frogs ahead of her.

Suppose Gilly starts as the last frog in line. After how many rounds will she be at the front of the line?

Solution: Gilly can never get to the front of the line in this version of the game. Imagine the nine frogs lined up from left to right with the numbers 1 through 9 painted on their backs, with 1 being the first frog and 9 being Gilly, the last frog:

1 -- 2 -- 3 -- 4 -- 5 -- 6 -- 7 -- 8 -- 9

After one round of the game the frogs have a new arrangement

7 -- 1 -- 2 -- 3 -- 4 -- 9 -- 5 -- 6 -- 8

Gilly has moved up to sixth place. After a second round of the game, the order of the nine frogs is now

5 -- 7 -- 1 -- 2 -- 3 -- 8 -- 4 -- 9 -- 6

with Gilly now second-to-last. Doing a third round of the game produces the new order

4 -- 5 -- 7 -- 1 -- 2 -- 6 -- 3 -- 8 -- 9

And now we see Gilly is back at the end of the line again. We don't need to play any more rounds to see what will happen to Gilly. She will never reach the front of the line with this pattern of leaps because her position will only cycle among three spots in the back half of the line. In fact, the frogs in places 6 and 8 will also never get to the front of the line either. Gilly and the frogs numbered 6 and 8 will rotate among the last, second-to-last, and sixth positions.

Ten toads are hopping in a rectangular formation. They marvel that their group has the ability to hop in a rectangular formation with exactly three more rows than columns, and the ability to hop in a triangular formation with one toad in front, where each row has one more toad than the previous row. See figure below.

Suddenly the ten toads come across a larger group of toads that also has the same ability, i.e., they can hop in a rectangular formation with three more rows than columns, and also in a triangular formation with one toad at the front where each row has one more toad than the row ahead of it. How many toads are in this new group?

Solution: There are actually infinitely many solutions to this problem. The smallest solution is that the new group has 28 toads. Twenty-eight toads can be grouped into a rectangle with four columns and seven rows. They can also be arranged into a triangle with one toad in front, two in the second row, three in the third row, ... and seven in the last row.

The next largest solution is 378 toads, which can form a 18-by-21 rectangle and a triangle with 27 rows. After that is 990 toads, which can form a 30-by-33 rectangle and a triangular formation with 44 rows. After that, the next is 12880 toads, which can form a 112-by-115 rectangle and a triangular formation with 160 rows.

Illustrations from Pixabay public domain stock.

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