Daily Test ~ {ERG}

[LoneRanger]

05 July 2008:

THE SOLUTION:

Here is one solution: (23 + 2) x 5 = 125. Are there other solutions?

Recent studies suggest that 8 possible solutions exist if we require that A, B, C, and D
are unique. These solutions are {A,B,C,D}
{8, 3, 2, 1}, {7, 9, 2, 1}, {2, 9, 4, 1}, {2, 3, 5,1}, {6, 0, 4, 2}, {6, 7, 5, 3}, {9, 0, 5, 4}, and {8, 9, 5, 4}.



The grid specifies 6 mathematical formulas, 3 horizontally and 3 vertically. (Recall that multiplication and division are performed before addition and subtraction.) Fill in the missing integer numbers.

[LoneRanger]

06 July 2008:

THE SOLUTION:

A wizard in a long white robe approaches you with a single card that has a 9 on one side and a 4 on the other. When the wizard tosses the card, it can land on either side.

The wizard can toss the card as often as you like, and each time he tosses the card, you add the number to the running sum. For example, he might toss a 9 and 4 and 4, which gives us a score of 17.

What is the highest whole number score that is impossible to obtain while playing this strange game?

[LoneRanger]

07 July 2008:

THE SOLUTION:

Number 23. In puzzles of this kind, if the two numbers have a common
divisor, then no numbers that are not a multiple of that divisor can be obtained. Otherwise, if one integer number is x and another is y, then the largest impossible score is xy - x - y.

So, in our case, we have 9 × 4 - 9 - 4 = 23.



Insert + or - between the numbers to find the total.

8 7 6 5 4 3 2 1 = 88

(If you don’t place a symbol between the numbers, they merge to form a multi digit number. For example, here is one possibility that, alas, turns out to be incorrect:

(8,765 - 4,321 = 88.)

[LoneRanger]

08 July 2008:

THE SOLUTION:

This was the first solution I came up with: 87 + 65 - 43 - 21 = 88. In fact, there are at least 9 solutions, including 87 - 6 + 5 - 4 + 3 + 2 + 1,
87 + 6 - 5 + 4 - 3 - 2 + 1,
8 + 7 + 65 + 4 + 3 + 2 - 1,
87 + 6 - 5 - 4 + 3 + 2 - 1,
87 - 6 + 5 + 4 - 3 + 2 - 1,
87 + 6 + 5 - 4 - 3 - 2 - 1,
8 - 7 + 65 + 4 - 3 + 21,
8 - 7 + 65 + 43 - 21, and
87 + 65 - 43 - 21.
At least 45 different solutions exist, if we were to allow the additional use of X and /. If we wish to study a related problem, 8 7 6 5 4 3 2 1 = 1, we find only one solution using + and - symbols: 8 - 76 + 5 + 43 + 21 = 1.



The missing numbers in the following grid are one-digit integers. The sums for each row and each column, and one diagonal, are listed outside the 4-by-4 array.

How quickly can you find the missing numbers?

[LoneRanger]

09 July 2008:

THE SOLUTION:

Here is one solution. Are there others?

My New Guinea sea turtle is four times older than I am. As it ages, its shell will turn a bright green. As I ponder its beautiful color, I suddenly realize that in 20 years, the turtle will be only twice as old as I will be then. How old is my turtle, and how old am I?

[LoneRanger]

10 July 2008:

THE SOLUTION:

I am 10 years old, and my turtle is 40 years old. Here’s one way to solve this. Let x = my age. Let 4x = the turtle’s age. Then,
4x + 20 = 2(x + 20)
4x + 20 = 2x + 40
2x = 20
x = 10



What are A and B in this mathematical expression? (BA is a two-digit number, and 176B is a four-digit number.)

(BA)^A = 176B

[LoneRanger]

11 July 2008:

THE SOLUTION:

The answer is (42)^2 = 1,764.


Suzy Samson is a world-champion weight lifter with arms strengthened by prosthetic ulnas shaped like helices.

Today, she is on a TV show, demonstrating her mental and physical strength. She gazes at some barbells, and a question forms in her mind. If she weighs 120 pounds plus a fourth of her own weight, how much does she weigh?

[LoneRanger]

12 July 2008:

THE SOLUTION:

She weighs 160 pounds. The equation is x = 120 + x/4.



Teja has a number of ladybugs in a jar. The number of ladybugs plus 10 grasshoppers is 2 less than 5 times the number of ladybugs. In addition, Teja has ten times the number of butterflies as she has grasshoppers.
If you wish, denote the number of ladybugs by L and the number of grasshoppers by G. How many ladybugs
does Teja have?

[LoneRanger]

13 July 2008:

THE SOLUTION:

The relevant equation is L + 10 = 5L - 2. The answer is 3. You don’t have to denote the number of grasshoppers by G.



Martian females. In a Martian crater, three-sevenths of the females are married to one-half of the males. What fraction of the crater’s Martians are married? (Assume that no Martian is married to more than one Martian.)

What is the least number of Martians who could live in the crater?

[LoneRanger]

14 July 2008:

THE SOLUTION:

In the crater, 3 females and 3 males are married. Let’s refer to the females as women and males as men, even if they don’t quite look like us. The number of married men and married women must be equal. Thus, we have 3/7 W = 1⁄2M. Multiply both sides of the equation by 14 to yield 6W = 7M, and thus the least number is W = 7, M = 6. This gives 13 Martians total, of which 6/13 of the crater’s inhabitants are married
(i.e., 3 wives of the 7 women and 3 husbands of the 6 men). There are exactly 3 married couples. For this problem, we assume that marriages are between male and female Martians. If I do not ask for the least number, there are an infinite number of solutions to 6W = 7M.



You are transported to a nearby dimension where half of 6 is 4, not 3, as you expected. What would a third of 12 be?

[LoneRanger]

15 July 2008:

THE SOLUTION:

A third of 12 would be 51⁄3. Assume that there is a strange factor that causes 6⁄2 to be equal to 4 instead of 3. This factor is 1.3333 . . . or 1 and 1/3. Therefore, (12/3) x 1.3333 . . . is 51⁄3. Of course, this is one of
those frustrating puzzles when more than one answer is possible. For example, we might consider the idea that 1 is added to each division operation so that 12⁄3 = 5. Platonists may disapprove of this puzzle and say that the eternal idea “6” is 6 in any dimension whatsoever.

If my wording bothered you, you can recast the question as the following when presenting it to friends. My calculator has been acting up lately. For instance, every time I divide 6 by 2, it gives the answer 4. What do you think it would give if I divided 12 by 3? That stops all of the philosophical arguments about what I might mean by “is” in “half of 6 is 4” or indeed “half,” because there is no reason that in this nearby dimension, “half ” should mean what you get if you divide by 2.



Monica walks along the New Jersey Turnpike, carrying ten 3-cup bottles of vinegar that are one-quarter full. Her friend William is carrying five 4- cup bottles of red wine that are one-quarter empty. How much more liquid does William carry than Monica?

[LoneRanger]

16 July 2008:

THE SOLUTION:

Monica carries 10 × 3 × (1⁄4) = 7.5 cups. William carries 5 × 4 × (3⁄4) = 15 cups. Thus, William carries 15 - 7.5 = 7.5 cups more than Monica does.


There are several humans and rabbits in a dirty cage (with no other types of animal). Perhaps they are trapped there for some kind of unholy experimentation.

All that we know is this: there are 70 heads and 200 feet inside the cage. Do you have the gut feeling that
there are more rabbits than humans? Exactly how many humans are there, and how many rabbits?

[LoneRanger]

17 July 2008:

THE SOLUTION:

There are 40 humans and 30 rabbits. Let h be the number of humans, and let r be the number of rabbits.
Thus, we have two equations with two unknowns.

r + h = 70
4r + 2h = 200

We can multiply the first equation by 2 and subtract it from the second:

2r + 2h = 140
4r + 2h = 200

Thus, 2r = 60; r = 30; h = 40.

Are there other answers to this problem?



America and Russia are in an important race from Earth to Saturn. Both spaceships start at Moscow and end at Saturn. They start the race traveling at the same speed and neither of them speeds up or slows down. The result is not a tie. How is this possible?

[LoneRanger]

18 July 2008:

THE SOLUTION:

The Americans and the Russians traveled different routes between the planets, and one route was shorter than the other. (You didn’t need much Maths Or General Knowledge for this one!) LOL!



NOT SO MUCH AS A TEST QUESTION BUT SOMETHING TO INTEREST YOU IN THE WORLD OF MATHEMATICS TODAY

The Norse god Odin tells you to pick any two-digit number. Multiply by 3, and use Odin’s mighty sword to sever the number so that you retain the last two digits of this result, and multiply by 3 again.

Repeat the process. For example, 13 becomes 39, then 117, which we cleave to 17. Thus, starting with 13, we produce 13 -> 39 -> 17 -> 51 -> 53 -> 59, . . .

How many steps does the starting number 13 take to return back to 13? Do such sequences always return to their starting numbers? If so, how many steps are usually needed?

[LoneRanger]

19 July 2008:

THE SOLUTION:

Indeed, every starting two-digit number in the Odin sequence returns to itself eventually. Number 13
requires 20 steps to return; that is, the 21st number is the same as the first. All other two-digit starting numbers require 20 steps to return, except for multiples of 5, which behave differently and require fewer steps.



Monica is visiting her zookeeper friend Bill in the rain forests of Tanzania. Bill loves long-necked animals,
and his zoo is a strange one, for it consists of just two types of animal: giraffes and ostriches. Monica gazes across the wooded area. “How many animals do you have?”

Bill replies, “Among my animals, I have 22 heads and 80 legs in all. The number of ostriches is less than the number of giraffes. From this little information, can you tell me how many giraffes and ostriches I have?”

For the second problem, consider Bill’s other zoo in Kenya.

This zoo is filled only with long-necked birds. Monica is strolling with Bill through the Kenya zoo. “How many birds do you have altogether?” she asks.

“In my vast collection of birds, all but two of them are geese, all but two of them are swans, and all but two of them are ostriches. From my meager information, you should be able to find the answer.”

“Are you some kind of nut?”

“Not at all. Tell me the answer, and we’ll have a fine goose for dinner.”