The sun god had a herd of cattle consisting of bulls and cows, one part of which was white, a second black, a third spotted, and a fourth brown. Among the bulls, the number of white ones was one half plus one third the number of the black greater than the brown; the number of the black, one quarter plus one fifth the number of the spotted greater than the brown; the number of the spotted, one sixth and one seventh the number of the white greater than the brown. Among the cows, the number of white ones was one third plus one quarter of the total black cattle; the number of the black, one quarter plus one fifth the total of the spotted cattle; the number of spotted, one fifth plus one sixth the total of the brown cattle; the number of the brown, one sixth plus one seventh the total of the white cattle. What was the composition of the herd?
Answer:
Step #1: The first task is to convert all of that into equations. Using W, X, Y, Z, w, x, y, z for white bulls, black bulls, spotted bulls, brown bulls, white cows, black cows, spotted cows, and brown cows, respectively, we get these equations:
W = (5/6)X + Z
X = (9/20)Y + Z
Y = (13/42)W + Z
w = (7/12)(X+x)
x = (9/20)(Y+y)
y = (11/30)(Z+z)
z = (13/42)(W+w)
There are seven equations with eight unknowns. So there are many possible solutions. And we want the smallest positive integer solution (positive integer because we don't want negative numbers or fractions of cattle). I will explain my reasoning below. Here is the solution:
W = 10,366,482
X = 7,460,514
Y = 7,358,060
Z = 4,149,387
w = 7,206,360
x = 4,893,246
y = 3,515,820
z = 5,439,213
Sometimes further restrictions are set on the numbers, in order to make the herd much larger. Such numbers were far beyond the limits of the Greek or Roman or Egyptian number systems. And that may have been the entire idea behind the problem. We can precisely describe numbers beyond the limits of our number systems.
Reasoning:
Combining the first three equations, we get W = (5/6)((9/20)Y+Z)+Z = (5/6)((9/20)((13/42)W+Z)+Z)+Z = (13/112)W + (3/8)Z + (5/6)Z + Z = (13/112)W + (53/24)Z, so (99/14)W = (53/3)Z, and 297W = 742Z. This is reduced to its smallest values, so W is divisible by 742 and Z is divisible by 297. By equation #3, W is divisible by 42. 2226 (3x742) is the smallest number divisible by both 742 and 42, so W is divisible by 2226. Let's try some W's: W = 2226 4452 6677 8903 ...
X = 1602 3204 4806 6408 ...
Y = 1580 3160 4740 6320 ...
Z = 891 1782 2673 3564 ...
The second column is twice the first; the third is three times the first, etc. For each of these columns, we can plug in the values of W, X, Y, and Z into equations #4 through #7. So we have four equations with four unknowns, and we should be able to solve for w, x, y, and z. Choosing the first column, we get these four equations:
w = (7/12)(1602+x)
x = (9/20)(1580+y)
y = (11/30)(891+z)
z = (13/42)(2226+w)
Solving these four equations for z, we get:
w = 7206360 / 4657
x = 4893246 / 4657
y = 3515820 / 4657
z = 5439213 / 4657
after reducing z to its lowest terms. So for these four number to be positive integers, W, X, Y, and Z must be 4657 times the first column above. W = 2226 x 4657 = 10366482. And the rest of the number follow. Problems which call for integer solutions are called Diophantine problems.
52. FIGHT FIRE.
A man was on a holiday last year on the island of Syracuse. Syracuse is a really peculiar shape for an island, it's rectangular, 20 miles long and only 15 feet wide, and totally covered in trees! It has two tinyvillages on it 'Start' and 'End', unfortunately, a fire has begun at Start and is travelling toward End at around 1 mile per hour. There is also a wind which is blowing towards End at around 2 miles perhour. Now, Billy, who is at End, has no way of getting off the island, he cannot pass through the fire, he cannot put the fire out and there is no one to help him. How can he save himself?
Answer :
Just light a new fire at about 1 mile from 'end'. This fire will destroy all the trees towards 'end'. because of wind. Now the old fire which is coming towards 'end' will stop at 1 mile from 'end'.
53. FIXING LIMIT.
Eventhough the odds are always in favour of the gambling house, why does the establishment insist on a house limit on stakes?
Answer:
Every casino in the world would go bankrupt without a house limit on stakes. Without it, gamblers would keep on doubling their stakes until they won. No matter how bad a losing streak they were on, they would eventually win.
54. THE SURVIVOR
In ancient Greece, it was decided that there were too many prisoners and many should be executed. One prisoner was given a sword and all 1000 prisoners were instructed to stand in a circle. The one with the sword was instructed to kill the man on his left and then pass the sword to the next man on the left, who would then do the same. The circle would continue to get smaller as this continued, and the last man left would be set free. Josephus, one of the prisoners, placed himself at the correct position in the lineup to be the one remaining man at the end of this elimination. At whatposition did he place himself on the circle?
Question 2: If the last two will be set free, where should Josephus direct his friend to stand so that his friend too can escape from the execution?
Answer for above question will be appearing later...
55. CUTTING EQUALLY
Given a rectangular cake within which a rectangular piece removed (any size or Orientation). How would you cut the remainder of the cake into two equal halves with one straight cut of a knife?
Answer:
If you cut a rectangular thing along the center (horizontally, vertically or at any angle), you will get two halves. Using this idea, find the centers of both - the original cake and the removed piece. Now, cut the reminder along the line connecting these two centers. This is true because this line cut both - the original cake and the removed piece - in half, thus the remainder into two halves too.
56. WHAT IS THE REMAINDER?
What is the remainder left after dividing 1! + 2! + 3! + � + 100! By7?
Answer:
From 7! onwards all the terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms. i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0. The only part to be consider is 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720= 873. The remainder left after dividing 873 by 7 is 5. Hence the remainder is 5.
57. BRING 100 BY ADDING.
Using the numbers 0 to 9 once each, what combination will add up to exactly 100? Note that you can use ONLY ADDITION and no other mathematical operations.
Answer:
It is impossible to get 100 using digits 0-9 once each and only through addition. All digits from 0 to 9 add up to 45. Now take any two digits (and make a two-digit number) and add it to the remaining digits. The result will always increase by a multiple of 9. e.g. Let's take 1 and 2 i.e. 12 and add all remaining digits to it, the total is 54 i.e. 45+ 9. Thus, whatever you do the answer will always be 45 + 9X, where X isan integer. In other words, we are trying to solve 45 + 9X = 100 forinteger value of X. Thus, it is impossible to get 100 using digits 0-9 once each and only through addition.
58. WHAT IS IT?
A teacher writes an integer less than 50,000 on the blackboard. One student states that the number is a multiple of 2; a second student states that the number is a multiple of 3; and so on and so forth until the twelfth student states that it is a multiple of 13. The teacher remarks that all except two of the students were right and moreover those two uttered wrongly spoke one after the other. What was the number that the teacher wrote on the blackboard?
Answer:
The teacher wrote 25,740 on the blackboard. Let N be the number that teacher wrote on the blackboard. Since all the students, except two who spoke one after the other, were correct; it can be deduced that N must be divisible by 1, 2, 3, 4, 5,6, 10, 11, 12 and 13. This is because if N is not divisible by 2, then it is not divisible by 4 also. If N is not divisible by 3, then it is not divisible by 6 also. If N is not divisible by 5, then it is not divisible by 10 also. And so on. All these leaves 7, 8 and 9 as the only possible numbers which do not divide N. Hence, there are 2 cases.
Case I: N is not divisible by 8 and 9. In this case, the smallest number divisible by all other numbers (i.e. Least Common Multiple) is 60,060. But the number is greater than the one written on the blackboard.
Case II: N is not divisible by 7 and 8. In this case, the smallest number divisible by all other numbers (i.e. Least Common Multiple) is 25,740. Thus, the teacher wrote 25,740 on the blackboard.
59. WHO IS THE RACE WINNER?
On a barnyard, there was a chicken, a turkey and a crow. The chicken could fly at 5 mph but, as chickens are poor flyers, he gets tired very soon. After only 2 minutes of flying, he had to walk at 1 mph for next three minutes before he could fly again. The turkey could fly continuously at 4 mph, but he had a habit of swerving back and forth. So for every 3 feet he flies, he actually only travels 2 feet. The crow could fly straight, at 2 mph. Which animal would win a one mile race?
Answer:
Chicken would win the one-mile race. If you calculate the weighted average, you'll see that the chicken's average speed, after 5 minutes, is 2.6 mph. The turkey's average speed is 2.67 mph. The crow, of course, has an average speed of 2 mph. From these figures, you might guess that the turkey would win. But NO !!! The catch is the fact that the one-mile race does not take an even multiple of five minutes to complete. The chicken's average speed is only accurate after the full five minutes of flying/walking is completed. At any time before that, the chicken's mean speed will be higher. After 20 minutes have passed, the chicken will have travelled 0.87 miles, and the turkey will have travelled 0.89 miles. However, the chicken will then begin flying at 5 mph, and will overtake the turkey. The final results will be: 1. Chicken - 21.6 minutes (21 minutes 36 seconds) 2. Turkey - 22.5 minutes (22 minutes 30 seconds) 3. Crow - 30 minutes.
60. CHILDREN PROBABILITY.
I have two children. At least one of them is a boy. What is the probability that both my children are boys? My sister also has two children. The older child is a girl. What is the probability that both her children are girls?
Answer:
The probability that both my children are boys is 1/3. And that mysister's both children are girls is 1/2. There are four possibilities of having two children (with the oldest listed first):
1. (Girl, Girl) 2. (Girl, Boy) 3. (Boy, Girl) 4. (Boy, Boy) As I have at least one boy, there are three possibilities i.e. 2, 3 and 4. And only in one case both are boys. Hence, the required probability is 1/3. Similarly, since my sister's older child is girl, there are just twopossibilities i.e. 1 and 2. And again, only in one case both are girls. Hence, the reqiured probability is 1/2.