Eight islands each have one or more air services. An air consists of Eights to a

Question

Eight islands each have one or more air services. An air consists of Eights to and from another island, and no two services link the same pair of islands. There are 17 air services in all between the islands. Show that it must be possible to use these air services to fly between any pair of islands. Note that 654/545, so that one can cancel the 54 in the numerator and denominator of 654/545 without changing the value of the the same number n of copies of 54 proceeding the final digit 5 of the denominator. Prove that this fraction is equal to 6/5 for every positive integer n, Cheryl chose three distance integers between 1 and 5, we would like to identify these numbers, We are allowed to make queries in the following form: we give Cheryl 3 distinct numbers a, b, c, and she tells us how many of these numbers are among the chosen ones. For example, if we guess 2, 3 and 4 the chosen number are 1, 2, and 5, we get the answer 1. Find the smallest number of queries needed to guarantee that we can always identify the three chosen numbers. Note that we do no need to actually name the three correct numbers in a quart, it is sufficient to identify them without a doubt from the replies to our queries. Some people are happy if their birthday occurs in a year for which the last two digits are the reverse of the digits on n, For example, someone born in 1945 would be happy on their 38th birthday, which occurs in 1983. Devise a formula that tells, for any year n (i greater than or equal to 0), how often a person who its born in year n will experience a happy birthday, Assume the person lives to have a 33th birthday, but not a 100th; and also count; the true date of birth as the 00th birthday. Thus is in particular, in what year should one be born to maximize happy birthday? Let n be an integer whose decimal digits from left to right are non-decreasing (e. g-113445889) and having digits as its two rightmost digits. Let 5 be the sum of the digits of 9n. Then 5 is divisible by 9, so 5 -9m. How large can m be? An irregularly shaped fault yard is surrounded by a tall fence built from connected straight pieces. We call location sightful if it is inside the yard and a surveyor (shorter than the fence) can see every part f the inside of the fence from that location.(The surveyor is allowed to turn around on the spot) If is a sightful location and B is a sightful location, show that every point on the straight path from A to B is sightful.(In the figure the point P is sightful, While the point Q is not).

Explanation / Answer

(According to Chegg policy, only four subquestions will be answered. Please post the remaining in another question)

5. Let us assume that it is not possible to fly between the islands using only 17 air services.

Since there is atleast one air service on every island, there must be another connected island with it.

This means there is one or more group of connected islands. These groups are disconnected from each other.

Since there are eight islands and each group has atleast two islands, there can be a maximum of three groups.

First let us consider, two groups.

Let the groups consist of a and 8-a islands.

The maximum number of air services can exist when every island has a service to all other islands in the same group.

If n is the number of islands in the group,

The first island can be connected to n-1 islands.

The second island can connect to the other n-2 islands and so on.

So the maximum number of air services = n-1 + n-2 + ...... 2 + 1

= n(n-1)/2

Thus the maximum number of air services in the two groups

= a(a-1)/2 + (8-a)(8-a-1)/2

= (a2-a +a2 -15a + 56)/2

= (2a2 -16a + 56)/2

= a2 -8a + 28

Since there are two groups, one group can have maximum of 6 islands.

So a <= 6

=> a2 <= 36

=> a2 - 8a < 36 - 8*6

=> a2 - 8a < -12

So the maximum number of air services < -12 + 28 or 16

But there are 17 airways. So we arrive at a contradiction.

Now let us consider three groups. Two will have two islands each and one will have three islands.

The two groups islands will have one air service between them each and the three islands will have 2+1 = 3 air services.

So the total is 1+1+3 = 5 air services but there are a total of 17 air services and we arrive at another contradiction.

So it is possible to fly between any two islands.

9. Let us consider the two rightmost digits first.

8 and 9 are the largest digits.

8+9 = 17

Consider 789 to be the last three digits.

7+8+9 = 24

If we consider 7+7+7+7+8+9 = 45, it is divisible by 9. So 777789 can be the biggest rightmost digits

But then we can stuff any 9*7s at the beginning

E.g 777777777777789

Since the number can be of any length, so can the sum of digits.

Hence m can be infinitely longer.

7. Since there are 5 numbers and three are to be guessed,

There are three possibilities:

(i) Cheryll replies 3. In this case all numbers guessed are correct. 1 query is enough.

(ii) Cheryll replies 1. In this case, only one of the numbers we guessed is right. So the two numbers we did not guess are the actual ones. To guess the remaining number, we say the two numbers we did not guess in the beginning followed by one of the three we initially guessed. So a maximum of three queries are needed.

(iii) Cheryll replies 2. In this case, two numbers guessed are right. We then say the two numbers we did not guess along with one of the numbers we guessed.

If Cheryll says 1, then the number we guessed and said was wrong and we discard it. We then say the other two numbers with one of the remaining two numbers one at a time to get the answer in max 3 queries.

If Cheryll says 2, then the number we guessed is right, we then repeat with second number, and repeat the step. If Cheryll again says 2, the second number is chosen, otherwise third is chosen. We then find the third number by taking one of the two numbers not guessed initially along with the two numbers we are sure about. So this takes a maximum of 4 queries.

So the maximum is 4.

8. Quite obviously only the last two digits of the year matter.

If the last two digits are a and b,

then the birthday is 10b+a

Also the birth year is (10a+b) - (10b+a)

= 9(a-b)

where a >=b

a>=9 and b>=9

So given a year we need to calculate 9-a.

So for n, we first calculate p = n mod 100 to get the last two digits as a number.

Then we calculate a-b = p/9

If 9 divides p we go forward.

x = p mod 10 to get first digit.

Then our answer will be 10-x

So the formula is 10 - (n mod 100)/9 mod 10 happy birthdays.

E.g if the birth year is 1935.

the number of happy birthdays = 10 - (1935 mod 100)/9 mod 10

= 10 - 35/9 mod 10

since 35 is not divisible by 9, there are no happy birthdays.

E.g : 1945

=> 10 - (1945 mod 100)/9 mod 10

=> 10 - (45/9 mod 10)

=> 10-5 = 5

The happy birthday's are 1950, 1961, 1972, 1983 and 1994.

For maximum number of birthdays, n should be a minimum and a multiple of 9 which is 00.

E.g if a person is born in 1900, the happy birthdays are 1900, 1911, 1922, 1933, 1944, 1955, 1966, 1977, 1988 and 1999.

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