Lesson Nice entertainment problems related to divisibility properties

Nice entertainment problems related to divisibility properties

Problem 1

Let    be Jason's age this year (= current age).


The last year his age was  = p, some prime number.


But   = (n+1)*(n-1).


Since p is a prime number, it implies that n-1 = 1.


Hence, n = 2, and Jason's current age is   =  = 4 years.      ANSWER

Problem 2

Let  x  and  y  be two positive integers John thinks about.


Then from the condition, you have this equation

    xy - x - y = 35.


Add  1  (one)  to both side. You will get

    xy - x - y + 1 = 36.


It is equivalent to

    (x-1)*(y-1) = 36


Thus the integer number 36 is presented as the product of integer positive factors  (x-1)  and (y-1).


The possible factors of the number 36 are

    1, 2, 3, 4, 6, 9, 12, 18, 36.


So, (x-1) may have these values  1, 2, 3, 4, 6, 9, 12, 18, 36

(and, correspondingly, (y-1) has the values  of the same list in reversed order).


So, all possible values for numbers x and y are these pairs

    (x,y) = (2,37), (3,19), (4,13), (7,7), (10,5), (13,4), (19,3) and (37,2).      ANSWER

Problem 3

In the given sequence, each term is divisible by  3 

(which is easy to prove, looking at the first term and using the rule of divisibility by 3).



In the given list, the number 5789 is the only number which IS NOT divisible by 3; 

so this number DEFINITELY does not belong to the given sequence.

Problem 4

First term, 457, has the sum of digits 4+5+7 = 16 = 1 (mod 3).

So, according to the divisibility rule by 3, the number 457 itself is 1 (mod 3).

    
    Therefore, the second term is 1+1 = 2 (mod 3).

    It means that the sum of the digits of the second term is 2 (mod 3).


It implies that the third term is 2+2 = 4 = 1 (mod 3)


It implies, in turn, that the fourth term is 2 (mod 3).


Thus we see that the given sequence of numbers is CYCLIC 1, 2, 1, 2 by modulo 3,

and it is true NOT ONLY for the first four terms, but IT IS TRUE for all other consecutive terms.


Now look at the given optional numbers.

Using the rule of divisibility by 3, you can easily check that the ONLY NUMBER of them 
which is DIVISIBLE by 3 is the number 4569.


So, we can conclude that this number, 4569, DOES NOT BELONG to the given sequence.

Problem 5

Let x be the number of baskets collected by some girl, and y be the number of basket collected by her brother.


Then for each such a pair, we have this equation

    x^2 - y^2 = 45.


Factor left side to get

    (x-y)*(x+y) = 45.


45 has these decompositions in the product of factors  1*45,  3*15  and  9*5.


These decompositions produce these systems of equations


a)  x - y =  1
    x + y = 45

        with the solution  x = 23, y = 22;


b)  x - y =  3
    x + y = 15

        with the solution  x = 9, y = 6;


c)  x - y =  5
    x + y =  9

        with the solution  x = 7, y = 2.



These are all possible solutions to the numbers of baskets collected by the children.

There is NO other solutions.



Hence, the total pay was   = 1183 dollars.    ANSWER

Problem 6

Let the number of blue marbles be x.

Then the number of green marbles is 5x.


Since there is a 35% chance of selecting a red marble first,
it means that the number of red marbles is 0.35 of the total number of marbles.


Since there are an equal number of red marbles and white marbles,
it means that the number of white marbles is 0.35 of the total number of marbles, too.


Thus we have red + white = 0.7 of total number of marbles;
hence, green + blue marbles = the rest 0.3 of the total number of marbles = 5x + x = 6x.


Thus 0.3 of the total marbles is the number multiple of 6;
hence, the minimum total number of marbles is   = 20 or multiple to it.


    +----------------------------------------------------------------+
    |    Now we check if this found total number of 20 marbles       |
    |                    does really work.                           |
    +----------------------------------------------------------------+


If the total number of marbles is 20, then the number of red + white is 0.7*20 = 14;
hence, the number of red marbles = number of white marbles = 7.


Then the rest 20-14 = 6 marbles are blue + green, 
and from it, we easily conclude that there is 1 blue marble and 5 green marbles.


Thus this reasoning gives the ANSWER : the fewest possible number of green marbles in the bag is 5.

Problem 7

With 3-cent and 7-cent stamps, the  ANSWER  is  11 cents.



    Indeed, it is easy to check that you can not 
    combine 11 cents using 3c and 7c stamps, only.


    From the other side, it is easy to prove that 
    you can combine any number of cents greater than 11, 
             using 3c and 7c stamps.



Indeed, 12 = 4*3;  13 = 7 + 2*3;  14 = 2*7.


Let N be the number of cents,  N >= 12.



    (a)  If N is a multiple of 3, you simply combine N as 
         the integer number    of 3c stamps.



    (b)  If N >= 12 gives the remainder 1 when divided by 3, then  (N-1)  
         is a multiple of 3, and you can combine N-1 cents using the 3c stamps, only.

         In this combination, you will have at least four 3c stamps.


         Now from this combination of 3c stamps, take off two 3c stamps and add 
         one 7c stamp - it will be your desired combination for N cents.



    (c)  If N >= 12 gives the remainder 2 when divided by 3, then  (N-2)  
         is a multiple of 3, and you can combine N-2 cents using 3c stamps, only.

         In this combination, you will have at least four 3c stamps.


         Now from this combination of 3c stamps, take off four 3c stamps and add 
         two 7c stamps - it will be your desired combination for N cents.


Thus I proved for you that you can combine everything above or equal  12 cents.

So, 11 cents is your  ANSWER  to this problem.

Problem 8

If  b+1  is a prime number, then obviously

     

is not an integer number.


If  b+1  is NOT a prime number, then  b+1 = p*q,  where p and q are integer numbers lesser than (b+1).

Hence, p and q are among the numbers in the numerator, and they can be canceled,
producing integer ratio.


Thus, the given expression is integer if and only if (b+1) is not a prime.


So, to answer the question, we need to count the number of primes in the interval 1 < b < 60.


These primes are  2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,  
and their number is 17.



At this point, we obtain the  ANSWER: the number of integers b, 1 < b < 60, such that      

      is integer 

is equal to (59-1)-17 = 58-17 = 41.


In this reasoning, there is only one weak point: what if in the decomposition b+1 = p*q
the integers p and q are equal ?


Then one of them cancels, and for the other equal factor in the denominator, there is still another 
number in the numerator, which is multiple to it, so in this case we AGAIN obtain an integer number. 


At this point, the solution is complete.

Problem 9

    +-------------------------------------------------------------------------------+
    |  To start solving, notice that taking a tablet every 10 days means            |
    |  that the process of taking tablets is periodical with the period of 10 days. |
    +-------------------------------------------------------------------------------+


Imagine that the days are coded in this natural way, using numbers from 1 to 7 
instead of the days standard naming

    Mo Tu We Th Fr Sa Su     (*)
     1  2  3  4  5  6  7


Next, imagine that you have long sequence of these numbers, repeating in this natural order

     1  2  3  4  5  6  7  1  2  3  4  5  6  7  1  2 . . . . 


Then your day in this sequence is this coding number  (10*24+1 mod 7)
(i.e. 24 intervals of 10 days plus the next day, and the sum is taken modulo 7).


Multiply 10*24 and add 1.  You will get 241.
Find the remainder after division 241 by 7. 
The division is 241 = 34*7 + 3.
Thus the remainder is 3.
In other words, (10*24+1 mod 7) = 3.
The remainder "3" is the code for the 3-rd day in the sequence (*).

Hence, Mike will take his last tablet on Wednesday.