Lesson Nice entertainment problems related to divisibility properties

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<H2>Nice entertainment problems related to divisibility properties</H2>

<H3>Problem 1</H3>Last year &nbsp;Jason's age was a prime number. &nbsp;This year it is a square number. &nbsp;How old is he this year?

<B>Solution</B>

<pre>
Let  {{{n^2}}}  be Jason's age this year (= current age).

The last year his age was {{{n^2-1}}} = p, some prime number.

But  {{{n^2-1}}} = (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  {{{n^2}}} = {{{2^2}}} = 4 years.      <U>ANSWER</U>
</pre>

<H3>Problem 2</H3>John thinks of two positive integers. &nbsp;He multiplies them together and then
subtracts each of the integers from the product, &nbsp;with a result of &nbsp;35. 
Find all possible pairs of numbers he could have chosen.

<B>Solution</B>

<pre>
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).      <U>ANSWER</U>
</pre>

<H3>Problem 3</H3>In the sequence &nbsp;456, &nbsp;471, &nbsp;483, &nbsp;498, ... each term is equal to the previous term 
added to the sum of its digits. &nbsp;Which number of the following list does not belong to the sequence ?
&nbsp;&nbsp;&nbsp;&nbsp;a) &nbsp;1869; &nbsp;&nbsp;b) &nbsp;1950; &nbsp;&nbsp;c) &nbsp;3477; &nbsp;&nbsp;d) &nbsp;4569; &nbsp;&nbsp;e) 5789.

<B>Solution</B>

<pre>
In the given sequence, each term is divisible by &nbsp;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.
</pre>

<H3>Problem 4</H3>In the sequence &nbsp;457, &nbsp;473, &nbsp;487, &nbsp;506, ... &nbsp;each term is equal to the previous term added 
to the sum of its digits. &nbsp;Which number of the following list does not belong to the sequence ?
&nbsp;&nbsp;&nbsp;&nbsp;a) &nbsp;1864; &nbsp;&nbsp;b) &nbsp;1949; &nbsp;&nbsp;c) &nbsp;3466; &nbsp;&nbsp;d) &nbsp;4569; &nbsp;&nbsp;e) 5767.

<B>Solution</B>

<pre>
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.
</pre>

<H3>Problem 5</H3>Three girls, &nbsp;Ann, &nbsp;Betty and &nbsp;Cynthia, &nbsp;each have a younger brother, &nbsp;Dylan, &nbsp;Ernie and &nbsp;Frank, &nbsp;respectively. 
All six children do some fruit picking for the local farmer. 
The farmer agrees to pay each child as many dollars per basket as the number of baskets of fruit collected by that child. 
Each of the girls earned &nbsp;$45 &nbsp;more than her brother, &nbsp;and all six children collected
a different number of baskets. &nbsp;How much did the farmer pay them all in total ?

<B>Solution</B>

<pre>
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  {{{23^2 + 22^2 + 9^2 + 6^2 + 7^2 + 2^2}}} = 1183 dollars.    <U>ANSWER</U>
</pre>

<H3>Problem 6</H3>A bag contains red, &nbsp;white, &nbsp;green, &nbsp;and blue marbles. 
There are an equal number of red marbles and white marbles, 
and five times as many green marbles as blue marbles. 
There is a &nbsp;35% &nbsp;chance of selecting a red marble first. 
What is the fewest possible number of green marbles in the bag?

<B>Solution</B>

<pre>
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  {{{6/0.3}}} = 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 <U>ANSWER</U> : the fewest possible number of green marbles in the bag is 5.
</pre>

<H3>Problem 7</H3>Assuming you have an unlimited supply of &nbsp;3-cent and &nbsp;7-cent stamps,
what is the largest amount of postage you cannot make?

<B>Solution</B>

<pre>
With 3-cent and 7-cent stamps, the  <U>ANSWER</U>  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  {{{N/3)}}}  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  <U>ANSWER</U>  to this problem.
</pre>

<H3>Problem 8</H3>Let &nbsp;&nbsp;A = {{{(b*(b-1)*(b-2)*ellipsis*(1))/(b+1))}}}, &nbsp;&nbsp;where &nbsp;b &nbsp;is an integer and &nbsp;1 < b < 60. 
For how many values of &nbsp;b &nbsp;is &nbsp;A &nbsp;an integer?

<B>Solution</B>

<pre>
If  b+1  is a prime number, then obviously

{{{(b*(b-1)*(b-2)*ellipsis*(1))/(b+1))}}}

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  <U>ANSWER</U>: the number of integers b, 1 < b < 60, such that

{{{(b*(b-1)*(b-2)*ellipsis*(1))/(b+1))}}}  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.
</pre>

<H3>Problem 9</H3>Mike takes a tablet every &nbsp;10 &nbsp;days. &nbsp;He has &nbsp;25 &nbsp;tablets. &nbsp;If &nbsp;Mike took 
his first tablet on &nbsp;Monday, &nbsp;on what day will he take the last tablet?

<B>Solution</B>

<pre>
    +-------------------------------------------------------------------------------+
    |  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.
</pre>

This problem has an underwater stone &nbsp;(a trap) &nbsp;which should be treated accurately.

The stone is to determine &nbsp;CORRECTLY &nbsp;that the day of interest is &nbsp;24*10+1 &nbsp;(not &nbsp;25*10).

After that, &nbsp;applying modular arithmetic leads to the answer quickly and straightforward.

My other lessons in this site on miscellaneous problems on divisibility of integer numbers are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Light-flashes-on-a-Christmas-tree-and--a-Least-Common-Multiple.lesson>Light flashes on a Christmas tree and a Least Common Multiple</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/The-number-that-leaves-a-remainder-1-when-divided-by-2-by-3-by-4-by-5-and-so-on-until-9.lesson>The number that leaves a remainder 1 when divided by 2, by 3, by 4, by 5 and so on until 9</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/The-number-rem4-mod7-rem5-mod8-and-rem6-mod9.lesson>The number which gives remainder 4 when divided by 7, remainder 5 when divided by 8 and remainder 6 when divided by 9</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/Introductory-problems-on-divisibility-numbers.lesson>Introductory problems on divisibility of integer numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Finding-Greatest-Common-Divisor-of-integer-numbers.lesson>Finding Greatest Common Divisor of integer numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/divisibility/lessons/Relativity-prime-numbers-help-to-solve-the-problem.lesson>Relatively prime numbers help to solve the problem</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Solving-equations-in-integer-numbers.lesson>Solving equations in integer numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Quadratic-polynomial-with-odd-integer-coefs-can-not-have-a-rational-root.lesson>Quadratic polynomial with odd integer coefficients can not have a rational root</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/Proving-the-equation-has-no-integer-solutions.lesson>Proving an equation has no integer solutions</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Composite-number-of-the-form-%284n%2B3%29-must--have-a-prime-divisor-of-the-form-%284n%2B3%29.lesson>Composite number of the form (4n+3) must have a prime divisor of the form (4n+3)</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Problems-on-divisors-of-a-given-number.lesson>Problems on divisors of a given number</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/How-many-three-digit-numbers-are-multiples-of-both-5-and-7.lesson>How many three-digit numbers are multiples of both 5 and 7?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/How-many-3-digit-numbers-are-not-dvsbl-by-2-not-dvsbl-by-3-notdvsbl-by-either-2-or-3.lesson>How many 3-digit numbers are not divisible by 2; not divisible by 3; not divisible by either 2 or 3</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/How-many-integer-numbers-in-the-range-1-300-are-divisible-by.lesson>How many integer numbers in the range 1-300 are divisible by at least one of the integers 4, 6 and 15 ?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-the-remainder-of-division.lesson>Find the remainder of division </A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Why-3%5En%2B7%5En-2-is-divisible-by-8-for-all-positive-integer-n.lesson>Why 3^n + 7^n - 2 is divisible by 8 for all positive integer n ?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/What-is--the-last-digit-of-the-number-a%5En.lesson>What is the last digit of the number a^n ?</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-the-last-three-digits-of-these-numbers.lesson>Find the last three digits of these numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-the-two-last-digits-of-the-number-3%5E123%2B7%5E123%2B9%5E123.lesson>Find the last two digits of the number 3^123 + 7^123 + 9^123</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/Find-the-last-two-digits-of-%281%21-%2B-2%21-%2B-3%21-%2B-%2B-2024%21%29%5E2024.lesson>Find the last two digits of (1! + 2! + 3! + ... + 2024!)^2024</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-n-th-term-of-a-sequence.lesson>Find n-th term of a sequence</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Solving-one-Diofantine-equation.lesson>Solving Diophantine equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Determine-the-number-of-integer-solutions-to-equation-n%5E2%2B18n%2B3=m%5E2.lesson>How many integers of the form n^2 + 18n + 13 are perfect squares</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Miscellaneous-problems-on-divisibility-numbers.lesson>Miscellaneous problems on divisibility numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-the-sum-of-digits-of-integer-numbers.lesson>Find the sum of digits of integer numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Two-digit-numbers-with-digit-9.lesson>Two-digit numbers with digit "9"</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-a-triangle-with-integer-side-lengths-and-integer-area.lesson>Find a triangle with integer side lengths and integer area</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-circle-level-problem-on-the-hundred-handed-monster-Briareus.lesson>Math circle level problem on the hundred-handed monster Briareus</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Math-Citcle-level-problem-on-lockers-and-divisors.lesson>Math Circle level problem on lockers and divisors of integer numbers</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Solving-problems-on-modular-arithmetic.lesson>Solving problems on modular arithmetic</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Using-the-little-Fermat-theorem-to-solve-a-problem-on-modular-arithmetic.lesson#google_vignette>Using the little Fermat's theorem to solve a problem on modular arithmetic</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/OVERVIEW-of-miscellanious-solved-problems-on-divisibility-of-numbers.lesson>OVERVIEW of miscellaneous solved problems on divisibility of integer numbers</A>