Question 1205868
.
A bag of lollies is shared. When shared equally among 4 people, there are 2 lollies left. 
When shared equally among 9 people, there are 5 lollies left. When shared equally among 7 people, 
there are 2 lollies left.
Determine the smallest possible number of lollies that could have been in the bag as well as the general solution.
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86 is the correct answer, as you can easily check it  ON  YOUR  OWN.


94 is incorrect answer, as you can easily check it  ON  YOUR  OWN.


I do not understand, why do you panic, having a correct answer in your hands.


You always can easily check it  ON  YOUR  OWN  if the answer is correct or not,
if you do understand correctly the meaning of the problem.



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        Edwin presented huge long solution, which may scare you.
        Therefore,  I decided to post here my solution,  which is a standard method 
        for such problems and is much shorter.



<pre>
Let N be our number.

Subtract 2 from N and consider the number N-2.


Since N gives the remainder 2 when is divided by 4 and by 7, we conclude that 
the number (N-2) is divided by 4 and by 7 with no remainder.


Hence, N-2 is divided by 4*7 = 28  (since 4 and 7 are relatively prime numbers).

So, we can write  N-2 = 28k,  where "k" is some integer number.
Hence,  N = 28k + 2.


Therefore, we will consider the numbers of the form 28k+2 to find value of "k" such 
that 28k+2 gives the remainder 5 when is divided by 9.


Try some values k= 1, 2, 3, 4, 5 . . . .


It will not take long time, since k=3 just provides you the number N= 28*3+2 = 86,
which gives the remainder 5 when is divided by 9.


So, the number 86 is the <U>ANSWER</U>.
</pre>

Solved.


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If you want to see many other similar &nbsp;(and different) &nbsp;solved problems, &nbsp;look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://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> 

in this site. 


These problems are typical for this subject, and reading these lessons,
you will get familiar with the basic methods.



Problems of this type are introductory into the subject on divisibility and remainders.


In advanced Math circles/classes, students start learning this material at 4th grade.
May be, it is too early, but learning it, let say, at 5th - 6th grade is just good time.



It is absolutely clear that 4th, 5th and even 6th grade student is not able to perceive long reasoning.


But such a student should be able to easy perceive what is written in my post - it is why I wrote it here.



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Since Edwin continue this discussion, I'd like to answer on his very first point.


Edwin writes that "Ikleyn believes trial and error is legitimate, but it isn't!".


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;- No.  &nbsp;&nbsp;It is not so.


What I state is another statement: 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"If trial and error produces the answer, which after checking 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;is proven to be correct, then the answer is correct 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and the solution is legitimate."



Edwin, it is not a good style to make a discussion
by attributing me statements that I did not make,
and then disproving these statements.


Such method of making discussion was known back in ancient Greece,
and it has the name - which I do not want to pronounce here.