# Homework Help: Pencil and circle

1. Oct 10, 2005

### momentum

i have a question . the question is as follows.

Question 1:

A company manufactures pencils in boxes of 6, 9, and 20. The boxes are sealed and the pencils
cannot be sold loose. What is the largest number of pencils that a wholesaler cannot purchase
using some combination of these boxes?

Ans : 43

How do i solve this problem ? Do i have to use LCM ? LCM does not give the correct answer though. i am clueless and confused. any tips how to solve this ?

question 2:

you know, i have a circle. inside circle there is an equilateral triangle. inside equilateral triangle there is another circle. is there any relationship between inner and outer cirle.

something like this ? radius of outer circle =2*radius of inner circle.

is that correct ?

i want to know what is the relationship between inner and outer circle ?

2. Oct 10, 2005

### Integral

Staff Emeritus
Somehow I feel like these are homework questions.

For the second question. Draw a picture, then find a triangle which has each radius as a side. Apply trigonometry.

3. Oct 10, 2005

### momentum

first of all , this is not a homework .

However , if you notice , i have posted some idea about how to solve this problem and have become unsuccessful. so, i asked for some hitns NOT the solution.

what i got ? i got accused of posting homework question !

i dont know why people are so impatient and not helpful in this board. there are 2 questions i have asked and i am stuck with that . i could not do it by myself and sought for help .

i posted the question from my notebook as it is , so that it help others to understand clearly what the question is without any ambiguity . what i got ? i got accused .

Its really very pathetic to get this type of discouraging answer.

even if you see , anybody is posting homework there are ways to help the guy out . may be you could provide some hints, some tutorial which is relevant to the question ... probabily this could be the quality answer .

i am very much upset and fed up with the forum . people are not helpful here. i must leave this board..

4. Oct 10, 2005

### ivybond

Q.1
Rephrase the question: there are numbers of pencils which can not be made up by adding up 6's, 9's, and 20's in different combinations. These "impossible" numbers can be arranged in a sequence in ascending order. The question states that there is the largest such number. That means the sequence stops at some point. It goes: 1, 2, 3, 4, 5, 7, 8, 10, 11, 13, ...
What is this sequence?

Or: find "possible" numbers that CAN be made up (6, 9, 12, 15, ...) and watch if a pattern emerges.

-----------------
Q.2
Non-trig way.
Connect the points of tangency of the inscribed circle.
Now you have two similar shapes: a larger triangle inscribed in a larger circle and a smaller ...
In a loose math-speak: "ratio of the circles" should equal "ratio of the triangles" (why? please let us know).

Last edited: Oct 10, 2005
5. Oct 11, 2005

### momentum

here is what i have understood

"...What is the largest number of pencils that a wholesaler cannot purchase
using some combination of these boxes..."

as you have given a sequence
It goes: 1, 2, 3, 4, 5, 7, 8, 10, 11, 13,

6 is already in the box, so it is excuded from the sequence . 15 i can make easily because 9+6=15 , so 15 is excluded from the sequence, similarly 20 ,26 =20+6 are also excluded from the sequence.

so, this sequence will gradually grow. and some of the elements will be missing from the sequence because they could be reproduced by elements{6,9,20} ...so where do i stop ? it seems no where to stop . it'll straight forward go to infinity . is not it ?

6. Oct 11, 2005

### momentum

here is what i have understood

"...What is the largest number of pencils that a wholesaler cannot purchase
using some combination of these boxes..."

as you have given a sequence
It goes: 1, 2, 3, 4, 5, 7, 8, 10, 11, 13,

6 is already in the box, so it is excuded from the sequence . 15 i can make easily because 9+6=15 , so 15 is excluded from the sequence, similarly 20 ,26 =20+6 are also excluded from the sequence.

so, this sequence will gradually grow. and some of the elements will be missing from the sequence because they could be reproduced by elements{6,9,20} ...so where do i stop ? it seems no where to stop . it'll straight forward go to infinity . is not it ?

7. Oct 11, 2005

### ivybond

Why are you afraid to continue? You got the idea right.
Since you are GIVEN the answer, you can trust that the sequence WILL NOT go straight forward go to infinity.
Be sure you don't miss "possible" numbers. 6+6=12, 9+9=18, 6+6+9=21 ...

8. Oct 11, 2005

### momentum

>Why are you afraid to continue?

of course, i could continue . but is this the only way to solve this problem ?

because i have to sit and take every number and have to think wheteher that number could be generated by the {6,9,20 } combination . again , i have to take next number and have to consider again the same . just think about the labour . we are not doing it in a cleverer way . we are doing it manually . you are right at some point of time , i'll get the solution . but the way we are doing it manually is not good . is not it ? what do you think ?

though here i know the answer is to be 43. that means 44,45,46...so on numbers can not be produced by {6,9,20} combination.

so, if i could continue the search then i would have stopped at 43 . but suppose the answer could have been say 101 ...that means i have to count upto 101 and for every number i have to check manually whether that could be produced by {6,9,20} or not ......huge labour.....this can not be the efficient way . is not it..... just think about it.

we need to be apply some intelligent logic/methods/formulas to achieve the solution.

thank you for your time and response.

9. Oct 11, 2005

### momentum

i may be wrong . just posted my view

10. Oct 11, 2005

### ivybond

momentum - there is no ground for you to act indignant.
That's what people do on this board.
It's not like "you've got questions, we've got answers".
You have to show some work.
Making a couple of guesses does not amount to much.
----------------------------
I suggested two approaches to you ("possible" and "impossible" numbers).
I did not say that was a solution, and you shouldn't have expected that.
It's a hint on how you can approach solving the question before coming up with the final and efficient solution.
======================
I am not sure if you made a typo here.
44,45,46...so on numbers CAN be produced by {6,9,20} combination.
How far did you go? Invest some labor and it'll pay.
One more hint: remainders from division by 6.
++++++++++++++++++++++
What often helps in questions like this is reducing them to easier ones.
How about combinations {5,7}? What's the largest "impossible" number? Why do the "impossibles" stop? What happens with the remainders of dividing by 5?
~~~~~~~~~~~~~~~~~~
Can't agree with you more.

Last edited: Oct 11, 2005