[wolfgang]

Hello...yesterday during the annual inventory in my factory,I faced a problem...!

I was to count howmany stickers are there in a roll( was already used..i.e. not original),and was the last one.

Each sticker is 10x10 cm,between each two stickers there is a (2mm)space.they are all carried side to side on a waxed paper roll,this roll has a hard carton center .

All what I know from my experience is that,there are exactly two stickers surrounding the carton center.

In this roll I can see exactly (6) stickers surrounding it from outside.

I was not allowed to count the stickers one by one!

can you find out howmany stickers were there in this roll?

Note: the Radius of an original roll is (12)cm,with 500 stickers.

Edited November 27, 2010 by wolfgang

[araver]

Just to clarify:

1) A full roll has 12 cm radius, 500 stickers, 2 stickers surrounding the carton cylinder in the center. Do the stickers in the center actually touch on the other side or there is a 2mm space between them on both sides? (making it rather symmetrical)

2) The unknown partially used roll that has 6 stickers on the outside: sticker 1 and sticker 6 actually touch or there is a 2mm space between them? (again, making it symmetrical).

[wolfgang]

Just to clarify:

1) A full roll has 12 cm radius, 500 stickers, 2 stickers surrounding the carton cylinder in the center. Do the stickers in the center actually touch on the other side or there is a 2mm space between them on both sides? (making it rather symmetrical)

2) The unknown partially used roll that has 6 stickers on the outside: sticker 1 and sticker 6 actually touch or there is a 2mm space between them? (again, making it symmetrical).

Thank you dear Araver...as I said...there are always 2mm space between each two Stickers.

so they are symmetrical.

[araver]

Thank you dear Araver...as I said...there are always 2mm space between each two Stickers.

so they are symmetrical.

Reason I asked if it is symmetrical is because I can find no solution with that symmetry in mind.

Assume the inner carton cylinder has radius r cm and the stickers are x cm thick.

The first layer has radius r1=r+x and a perimeter of 2*Pi*r1.

You said it's symmetrical so its perimeter equals 2 stickers(10cm each) + 2 interleaving portions(.2cm each) so 10+0.2+10+0.2=20.2cm.

(Eq1) 2*Pi*(r+x)=20.2

The unknown used roll has an unknown radius r2 and a perimeter 2*Pi*r2. Now since its perimeter has 6 stickers with 6 interleaving portions we get the second equation (where m is the number of times you pass the original point where the first sticker is glued to the inner carton).

(Eq2) 60.6=2*Pi*r2=2*Pi*(r+m*x)

A full roll has radius r3=12. The length of 500 stickers in a line is 10+.2+10+.2+...+10+.2+10 = 500*10.2-.2 =5099.8. First, the full roll has an unknown number n of full sheets of stickers (n=the number of times you pass the original point where the first sticker is glued to the inner carton). Since each "circle" is x cm thick, we get:

(Eq3) r3=r+n*x=12

From Eq1 and Eq3 we get:

(Eq4) (n-1)*x=12-10.1/Pi (approx 3.21 cm)

The length of n full sheets of stickers can be expressed as the sum of perimeters of the n circles, each circle having a radius with xcm bigger than the previous circle:

(Eq5) Length=2*Pi*(r+x) +2*Pi*(r+2*x)+ ... +2*Pi*(r+n*x) 

= 2*Pi*n*(r+x)+2*Pi*(0+1+...+n-1)*x

= 2*Pi*n*(r+x)+2*Pi*n*(n-1)/2*x

= 2*Pi*n*(r+x)+Pi*n*(n-1)*x

Using Eq1 and Eq4 in Eq5 we get:

(Eq6) Length = 2*Pi*n*10.1/Pi + Pi*n*(12-10.1/Pi)

= n*(20.2+12*Pi-10.1)

= n*(10.1+12*Pi)

Assuming the rolling of a full roll finishes right on top of the starting position then Length in Eq6 is equal to the length of 500 stickers in a line (5999.2) (Eq6) Length = 2*Pi*n*10.1/Pi + Pi*n*(12-10.1/Pi) = n*(20.2+12*Pi-10.1) = n*(10.1+12*Pi)

Length(m)=2*Pi*(r+x) +2*Pi*(r+2*x)+ ... +2*Pi*(r+m*x) 

= 2*m*Pi*r+2*Pi*m*(m+1)/2*x

which yields Length(m) approx. 3185.199 cm If one completes the next full circle, one gets:

Length(m+1)=Length(m+1)+2*Pi*(r+(m+1)*x)

making Length(m+1)=3246.153 cm. Between these two lengths we have the actual sticker tape length which is 10.2*N-0.2 (where N is the number of stickers and we substract 0.2 since after the last sticker there is no interleaving space). The following table list appropriate N values which fall in the [3185.199,3246.153] interval.

 N   Length of sticker tape

313	 3190.6

314	 3200.8

315	 3211

316	 3221.2

317	 3231.4

318	 3241.6

Which would make the number of stickers in the mystery roll somewhere between 313 and 318.

Yet something makes no sense - 6 stickers on the outside with symmetrical interleaving spaces = 61.2 cm perimeter.

And the actual perimeter of the m-th roll (2*Pi*(r+m*x)) is 60,43795739 and of the m+1-th roll is 60,95382864.

Which means something is wrong and I cannot figure out what.

which yields n = 106.6923591. Since n is clearly an integer, we conclude that ending position does not get over the starting position in the n-th circle. So there are 106 full circles and an incomplete outer circle, n = 107. From Eq4 we get x=0.082103459 cm. From Eq1 we get r=3.132826395 cm. From Eq2 we get m=79,31411444 cm, which means 79 full circles around the origin point and then an incomplete circle. The same length formula (as in Eq5) for m consecutive circles gives:

[Guest]

(Eq1) 2*Pi*(r+x)=20.2

I found nothing wrong with your formula at first glance although I'm wondering if the first line should read:

2*Pi* (r+x) = 20.4

Just wondering if there is a so called zero space of 2mm? Although it still didn't work out for me that way either. Of course I may have missed something.

[araver]

(Eq1) 2*Pi*(r+x)=20.2

I found nothing wrong with your formula at first glance although I'm wondering if the first line should read:

2*Pi* (r+x) = 20.4

Just wondering if there is a so called zero space of 2mm? Although it still didn't work out for me that way either. Of course I may have missed something.

thanks. It's 20.4.

Can't believe I missed changing that after wolfgang's response that there is a space there.

But it's not enough - it doesn't save the whole thing.

[wolfgang]

I took the case in a very easy way...

I am not so a (complex)thinking man...take it easy!

[araver]

I took the case in a very easy way...

I am not so a (complex)thinking man...take it easy!

I understand that you're expending a more logical/intuitive solution.

However I felt inclined to solve it as an (amateur) engineer would. I agree, I made a lot of errors yet managed to narrow it down to 6 possibilities.

Assume the inner carton cylinder has radius r cm and the stickers are x cm thick.

The first layer has radius r1=r+x and a perimeter of 2*Pi*r1. Its perimeter equals 2 stickers(10cm each) + 2 interleaving portions(.2cm each) so 10+0.2+10+0.2=20.4cm.

(Eq1) 2*Pi*(r+x)=20.4

The unknown used roll has an unknown radius r2 and a perimeter 2*Pi*r2. Now since its perimeter has 6 stickers with 6 interleaving portions we get the second equation (where m is the number of times you pass the original point where the first sticker is glued to the inner carton).

(Eq2) 61.2=2*Pi*r2=2*Pi*(r+m*x)

A full roll has radius r3=12. The length of 500 stickers in a line is 10+.2+10+.2+...+10+.2+10 = 500*10.2-.2 =5099.8. First, the full roll has an unknown number n of full sheets of stickers (n=the number of times you pass the original point where the first sticker is glued to the inner carton). Since each "circle" is x cm thick, we get:

(Eq3) r3=r+n*x=12

From Eq1 and Eq3 we get:

(Eq4) (n-1)*x=12-10.2/Pi (approx 8.75 cm)

The length of n full sheets of stickers can be expressed as the sum of perimeters of the n circles, each circle having a radius with xcm bigger than the previous circle:

(Eq5) Length=2*Pi*(r+x) +2*Pi*(r+2*x)+ ... +2*Pi*(r+n*x) 

= 2*Pi*n*(r+x)+2*Pi*(0+1+...+n-1)*x

= 2*Pi*n*(r+x)+2*Pi*n*(n-1)/2*x

= 2*Pi*n*(r+x)+Pi*n*(n-1)*x

Using Eq1 and Eq4 in Eq5 we get:

(Eq6) Length = 2*Pi*n*10.2/Pi + Pi*n*(12-10.2/Pi)

= n*(20.4+12*Pi-10.2)

= n*(10.2+12*Pi)

Assuming the rolling of a full roll finishes right on top of the starting position then Length in Eq6 is equal to the length of 500 stickers in a line (5999.8) (Eq6) Length = 2*Pi*n*10.2/Pi + Pi*n*(12-10.2/Pi) = n*(20.4+12*Pi-10.2) = n*(10.2+12*Pi)

Length(m)=2*Pi*(r+x) +2*Pi*(r+2*x)+ ... +2*Pi*(r+m*x) 

= 2*m*Pi*r+2*Pi*m*(m+1)/2*x

which yields Length(m) approx. 3256.246604 cm If one completes the next full circle, one gets:

Length(m+1)=Length(m+1)+2*Pi*(r+(m+1)*x)

making Length(m+1)=3317.766771 cm. Between these two lengths we have the actual sticker tape length which is 10.2*N-0.2 (where N is the number of stickers and we substract 0.2 since after the last sticker there is no interleaving space). The following table list appropriate N values which fall in the [3256.246604,3317.766771] interval.

 N   Length of sticker tape

320	3262

321	3272,2

322	3282,4

323	3292,6

324	3302,8

325	3313

The actual perimeter of the m-th roll (2*Pi*(r+m*x)) is 61.00616509 and of the m+1-th roll is 61.52016718. 6 stickers on the outside with symmetrical interleaving spaces = 61.2 cm perimeter.

which yields n = 106.4696152. Since n is clearly an integer, we conclude that ending position does not get over the starting position in the n-th circle. So there are 106 full circles and an incomplete outer circle, n = 107. From Eq4 we get x=0.081805973 cm. From Eq1 we get r=3.164954869 cm. From Eq2 we get m=80.37710919 cm, which means 80 full circles around the origin point and then an incomplete circle. The same length formula (as in Eq5) for m consecutive circles gives:

Not sure how to narrow it further, but I will think further on this.

[Guest]

Is each roll only 10cm wide? ie. One sticker only?

[araver]

Is each roll only 10cm wide? ie. One sticker only?

Nice

There are at most two, so it's a good question.

[plainglazed]

about 316 stickers

[Guest]

I get 371 if the rolls are 1 sticker wide. I will add logical reasoning if this answer is correct.... If not, back to the drawing board

Edited November 29, 2010 by Oboe Passion

[Guest]

156 if it is two stickers wide.

[wolfgang]

Nice

There are at most two, so it's a good question.

[wolfgang]

about 316 stickers

No....more..!

[araver]

Seriously, OboePassion's question is valid: "Is each roll only 10cm wide? ie. One sticker only?".

There is nothing in your Original Post(OP) that forbids a tape with two stickers one on top of the other (equivalent to two parallel sticker tapes), so if you want to exclude this case, you need to provide a Yes/No answer as clarification.

Thank you.

Edited November 29, 2010 by araver

[k-man]

370 or 371 stickers.

The total circumference of the full new roll containing 500 stickers is 2*pi*120mm = 754mm (rounded). The circumference of the empty roll is 204mm (2 stickers and 2 spaces). Since the thickness of the sticker tape is constant the circumference of each layer of sticker tape changes in the linear fashion from 754mm to 204mm as the roll is unwound. The roll in question has the circumference of 612mm (6 stickers and 6 spaces). From this we can determine that there is approximately 74% of the roll remaining (74.18181818% is more accurate number).

500 * 74.18(18)% = 370.90(90)

[araver]

370 or 371 stickers.

The total circumference of the full new roll containing 500 stickers is 2*pi*120mm = 754mm (rounded). The circumference of the empty roll is 204mm (2 stickers and 2 spaces). Since the thickness of the sticker tape is constant the circumference of each layer of sticker tape changes in the linear fashion from 754mm to 204mm as the roll is unwound. The roll in question has the circumference of 612mm (6 stickers and 6 spaces). From this we can determine that there is approximately 74% of the roll remaining (74.18181818% is more accurate number).

500 * 74.18(18)% = 370.90(90)

The circumference grows linearly, but the number of stickers does not grow linearly (quadratic would be a more accurate polynomial). With each layer you get more and more stickers/circumference (e.g. 1+2+3+... - quadratic growth).

[plainglazed]

The length of the complete roll can be calculated exactly ==> 500 stickers * 10.2cm/sticker = 5100 cm

A very good estimate for the length of this complete roll is the sum of the circumferences of (n) concentric circles

a distance of the thickness (t) of the stickers apart; starting from the core diameter (D1) and going to

the outer diameter of the full roll (Dn):

L ~ pi(D1+(D1+t)+(D1+2t)+(D1+3t)+...+(Dn-2t)+(Dn-t)

this is an arithmetic series equal to pi*n(D1+Dn)/2 and

We know two stickers each of length 10.2cm is appoximately the circumference of the core:

pi(D1) = 20.4cm ==> D1 = 6.5cm

and the diameter of the full roll is 24cm so:

5100 cm ~ pi*n(24cm + 6.5cm)/2 ==> n ~ 106.45 layers in the full roll

Calculating the number of layers of stickers in the incomplete roll (n2) of diameter D2:

(D2-D1)/(Dn-D1) = n2/106.45

In this case D2 is 19.5cm (6 stickers * 10.2 cm/sticker / pi) so n2 = 79.08 layers in the incomplete roll

We can plug this n2 into our reduced arithmetic series formula to find the approximate length of the partial roll (L2)

L2 ~ 79.08 pi(19.5cm + 6.5cm)/2 = 3226.4cm

And finally if each sticker is 10.2cm long ==> 3226.4cm / 10.2cm/sticker = ~316 stickers in the partial roll

Where the approximation comes into play is the slight difference between the true spiral length and the average of the inside

and outside diameters of one layer of stickers. This difference is very slight when the thickness of the sticker is small as

it is in this case (.0822cm) and the number of layers is also relatively small (less than 80) so would not think the difference

from this approximation and the true length would be more or less than one sticker (10cm).

[wolfgang]

Seriously, OboePassion's question is valid: "Is each roll only 10cm wide? ie. One sticker only?".

There is nothing in your Original Post(OP) that forbids a tape with two stickers one on top of the other (equivalent to two parallel sticker tapes), so if you want to exclude this case, you need to provide a Yes/No answer as clarification.

Thank you.

Its only one sticker..i.e. the widith is 10.4cm