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The first challenge to complete is the Penny Problem. The radio station is giving the winner of this challenge a prize pack that includes tickets to see his or her favorite band in concert. To start off the challenge, the radio station has placed pennies in a cylindrical glass jar. Each penny is 0.75 inches in diameter and 0.061 inches thick. If the cylindrical glass jar containing the pennies has a diameter of 6 inches and a height of 11.5 inches, how many pennies can fit inside the jar? You must show all work to receive credit.

Guest Jun 23, 2015

#1**+11 **

I'm assuming the person putting the pennies in is slightly obsessive and is totally okay with spending possibly hours measuring exactly where to place their pennies into perfect stacks.

Also I am not going to round any numbers, so apologies for some of the crazy values I get later.

Firstly, let's place a penny in the dead center of the jar, giving it a radius of ^{0.75}/_{2} inches.

Now let's place some pennies around it in a ring!

I want to know how much angular space each surrounding penny will take. We know exactly how far their centres are from the jar's centre (assuming we place them perfectly in contact), and we know how far they are from the next penny who is just as far from the centre (again assuming they are perfectly in contact).

We can give a fancy forumla, but for the first ring it's actually exactly 60° per penny. Reason being is the distance of each of their centres is the same, forming and equilateral, and the radius they take is (2 * the angle from their centre to the line of contact they form with the next penny in the ring). Since this angle is just 60°/2 it ends back where we started.

Below is a diagram of this. The pink dot is the jar's centre. "r" is the distance from the centre to another coin's centre (Which in this case is (^{0.75}/_{2})*2= 0.75 inches). The blue angle is the angular space the penny takes up, and the green angles are the angles formed by the triangle.

So in this particular case, we can fit 360/60 = **6 pennies** into this ring.

Current total = 1 + 6 = **7 pennies**

Now let's go for the next ring!

But this time they won't form our perfect equalateral. No, but we still know the lengths formed!

r = distance from centre.

Using the cosine rule we can find the angle needed (Note how as previously explained, the angle from centre-to-centre of the pennies is the same as the angle from edge-to-edge of the pennies.)

a^{2} = b^{2} + c^{2} - 2bc*cosine(A)

cosine(A) = [a^{2} - (b^{2} + c^{2})] / [-2bc]

A = cos^{-1}([a^{2} - (b^{2} + c^{2})] / [-2bc])

**The length 'a' is 0.75, 'b' and 'c' are (0.75/2 + 0.75 + 0.75/2) = 1.50**

Angle θ = cos^{-1}([0.75^{2} - (1.5^{2} + 1.5^{2})] / [-2 * 1.5 * 1.5])

= cos^{-1}([0.5625 - 4.5] / [-4.5])

= cos^{-1}(0.875)

= 28.95502437186°

Number of pennies in ring 2 = 360 / 28.955 = 12.433085822828527

Since we can't have parts of pennies, this ring will have **12 pennies**

Current total = 7 + 12 = **19 pennies**

I shall now continue this process until (r+0.75) > 3 inches

*(When the next ring would exceed our limit)*

**Ring 3:**

a = 0.75, b and c = 1.5+0.75 = 2.25

Angle θ = cos^{-1}([0.75^{2} - (2.25^{2} + 2.25^{2})] / [-2 * 2.25 * 2.25])

= cos^{-1}([0.5625 - (10.125)] / [-10.125])

= cos^{-1}(0.9444444444444444)

= 19.188136453721°

Number of pennies in ring 3 = 360 / 19.188 = 18.7617260787992495

Again, round down to **18 pennies**

Current total = 19 + 18 = **37 pennies**

**Ring 4:**

a = 0.75, b and c = 2.25+0.75 = 3.0

Angle θ = cos^{-1}([0.75^{2} - (3.0^{2} + 3.0^{2})] / [-2 * 3.0 * 3.0])

= cos^{-1}([0.5625 - (18)] / [-18])

= cos^{-1}(0.96875)

= 14.361511562917

Number of pennies in ring 4 = 360 / 14.362 = 25.0661467762150118

Again, round down to **25 pennies**

Current total = 37 + 25 = **62 pennies**

**Ring 5:**

a = 0.75, b and c = 3.00+0.75 = 3.75

This ring won't fit, so we have now fitted the jar with 4 rings.

The total number of pennies without stacking on a 6-inch diameter is **62 pennies**

Now, we have a height of 11.5 inches, and each penny is 0.061 inches thick. Let's see how many layers of pennies we can fit! 11.5 / 0.061 = 188.5245901639344262

We can't have partial layers, so that is **188 layers of pennies**

The total number of pennies we can fit is: pennies per layer * number of layers

= 62 * 188

= **11656 Pennies.**

**But wait: THERE'S MORE!**

Remember those gaps we left on rings are actually wide enough to fit a **penny on it's side**

So how do we figure these out? Well we know how much angular was space left per ring, and from there we can figure out the chord formed which would be how much space a coin has to fit (Even on it's side, the coin is the same width, so that works for us perfectly). Of course we need to look at the inner-most ring they form since they are not circular and will make contact at the inner-most point.

So for each ring, let's repeat the process and find the angular space each coin will need. Of course since they have a flat end, it's extremely easy to look at a single coin to find it's angular space-taking.

**Ring 1:**

a = 0.061, b and c = 0.75/2 = 0.375

Angle θ = cos^{-1}([0.061^{2} - (0.375^{2} + 0.375^{2})] / [-2 * 0.375 * 0.375])

= cos^{-1}([0.00374544 - (0.28125)] / [-0.28125])

= cos^{-1}(0.98668288)

= 9.361079434368

Side-ways pennies we can fit is (360 - 60*6) / 9.361079434368 = 0

Current total = 0

**Ring 2:**

a = 0.061, b and c = 0.375 + 0.75 = 1.125

Angle θ = cos^{-1}([0.061^{2} - (1.125^{2} + 1.125^{2})] / [-2 * 1.125 * 1.125])

= cos^{-1}([0.00374544 - (2.53125)] / [-2.53125])

= cos^{-1}(0.99852032)

= 3.11727486689

Side-ways pennies we can fit is (360 - 28.95502437186*12) / 3.11727486689 = 4.0226505756261537

Round down: **4 pennies**

Current total = 0 + 4 = **4 pennies**

**Ring 3:**

a = 0.061, b and c = 1.125 + 0.75 = 1.875

Angle θ = cos^{-1}([0.061^{2} - (1.875^{2} + 1.875^{2})] / [-2 * 1.875 * 1.875])

= cos^{-1}([0.00374544 - (7.03125)] / [-7.03125])

= cos^{-1}(0.9994673152)

= 1.870217269266

Side-ways pennies we can fit is (360 - 19.188136453721*18) / 1.870217269266 = 7.8138214597694015

Round down: **7 pennies**

Current total = 4 + 7 = **11 pennies**

**Ring 4:**

a = 0.061, b and c = 1.875 + 0.75 = 2.625

Angle θ = cos^{-1}([0.061^{2} - (2.625^{2} + 2.625^{2})] / [-2 * 2.625 * 2.625])

= cos^{-1}([0.00374544 - (13.78125)] / [-13.78125])

= cos^{-1}(0.999728222040816326530612244897959183673469387755102)

= 1.335840429288

Side-ways pennies we can fit is (360 - 14.361511562917*25) / 1.335840429288 = 0.7203037922634639

Round down: **0 pennies**

Current total = 11 + 0 = **11 pennies**

**So our new grand total is:**

11656 + 11 = **11667****pennies**

**NOTE:**

It is unlikely this is perfectly accurate. There are small gaps my calculations will leave behind, so I doubt this is perfect, however it's as close to it as I can get with what I came up with :)

**Edit:**

Just noticed I forgot to mention how many stacks of side-ways pennies you can fit. This will be:

11.5 / 0.75 = 15.3333333333333333

Round down: **15 layers**

Total number of side-ways pennies = 11 * 15 = **165 pennies**

**Actual grand total:**

165 + 11656 = **11821 pennies**

Sir-Emo-Chappington Jun 23, 2015

#1**+11 **

Best Answer

I'm assuming the person putting the pennies in is slightly obsessive and is totally okay with spending possibly hours measuring exactly where to place their pennies into perfect stacks.

Also I am not going to round any numbers, so apologies for some of the crazy values I get later.

Firstly, let's place a penny in the dead center of the jar, giving it a radius of ^{0.75}/_{2} inches.

Now let's place some pennies around it in a ring!

I want to know how much angular space each surrounding penny will take. We know exactly how far their centres are from the jar's centre (assuming we place them perfectly in contact), and we know how far they are from the next penny who is just as far from the centre (again assuming they are perfectly in contact).

We can give a fancy forumla, but for the first ring it's actually exactly 60° per penny. Reason being is the distance of each of their centres is the same, forming and equilateral, and the radius they take is (2 * the angle from their centre to the line of contact they form with the next penny in the ring). Since this angle is just 60°/2 it ends back where we started.

Below is a diagram of this. The pink dot is the jar's centre. "r" is the distance from the centre to another coin's centre (Which in this case is (^{0.75}/_{2})*2= 0.75 inches). The blue angle is the angular space the penny takes up, and the green angles are the angles formed by the triangle.

So in this particular case, we can fit 360/60 = **6 pennies** into this ring.

Current total = 1 + 6 = **7 pennies**

Now let's go for the next ring!

But this time they won't form our perfect equalateral. No, but we still know the lengths formed!

r = distance from centre.

Using the cosine rule we can find the angle needed (Note how as previously explained, the angle from centre-to-centre of the pennies is the same as the angle from edge-to-edge of the pennies.)

a^{2} = b^{2} + c^{2} - 2bc*cosine(A)

cosine(A) = [a^{2} - (b^{2} + c^{2})] / [-2bc]

A = cos^{-1}([a^{2} - (b^{2} + c^{2})] / [-2bc])

**The length 'a' is 0.75, 'b' and 'c' are (0.75/2 + 0.75 + 0.75/2) = 1.50**

Angle θ = cos^{-1}([0.75^{2} - (1.5^{2} + 1.5^{2})] / [-2 * 1.5 * 1.5])

= cos^{-1}([0.5625 - 4.5] / [-4.5])

= cos^{-1}(0.875)

= 28.95502437186°

Number of pennies in ring 2 = 360 / 28.955 = 12.433085822828527

Since we can't have parts of pennies, this ring will have **12 pennies**

Current total = 7 + 12 = **19 pennies**

I shall now continue this process until (r+0.75) > 3 inches

*(When the next ring would exceed our limit)*

**Ring 3:**

a = 0.75, b and c = 1.5+0.75 = 2.25

Angle θ = cos^{-1}([0.75^{2} - (2.25^{2} + 2.25^{2})] / [-2 * 2.25 * 2.25])

= cos^{-1}([0.5625 - (10.125)] / [-10.125])

= cos^{-1}(0.9444444444444444)

= 19.188136453721°

Number of pennies in ring 3 = 360 / 19.188 = 18.7617260787992495

Again, round down to **18 pennies**

Current total = 19 + 18 = **37 pennies**

**Ring 4:**

a = 0.75, b and c = 2.25+0.75 = 3.0

Angle θ = cos^{-1}([0.75^{2} - (3.0^{2} + 3.0^{2})] / [-2 * 3.0 * 3.0])

= cos^{-1}([0.5625 - (18)] / [-18])

= cos^{-1}(0.96875)

= 14.361511562917

Number of pennies in ring 4 = 360 / 14.362 = 25.0661467762150118

Again, round down to **25 pennies**

Current total = 37 + 25 = **62 pennies**

**Ring 5:**

a = 0.75, b and c = 3.00+0.75 = 3.75

This ring won't fit, so we have now fitted the jar with 4 rings.

The total number of pennies without stacking on a 6-inch diameter is **62 pennies**

Now, we have a height of 11.5 inches, and each penny is 0.061 inches thick. Let's see how many layers of pennies we can fit! 11.5 / 0.061 = 188.5245901639344262

We can't have partial layers, so that is **188 layers of pennies**

The total number of pennies we can fit is: pennies per layer * number of layers

= 62 * 188

= **11656 Pennies.**

**But wait: THERE'S MORE!**

Remember those gaps we left on rings are actually wide enough to fit a **penny on it's side**

So how do we figure these out? Well we know how much angular was space left per ring, and from there we can figure out the chord formed which would be how much space a coin has to fit (Even on it's side, the coin is the same width, so that works for us perfectly). Of course we need to look at the inner-most ring they form since they are not circular and will make contact at the inner-most point.

So for each ring, let's repeat the process and find the angular space each coin will need. Of course since they have a flat end, it's extremely easy to look at a single coin to find it's angular space-taking.

**Ring 1:**

a = 0.061, b and c = 0.75/2 = 0.375

Angle θ = cos^{-1}([0.061^{2} - (0.375^{2} + 0.375^{2})] / [-2 * 0.375 * 0.375])

= cos^{-1}([0.00374544 - (0.28125)] / [-0.28125])

= cos^{-1}(0.98668288)

= 9.361079434368

Side-ways pennies we can fit is (360 - 60*6) / 9.361079434368 = 0

Current total = 0

**Ring 2:**

a = 0.061, b and c = 0.375 + 0.75 = 1.125

Angle θ = cos^{-1}([0.061^{2} - (1.125^{2} + 1.125^{2})] / [-2 * 1.125 * 1.125])

= cos^{-1}([0.00374544 - (2.53125)] / [-2.53125])

= cos^{-1}(0.99852032)

= 3.11727486689

Side-ways pennies we can fit is (360 - 28.95502437186*12) / 3.11727486689 = 4.0226505756261537

Round down: **4 pennies**

Current total = 0 + 4 = **4 pennies**

**Ring 3:**

a = 0.061, b and c = 1.125 + 0.75 = 1.875

Angle θ = cos^{-1}([0.061^{2} - (1.875^{2} + 1.875^{2})] / [-2 * 1.875 * 1.875])

= cos^{-1}([0.00374544 - (7.03125)] / [-7.03125])

= cos^{-1}(0.9994673152)

= 1.870217269266

Side-ways pennies we can fit is (360 - 19.188136453721*18) / 1.870217269266 = 7.8138214597694015

Round down: **7 pennies**

Current total = 4 + 7 = **11 pennies**

**Ring 4:**

a = 0.061, b and c = 1.875 + 0.75 = 2.625

Angle θ = cos^{-1}([0.061^{2} - (2.625^{2} + 2.625^{2})] / [-2 * 2.625 * 2.625])

= cos^{-1}([0.00374544 - (13.78125)] / [-13.78125])

= cos^{-1}(0.999728222040816326530612244897959183673469387755102)

= 1.335840429288

Side-ways pennies we can fit is (360 - 14.361511562917*25) / 1.335840429288 = 0.7203037922634639

Round down: **0 pennies**

Current total = 11 + 0 = **11 pennies**

**So our new grand total is:**

11656 + 11 = **11667****pennies**

**NOTE:**

It is unlikely this is perfectly accurate. There are small gaps my calculations will leave behind, so I doubt this is perfect, however it's as close to it as I can get with what I came up with :)

**Edit:**

Just noticed I forgot to mention how many stacks of side-ways pennies you can fit. This will be:

11.5 / 0.75 = 15.3333333333333333

Round down: **15 layers**

Total number of side-ways pennies = 11 * 15 = **165 pennies**

**Actual grand total:**

165 + 11656 = **11821 pennies**

Sir-Emo-Chappington Jun 23, 2015