## ________________

## ________________

### In The Groove

On a typical LP record, the spiral groove runs from an outside radius of 15 cm to an inside radius of 6 cm, with equal radial spacing between successive turns of the spiral. The record spins at a constant rotational velocity of 33 1/3 rotations per minute. The record plays for 22 1/2 minutes before you have to flip it over to play the other side.

*What is the radial spacing between successive turns of the groove on the record? (Ignore the width of the groove; assume that in effect, it has zero width.) *

### On The Path

A typical CD has an outer radius of 6 cm, and a 45-minute track is recorded on a spiral path with an inner radius of 4 cm. That spiral path is read by an optical sensor, and there is again equal radial spacing between successive turns of the path. However, unlike a record, the rotational speed of the CD varies as the sensor moves along the path so that the sensor is traveling at a constant velocity of 120 cm (of path length) per second.

*What is the radial spacing between successive turns of the recording path? (For this one, you may just want to find an approximate answer; and again, ignore the width of the path.)*

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## Solutions to week 69

**Not Dozens**. It’s easiest to do this by adding up *all* of the numbers from 1 to 1,728 and then subtracting the ones that are multiples of twelve. Of course, we won’t really add up all those numbers, but rather use the formula (attributed to Gauss, probably apocryphally) that the sum of the first *k* positive whole numbers is *k*(*k*+1)/2. So the sum from 1 to 1,728 is 864×1729. How do we subtract all the multiples of twelve? Well, the multiples of twelve in question start from twelve itself and run all the way up to 1,728, which turns out to be 12×144. By the distributive property, the sum of 12×1 + 12×2 + … + × 12×144 is just twelve times the sum of 1 + 2 + … + 144. So we can re-use the same formula to get that the sum of the multiples of twelve up to 1,728 is 12×72×145. Thus Chris’s answer is 864×1729 – 12×72×145 = **1,368,576**.

Did you notice that 144 is the square of twelve? In other words, we’re being asked to add up all of the numbers up to the cube of a number

*n*except for multiples of

*n*, and that is something that we can do all at once, for any possible number, using the so-called Gauss formula. In symbols, it is

*n*³(

*n*³+1)/2 –

*n*×

*n*²(

*n*²+1)/2 =

*n*³(

*n*³-

*n*²)/2 =

*n*

^{5}(

*n*-1)/2, for any

*n*; if you plug in twelve to this expression, you will see the result matches the answer above.

**Festive Factorial**. The critical observation needed to solve this problem is that if you have a product the last digit of which is non-zero, then the last digit of the product is the last digit of the product of all the last digits of the factors. For example, the last digit of 34×13 = 442 is two, which is also the last digit of 4×3. Furthermore, if you have a product with last digit zero, then there is always at least one factor of five and one factor of two, and you can cancel them out to get a product with the same last non-zero digit. For example, 15×8 is 120, with the same last digit as 3×4.

These facts suggest that we should break the product of 2017! into groups of five numbers, of which one will be a multiple of five and the others will have no factor of five. There will be exactly 403 such groups, with 2016 and 2017 left over to include in the final product. Each group consists of one of the first 403 multiples of five, together with a product of the form (5*k*+1)(5*k*+2)(5*k*+3)(5*k*+4) for some non-negative integer *k*. Multiplying that product out, it is 24 + 250*k* + 875*k*² + 1250*k*³ + 625*k*^{4}, which can be re-written as 24 + 125*k*(2+7*k*+10*k*²+5*k*³). It turns out the second term of that sum is always a multiple of 1,000 for any *k*. That means that 2017! is the product of 2017 and 2016 and the first 403 multiples of five and 403 numbers each of the form 1000*j*+24.

Now, we can cancel out one factor of five from each of those multiples of five, and one factor of two from each of the other numbers, to get a product with the same last non-zero digit. In other words, 2017! has the same last non-zero digit as 2017 times 2016 times 403! times 403 numbers each of the form 500*j*+12.

To finish off the problem, we proceed in the same way: 403! is the product of 403, 402, 401, and the first 80 multiples of 5, and 80 numbers of the form 1000*j*+24. So 2017! has the same last non-zero digit as the product of 2017, 2016, 403, 402, 401, 80!, and 403 + 80 = 483 numbers of the form 500*j*+12. Taking another step, it has the same last non-zero digit as the product of 2017, 2016, 403, 402, 401, 16!, and 483+16 = 499 numbers of the form 500*j*+12. And we only need to do this one last time: 2017! has the same last non-zero digit as the product of 2017, 2016, 403, 402, 401, 16, 3!, and 499+3 = 502 numbers of the form 500*j*+12.

Now, none of the numbers in this last product has any factor of five, so its last digit is non-zero. Hence, the last digit of this product is the same as the last digit in the product of the last digits of each of the factors. So we have reduced the last non-zero digit of 2017! to the last digit of 7×6×3×2×1×6×3×2×1×2^{502}. Multiplying out, we want the last digit of 9072×2^{502}. Since there are no factors of five in this product, this is the same as the last digit of 2×2^{502} = 2×4^{251} = 2×4×4^{250} = 2×4×16^{125}. Again, we can take just the last digits of the factors, so the last non-zero digit of 2017! is the same as the last digit of 2×4×6^{125}. But it is well known that the last digit of every power of six is just six, so the last non-zero digit of 2017! is the same as the last digit of 2×4×6 = 48, that is, **eight**.

Actually, because each of those 502 numbers is of the form 500

*j*+12, we can get the last two non-zero digits of 2017! without any additional work. In the absence of factors of five, the last two digits of a product is the last two digits of the product of the last two digits of each factor. So the last two non-zero digits of 2017! are the same as the last two digits of 17×16×3×2×1×16×3×2×1×12

^{502}= 156672×12

^{502}. That’s the same as the last two digits of 72×12

^{502}= 72×144

^{251}, which is the same as the last two digits of 72×44×44

^{250}= 3168×1936

^{125}. So that’s the same as the last two digits of 68×36×36

^{124}= 2448×1296

^{62}, which is that same as the last two digits of 48×96

^{62}= 48×9216

^{31}. Continuing in this fashion, let’s write –> to mean “has the same last two digits as”. Then 2017! has the same last two non-zero digits as 48×16×16

^{30}= 768×256

^{15}–> 68×56×56

^{14}= 3808×3136

^{7}–> 8×36×36

^{6}= 288×1296³ –> 88×96³ = 77856768 –> 68. To get the last

*three*digits, though, we’d have to be more careful, in that we’d need to know which of those 502 numbers ended in 512 and which in 012.

## Recent Weeks

**Week 69**: Not Dozens & Festive Factorial, solutions to Snug Circle & Congruence Time

**Week 68**: Snug Circle & Congruence Time, solutions to Card 61 & Card T

**Week 67**: Card 61 & Card T, solutions to Long Haul & Postal Pathways

**Week 66**: Long Haul & Postal Pathways, solutions to Card 129 & Card 94

**Week 65**: Card 129 & Card 94, solutions to Skimpy Schedule & Passing Pennies

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

**Come back next week** for answers and more puzzles.

[asciimathsf]