Glamor Geometric Series

Agnishom · 2013-06-18 14:15:58

Integers a, b, c, d and e satisfy 50<a<b<c<d<e<500, and a,b,c,d,e form a geometric sequence. What is the sum of all possible distinct values of a?

bobbym · 2013-06-18 15:53:49

Hi;

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Agnishom · 2013-06-18 17:01:14

That seems to be correct!

But how isn't a = 50*(10^(1/6))

bobbym · 2013-06-18 17:56:52

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Post cleared.

Bob · 2013-06-18 19:59:47

hi

I just reasoned it out from the integer property. Useful to have a spreadsheet but, I suppose I could have worked these numbers out anyway.

Bob

Agnishom · 2013-06-19 00:12:45

Help! I did not understand anything at all

BTW This is a Level 4 problem

bobbym · 2013-06-19 00:25:45

Hi;

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Bob · 2013-06-19 00:49:44

hi Agnishom,

Are you wanting an outline of my answer? (note; it was mainly trial and improvement)?

Bob

Agnishom · 2013-06-19 00:54:17

Yep and please explain it. I don't understand it at all

Bob · 2013-06-19 01:09:18

hi Agnishom,

I made a spreadsheet just to make the calculations easy.

r is the common ratio.

r>1 or you won't get b>a etc.

I tried r = 2 and even with a = 51 the numbers quickly get out of range, so

1 < r < 2

So I tried r = 1.1

The numbers become decimals not integers.

Tried r = 1.5

Term will be the previous term plus half of it so 'a' needs to be even.

Then I realised that unless a is a multiple of 16, the successive halving will generate a decimal.

eg If a = 100 then b = 150 and c = 225 and then we get decimals

So I tried 64 and got an answer, then 80, then 96 and then I was out of range.

So what else might work.

r = 1.25 No because successive quartering means I have to start with a large power of 4, out of range.

r = 1.333333 Only if I start with a multiple of 3^4

That gave me one more solution.

r = 1.666666 goes out of range.

Any other (1 plus fraction) won't work because

say r = 1 + 1/n

a must be a fourth power of a multiple of n. Out of range for all n > 3

And that's about it.

As you can see I sort of discovered what to do as I tried numbers and gradually developed a rule.

Bob

got to go out so any clarification will have to wait 'til later.

Agnishom · 2013-06-19 02:59:24

Just passed like a tangent over my tiny brain. After my examination stress is reduced, I'll trying reading that again

bobbym · 2013-06-19 03:03:58

Did you try programming it?

Agnishom · 2013-06-19 03:42:46

No.

I am not at all sure what the problem means.
Doesn't it mean that a = 50x, b = 50x*x, c = 50x*x*x, d = 50x*x*x*x, e = 50x*x*x*x*x
The ratio needs to be fixed, right?

bobbym · 2013-06-19 03:45:49

Yes, exactly!

phanthanhtom · 2013-06-19 03:51:02

a min = 51.
e max = 499.
==> ratio max ~ 1.768
Now if ratio was expressed m/n then a is divisible by n^4 for e to be integral. Therefore n < 5, if n > 4 => a > 495 contradiction.
Now try ratios m/n = 3/2; 4/3; 5/3; 5/4; 7/4.
E.g. m/n = 3/2. ==> e = 81a/16 < 500
==> a < 99. But a is divisible by 2^4 therefore 3 values 64, 80 and 96.

bobbym · 2013-06-19 03:56:52

There is one more.

Agnishom · 2013-06-19 03:57:50

So 500 = 50x^6 => 10 = x^6

x has exactly 6 solutions the sum of them is zero.

Since, a = 50x, the sum of all a should be zero too.
How can it be an integer?

Bob · 2013-06-19 04:42:38

hi Agnishom,

You are still misunderstanding the question.

The number 'r' is called the common ratio of the geometric series. This means that

a is the first term

b = ar is the second

c = ar^2 is the third

d = ar^3 is the fourth

e = ar^4 is the fifth (and, for our purposes, the last).

Here's what I tried first.

r = 1.1 a = 51 gives b = 56.1 c = 61.71 d = 67.881 e = 74.6891

That won't do because they are supposed to be integers.

So I tried r = 1.5 a = 62

Better because b is also now an integer but c isn't.

So I started thinking ......

For e to be an integer, d must be even. (1.5 x even is bound to be an integer)

So c must be a multiple of 4

eg. c = 84 => d = 126 => e = 189

Reasoning backwards, b must be a multiple of 8 and a must be a multiple of 16

Tried r = 1.5 a = 64 => b = 96 => c = 144 => d = 216 => e = 324

Success!

So keep r = 1.5 and try other multiples of 16

80,120,180,270,405
96,144,216,324,486

This last is only just under 500 so there's no point trying any more.

Now what else could r be ?

r = 1 and a third ??

If any term is multiplied by r you'd get that term + 1/3 of that term

So for e to be an integer, d must be divisible by 3 => c must be divisible by 9 => b must be divisible by 27 => a must be divisible by 81

So try 81,108,144,192,256

Could r be 1 and two thirds.

You get integers but it goes above 500.

Could r be 1 and a quarter ?

For e to be an integer, d must be divisible by 4 => c by 16, => b by 64 => a by 256

It'll work but be out of range. Same goes for any other r of the form 1 + a proper fraction.

Could r be irrational ?

No. Fails at the first hurdle because integer x irrational gives an irrational and not an integer.

So that's the complete list.

Sum of a = 64 + 80 + 96 + 81 = 321

Bob

Agnishom · 2013-06-19 14:55:36

Then I realised that unless a is a multiple of 16, the successive halving will generate a decimal.

Why and How?

phanthanhtom · 2013-06-19 17:08:26

a has to be divisible by 16 so that it could cancel with 16 in the denominator.

Bob · 2013-06-19 19:16:34

hi Agnishom,

Let's take as an example the sequence that ends e = 405

That required that d was even. And it was d = 270.

But if c is 'just' even it will mean that d is an integer, but it won't guarantee that e is an integer.

From c to e is

Similarly

and finally

so we have c = 180. This is divisible by 4

and b = 120. This is divisible by 8

and a = 80. This is divisible by 16

Try out some other numbers and you'll begin to see why it has to be 16.

(and for r = 1 and a third a has to be divisible by 3 x 3 x 3 x 3 .)

Bob

Agnishom · 2013-07-02 04:47:14

The Featured Solution: https://brilliant.org/i/b6ozMJ/

bobbym · 2013-07-02 04:59:37

Yes, if you are signed in as a member.

Agnishom · 2013-07-02 05:10:27

New Problem There are 5 kinds of dishes i, ii, iii, iv, and v. Four Customers A, B, C and D choose a dish at random. What is the probability that person D chooses a previously unordered dish?

bobbym · 2013-07-02 05:16:58

Although I have the exact answer by experimental methods I am still working on a good solid method that would please an olympic dude.

Math Is Fun Forum

#1 2013-06-18 14:15:58

Glamor Geometric Series

#2 2013-06-18 15:53:49

Re: Glamor Geometric Series

#3 2013-06-18 17:01:14

Re: Glamor Geometric Series

#4 2013-06-18 17:56:52

Re: Glamor Geometric Series

#5 2013-06-18 19:59:47

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#6 2013-06-19 00:12:45

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#8 2013-06-19 00:49:44

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Re: Glamor Geometric Series

#10 2013-06-19 01:09:18

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#15 2013-06-19 03:51:02

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#18 2013-06-19 04:42:38

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#19 2013-06-19 14:55:36

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#20 2013-06-19 17:08:26

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