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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?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

Hi;

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**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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That seems to be correct!

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

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**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,752

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Help! I did not understand anything at all

BTW This is a Level 4 problem

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

Hi;

[removed by administrator]

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,752

hi Agnishom,

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

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,752

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.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

Did you try programming it?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

Yes, exactly!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

There is one more.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,752

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

Why and How?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**phanthanhtom****Member**- Registered: 2012-06-22
- Posts: 290

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,752

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

Yes, if you are signed in as a member.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**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?

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,685

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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