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#1 2006-10-03 08:02:06

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Algebra....

Please help me...

Amy has fewer than 100 computer disks. When she stacks them by elevens, ten are left over. When she stacks them by tens, seven are left over, and three are left over when she stacks them by sixes. How many disk does she have?

Advance Thanks up


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#2 2006-10-03 08:25:11

Devantè
Real Member
Registered: 2006-07-14
Posts: 6,400

Re: Algebra....

Amy has 87 disks.

Stacking them by 11s = 77 with 10 left over
Stacking them by 10s = 80 with 7 left over
Stacking them by 6s = 84 with 3 left over

But how is the question asking you to express the answer? You've just written 'algebra'.

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#3 2006-10-03 08:37:20

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: Algebra....

Can you give me a method and steps?

Last edited by Prakash Panneer (2006-10-03 08:38:09)


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#4 2006-10-03 09:25:23

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Algebra....

Here is my procedure for solving it.. which is probably not too good but it is easy to follow:

let N = number of comp disks.

N < 100

By the descriptions:

11A + 10 = N   .. eqn 1

10B + 7 = N     .. eqn 2

6C + 3 = N       .. eqn 3

tweak these starting with maximums for A, B, C:

A = 8  -> N = 98
B = 9  -> N = 97
C = 16 -> N = 99

They are not equal, try again.
There is no way to make eqn 2 produce 98, so lower A to 7.

A = 7 -> N = 77 + 10 = 87

Try to make eqn 2 produce 87:

B = 8 -> N = 80 + 7 = 87

Yay it worked! Now try to produce 87 with eqn 3

C = 15 -> N = 93 ,
C = 14 -> N = 87 ; works!

So it seems 87 is the highest number we can get out of all 3 equations at once.

But perhaps there are other combinations that work?

I don't really like this question, it seems to be all about trial and error.


I guess Devanté got it by just reasoning it out.. impressive! smile

Last edited by polylog (2006-10-03 09:26:10)

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#5 2006-10-03 09:57:32

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Algebra....

No other combinations less than 100 work. Dividing by 10 leaves 7, so your number has to end in a 7.

Also, the number is ten more than a multiple of 11, so the first digit must be one more than the second. Therefore, 87 is the only solution.


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#6 2006-10-03 10:33:30

All_Is_Number
Member
Registered: 2006-07-10
Posts: 258

Re: Algebra....

mathsyperson wrote:

No other combinations less than 100 work. Dividing by 10 leaves 7, so your number has to end in a 7.

Also, the number is ten more than a multiple of 11, so the first digit must be one more than the second. Therefore, 87 is the only solution.

Impressive logic.


You can shear a sheep many times but skin him only once.

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#7 2006-10-03 10:34:17

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Algebra....

Indeed, a wonderful solution. smile

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#8 2006-10-03 12:58:10

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Algebra....

Actually, there is an easier solution.  We know that n < 100.  We also know that:

n = 10 mod 11.

All this means is that when we divide n by 11, we are left with a remainder of 10.

This means that the possible value of n is:

10, 21, 32, 43, 54, 65, 76, 87, and 98.

Simply start with 10 and keep adding 11.

Now we can use the same logic with 7:

n = 7 mod 10.

Thus, n can be 7, 17, 27, .... 97.  But the only one with a 7 in the one's place above is 87.  Thus, that is the only possible solution, if there is one.

Finally, we try n = 3 mod 6.

87 / 6 = 14.5

14 * 6 = 84

87-84 = 3

Thus, 87 = 3 mod 6 as required.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2006-10-04 00:39:04

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: Algebra....

Thanks for all up

Is there any other method apart from trial and error method?

Please........


Letter, number, arts and science
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#10 2006-10-04 02:30:42

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Algebra....

I wouldn't say that my or Ricky's methods were trial and error.

Ricky made exhaustive lists of all the numbers that worked for each individual rule and then cross-referenced them to get a solution.

I kind of did the same, but with less list-making and more reasoning.


Why did the vector cross the road?
It wanted to be normal.

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#11 2006-10-04 02:53:50

polylog
Member
Registered: 2006-09-28
Posts: 162

Re: Algebra....

Yes I agree, mathsyperson's and Ricky's excellent solutions were not trial and error!

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#12 2006-10-04 03:38:15

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Algebra....

There is a way to do it without lists, but it involves special properties of mod and would be too long winded for me to explain.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#13 2006-10-04 05:56:31

Prakash Panneer
Member
Registered: 2006-06-01
Posts: 110

Re: Algebra....

Thanks for all.

Finally, I have found the solution in anotyher method.

Here is the solution.

Problem:

Amy has fewer than 100 computer disks. When she stacks them by elevens, ten are left over. When she stacks them by tens, seven are left over, and three are left over when she stacks them by sixes. How many disk does she have?

Solution: Using Chinese Remainder Theorem:

When she stacks them by elevens, ten are left over.

It can be written as, x = 10 (mod 11) => x = 11t +10

When she stacks them by tens, seven are left over.

x = 7 (mod 10) => x = 10 t + 7

when she stacks them by sixes, three are left over.

x = 3 (mod 6)=> x = 6 t + 3.

x = 11t + 10.---------(A)

x = 10t + 7.

x = 6t + 3.

Substitute x = 11t+10 in the second statement.

We get,

11t+10 = 7 (mod10)

Casting out 10, we get,

t + 0 = 7 (mod 10)

It can be written as,

t = 10s+7.

Substitute this in (A), we get,

x = 11(10s+7)+10.

x = 110s + 77 + 10

x = 110s +87

From the third statement,

x = 3 (mod 6)

Substitute x = 110s+87 in the above,

110s + 87 = 3 (mod 6)

Casting out 3 we get,

2s + 3 = 3 (mod6)

2s = 0 (mod 6)

It can be written as,

2s = 6t.

Therefore, s = 3t

We found x = 110s +87

x = 330s + 87.

Therefore,the solutions 87, 417, 747,. . .


Thanks for all.up


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