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Divide 136 into two parts, one of which when divided by 5 [#permalink]
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26 Aug 2004, 18:50
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Divide 136 into two parts, one of which when divided by 5 leaves a remainder of 2 and the other when divided by 8 leaves a remainder of 3.



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Joined: 19 May 2004
Posts: 291

I got 77 and 59.
If i'm right, i will explain.
Difficult one!



Manager
Joined: 02 Apr 2004
Posts: 222
Location: Utrecht

I got 57 and 59 too.
I got this answer by writing down all possible values.
Pretty time consuming, so if there is anyone who has an formula for this kind of question, please let us know.
Regards,
Alex



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Joined: 07 Jul 2004
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Location: Singapore

I'm trying to get a formula too !



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Joined: 19 May 2004
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well, here is what i did.
I hope you guys can improve this.
One number is X, and the other is 136X.
1) X=5K+2
2) 136X=8J+3
1,2) 1365k2 = 8j+3
Therefore,
131 = 5k + 8j.
Here i got stuck, so i picked numbers and got the correct k and j.
The only thing left was to calculate X accordingly.
If you have any idea how to escape picking numbers i would love to hear it



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Joined: 07 Jul 2004
Posts: 5043
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that's exactly what I did. I didn't enjoy picking numbers. on an actual gmat, i think if we needed to do that, it's too time consuming. but on the other hand, in a actual gmat, you would have 5 choices to pick from, so maybe picking numbers won't take that long afterall.



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Joined: 15 Jul 2004
Posts: 75
Location: London

Dookie wrote: well, here is what i did. I hope you guys can improve this. One number is X, and the other is 136X. 1) X=5K+2 2) 136X=8J+3 1,2) 1365k2 = 8j+3 Therefore, 131 = 5k + 8j. Here i got stuck, so i picked numbers and got the correct k and j. The only thing left was to calculate X accordingly. If you have any idea how to escape picking numbers i would love to hear it
After obtaining 131 = 5k + 8j, I wrote down two multiplication tables:
x5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65...
and
x8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136
From this it takes a few seconds to realise that 96 + 35 = 131, or 75 + 56 = 131 (picking numbers from the x8 multiplication table finishing in either 1 or 6)
Therefore number sets 99 + 37 and 59 + 77 are valid results.



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An alternative solution [#permalink]
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30 Aug 2004, 12:10
Hi Everyone, I figured out a different way of solving this that gave me all the possible solutions to this problem Pls let me knows your comments about my solution.
Since we need a number that divided by 5 has a reminder of 2
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,
115,120,125,130,135.
For a number in this table we are going to add 2 in order to get it as a reminder.
{[n(5)]+[n+1(5)]+[n+...(5)]}+2
The same for the table of 8
8,16,24,32,40,48,56,64,72,88,96,104,112,128,136
{[n(8)]+[n+1(8)]+[n+...(8)]}+3
If we add the two figures we get 2+3 = 5
So now we just need 131 instead of 136
Of all the possibles aoutcomes from the table of 8 added to the table of 5 that could help me to get this results,are the ones bolded (we just take the one that finish with 6
8, 16,24,32,40,48, 56,64,72,88, 96,104,112,128
now we just need to find the complement to these numbers to get 131 from the table of 5
for 16 +115 = 131 for 56+75 = 131 for 96+35 = 131
Finally if we add the numbers that we took it before to each one of these
(115+2)+(16+3) = 136
(75+2) + (56+3) = 136
(35+2) + (96+3) = 136
To chek this every single divide the figures 37, 77 and 117 by 5
and 19,59,99 by 8.
I give you an apologize for my poor english, but I hope this way could help to find a different solution.
I would like to hear your comments since this is my first post
Thanks / ArturoC



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Re: An alternative solution [#permalink]
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30 Aug 2004, 14:15
ArturoC wrote: Hi Everyone, I figured out a different way of solving this that gave me all the possible solutions to this problem Pls let me knows your comments about my solution. Since we need a number that divided by 5 has a reminder of 2 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110, 115,120,125,130,135. For a number in this table we are going to add 2 in order to get it as a reminder. {[n(5)]+[n+1(5)]+[n+...(5)]}+2 The same for the table of 8 8,16,24,32,40,48,56,64,72,88,96,104,112,128,136 {[n(8)]+[n+1(8)]+[n+...(8)]}+3 If we add the two figures we get 2+3 = 5So now we just need 131 instead of 136 Of all the possibles aoutcomes from the table of 8 added to the table of 5 that could help me to get this results,are the ones bolded (we just take the one that finish with 6 8, 16,24,32,40,48, 56,64,72,88, 96,104,112,128 now we just need to find the complement to these numbers to get 131 from the table of 5 for 16 +115 = 131 for 56+75 = 131 for 96+35 = 131 Finally if we add the numbers that we took it before to each one of these (115+2)+(16+3) = 136 (75+2) + (56+3) = 136 (35+2) + (96+3) = 136 To chek this every single divide the figures 37, 77 and 117 by 5 and 19,59,99 by 8. I give you an apologize for my poor english, but I hope this way could help to find a different solution. I would like to hear your comments since this is my first post Thanks / ArturoC
I have not seen anyone making so much of an effort in first post. Welcome ArturoC and don't be a guest next time.



Intern
Joined: 03 Aug 2004
Posts: 26

1. the first number must be X, second number 136  X...
2. if the second number must have a remainder 3 after dividing by 8, then it is very easy to write down a list of those numbers:
11,19,27,35,43,51,59,67,75,83,91,99,107,115,123,131
3. since the other number must have a remainder 2 after dividing by 5, then the number must end in 2 or 7
4. take 136 and subtract each number in item #2...but you only have to find the ones digit (because the second number must end in either 2 or 7), so you can go through the list very quickly to eliminate the ones that don't leave a 2 or 7 in the ones digit after subtracting
5. that leaves three sets of numbers (117,19) (77,59) (37,99) which all fit the problem










