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Praetorian
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[#34] 2 Min. Challenge: Number 26 Mar 2004, 17:46
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132.

1. Time yourself
2. Solve as fast as you can
3. Write your solution , and your time please.

What are the total number of positive integer solutions of 4x + 3y = 63.

a. 4
b. 5
c. 6
d. 7
e. 8



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kpadma
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[#34] 2 Min. Challenge: Number 26 Mar 2004, 18:32
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My Ans: B) 5

Time: Took more than the time required to complete the quant section.
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Paul
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[#34] 2 Min. Challenge: Number Updated on: 27 Mar 2004, 06:57
Originally posted by Paul on 26 Mar 2004, 19:16.
Last edited by Paul on 27 Mar 2004, 06:57, edited 1 time in total.
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Wow, this one looks simple at first but it took me 8 min or so. I just layed down the different solutions. My take is C) 6
Corresponding X-Y values are:
X: 0 - 3 - 6 - 9 - 12 - 15
Y: 31-17-13-11 - 5 - 3
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Paul
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[#34] 2 Min. Challenge: Number 26 Mar 2004, 21:45
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B) 5
In my previous answer, 0 would have to be removed
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shubhangi
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 00:02
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i took nearly 10 mins.. wonder :roll: if there must be some shortcut..
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hallelujah1234
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 00:25
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solve 4x+3y = 1 for integers

3y = 1 - 4x = (1-x)-3x
y = [(1-x)/3] - x

x = 1-3k, y = 4k-1

for 4x+3y = 63, x = 63-3k, y = 4k-63

x, y > 0 63-3k> 0, k < 21
4k-63 > 0, k >= 16

16 <= k < 21

answer = |{16, 17, 18, 19, 20}| = 5
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 01:31
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Ans) 5...

Took me 2.5 minutes

4x + 3y = 63
4x = 63-3y
x = 3(21-y)/4

Now All we have to do to get X as an integer is to get 21 -y as a multiple of 4. So either by Just picking the values for y that yield multiples of 4 we get integer multiples of X. So basically the values for Y would be 17, 13, 9, 5 and 1. Totally 5.
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 07:06
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from equation: y = 21-4x/3
therefore for positive values of Y, X can be multiple of 3 which are, 3, 6, 9,12,15..........Hence 5 Solutions!!

took me around 2 minutes..........!!
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 07:55
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cbrf3
from equation: y = 21-4x/3
therefore for positive values of Y, X can be multiple of 3 which are, 3, 6, 9,12,15..........Hence 5 Solutions!!

took me around 2 minutes..........!!
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5 solutions is correct.

though everyone got this, Cbfr3 did it the fastest and thus wins. The solution i have is exactly like cbrf3.

These problems look easy. but the important thing is to get the easy problems out of the way as quickly as you can. The challenge series serves this dual purpose of testing your time and your accuracy.

good luck
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[#34] 2 Min. Challenge: Number 27 Mar 2004, 08:08
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cbrf3
from equation: y = 21-4x/3
therefore for positive values of Y, X can be multiple of 3 which are, 3, 6, 9,12,15..........Hence 5 Solutions!!

took me around 2 minutes..........!!
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Nice one cbrf3! :P



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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