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Guest Message by DevFuse
 

OMGTHATSIT

Member Since 06 Jun 2012
Offline Last Active Jun 08 2012 01:40 AM
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Posts I've Made

In Topic: Help! A remainder is chasing me

06 June 2012 - 12:24 AM

if you check out each number out each number's posiblities and put it into a formula, you will obtain:

for the number 2: (x is any natural number, and it's diferent for every equation)

1+ 2x

Number 3

29+30x

Number 4

19+ 20x

Number 5

9+ 10x

Number 6

29 + 30x

Number 7

69 + 70x

Number 8

39 + 40x

Number 9

89+90x

Number 10

9 + 10X

So we need a number that is equivalent to all those equations. we have:

1+2X

29+30X

19+20X

9+10X

69+70X

39+40x

89+90X

So, we go and analyse the equations who give bigger products for the smallest number of X. those are: 89+90x and 69 + 70x. Replacing one of the x for another letter (y) we get:

89+90y = 69 + 70x <=> 20 = 70x - 90y

With that, we can now replace the "x" and the "y". for that formula to work, after some calculations, I notice that the x must go for 8 + 9z and the y must go 6+7z

So, starting for the 1st, if z = 0, x=8 and y=6. we replace them in the equation

89 + 90 x 6 = 629.

However, this number doesnt exist on the equation 39+40x, so its invalid.

You keep going,

z=1; x= 17 and y = 13, the final =1259, also doesnt exist on the equation 39+40x, so its invalid.

z=2, the final is 1889, also invalid.

Then, finally, z=43 y = 6+(7x3) =27........ 89 + 90x27 = 2519; which is valid for all the equations.