Solving for a: A Constant Value Equation

[jalaldn]

answer = "a" is constant
a / 9 = x + 8
a / 8 = x + 7
a / 7 = x + 6
a / 6 = x + 5
a / 5 = x + 4
a / 4 = x + 3
a / 3 = x + 2
a / 2 = x + 1


a = ?

 

[Data]

Why do you think there's a solution? There isn't one.

Pick any two of the equations and solve them simultaneously for a, x. You will have a unique solution. Plug that solution into any other equation and see if it works (it won't).

 

[neurocomp2003]

is this a modulus question?

 

[jalaldn]

i got it by microsoft excel if you can't fint it answer is 254+254+1245+766

 

[eljose]

I got a=-n(n+1) perhaps he missed something...

\frac{a}{n}=x+(n-1)

 

[jalaldn]

yes that is wrong
answer = "a" is constant
a / 9 = x9 + 8
a / 8 = x8 + 7
a / 7 = x7 + 6
a / 6 = x6 + 5
a / 5 = x5 + 4
a / 4 = x4 + 3
a / 3 = x3 + 2
a / 2 = x2 + 1

a = ? a = 2519
i don't know how to explain

 

[3trQN]

2519 strikes a bell, it is close to 1/2 7!, which should help you find the answer.

 

[HallsofIvy]

yes that is wrong
answer = "a" is constant
a / 9 = x9 + 8
a / 8 = x8 + 7
a / 7 = x7 + 6
a / 6 = x6 + 5
a / 5 = x5 + 4
a / 4 = x4 + 3
a / 3 = x3 + 2
a / 2 = x2 + 1

a = ? a = 2519
i don't know how to explain

You've still written that completely wrong. You must mean:
a= 9x₁+ 8
a= 8x₂+ 7
a= 7x₃ + 6
a= 6x₄ + 5
a= 5x₅ + 4
a= 4x₆ + 3
a= 3x₇ + 2
a= 2x₈ + 1
where the "x"s are all (possibly different) integers.

That's a "Chinese remainder theorem" problem and can also be stated as
a= 8 (mod 9)
a= 7 (mod 8)
a= 6 (mod 7)
a= 5 (mod 6)
a= 4 (mod 5)
a= 3 (mod 4)
a= 2 (mod 3)
a= 1 (mod 2)

The smallest number divisible by 2, 3, 4, 5, 6, 7, 8, 9 is 9*8*7*5= 2520 (using the highest power of each prime).
2519 is one less than that: 2519= 2520-1 so 2519= 9(280)- 1= 9(279)+ 9- 1= 9(279)+ 8. 2519= 8(315)- 1= 8(314)+ 8- 1= 8(314)+ 7, etc.

 

[eljose]

Ah.."Hallsoftivy" but the problem with "mod" isn't the same posted above..i will try this last one that seems to be clearer...good luck.