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(1) X = k*(m^3 - m), where m and k are both integers > 9 (2) X = n^5 - n, where n is an integer > 9

B
factors of 30 = 2*3*5

1
k*(m^3 - m) = k*m*(m^2-1) = k*m*(m+1)*(m-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5
for eg,
m=21, k= 13 do not allow for divisibility by 5
m=23, k= 24 allows for divisibility by 5

insuffi

2
x = n^5 - n = n*(n^4-1) = n*(n^2+1)*(n+1)*(n-1)
any 3 consecutive integers > 9 will allow for divisibility by 2 and 3
but they may or maynot allow for for divisibility by 5,

the new expression here is (n^2+1)
for n= 11 , (n^2+1) = 122 is not divisible by 5, but 10=n-1 is
for n= 12 , (n^2+1) =145 is divisible by 5

suffi

Last edited by muddypaws on 27 Sep 2006, 18:29, edited 1 time in total.

(1) X = k*(m^3 - m), where m and k are both integers > 9 (2) X = n^5 - n, where n is an integer > 9

statement 1: X= K*(m^3-m)

i.e X= K*(m-1)(m)(m+1).

Product of any n consecutive positive numbers is always divisible by n!.

So clearly (m-1)(m)(m+1) is divisible by 3! =6 Hence X is divisible by 6. We are not sure whether x is divisible by 5 or not. So 1 is not sufficient.

Statement 2: X= n^5-n This will always end with 0. So X is clearly divisible by 5. X=n^5-n = n(n^4-1) = n[((n^2)^2) - 1^2] ie X= n(n^2+1)(n^2-1) X= n(n^2+1)(n+1)(n-1) X= (n-1)(n)(n+1)(n^2+1) Clearly X is divisible by 6.

N^5-n=n(n^2+1)(n^2-1)=n(n^2+1)(n+1)(n-1)
We already know n(n+1)(n-1) is divisible by 6, just need to know if the whole thing is divisible by 5.

if n=5k then it is divisible by 5.
if n=5k+1, n^2-1=25k^2+10k, divisible by 5.
if n=5k-1, n^2-1=25k^2-10k, divisible by 5.
if n=5k+2, n^2+1=25k^2+20k+5, divisible by 5.
if n=5k-2, n^2+1=25k^2-20k+5, divisible by 5.

I went the long way to find out that it is also divisible by 5. Cicerone, can you explain why n^5-n is always multiples of 10? Did you have to go through this long derivation or if there's a more straight forward way that we can see the result immediately?
_________________

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