number theory problem?

Question

If 125 is appended to the right end of the number

5^(n-1) - 1
------------------ , the number 5^n is obtained . which of the
8
following is true of n ?

(a)n=4k , k belongs to N

(b)n=4k+2 , k belongs to N

(c)n belongs to N and n >2

(d)n=2k+1 , k belongs to N

PLZ note -1 in the expression is NOT in the power ....dont do mistake by assuming that.

Answer 1

All these statements are true, because
there is no such n. (!!)
(Recall that a false hypothesis implies a
true conclusion.)
Let's show no n exists that satisfies
the conditions of the problem.
First,
We must have n > 2
because
(5^(n-1) -1)/8 is not a whole number if n = 0,1 or 2.
Next, n must be odd.
Let me use congruences:
5 = 5(mod 8)
5^2 = 1(mod 8).
So, 5^(n-1) -1 = 4 mod 8 if n is even
and is 0 mod 8 if n is odd.
So (5^(n-1) -1)/8 is not a whole number if n is even.
Now It always happens that if
n is odd and
we append 125 to the end of (5^(n-1) -1)/8
we always get 5^(n+2).
Why?
Well, appending 125 to the end of a number
is equivalent to multiplying that number by 1000
and adding 125.
So we get
1000((5^(n-1) -1)/8) + 125
= 125(5 ^(n-1) - 1) + 125 = 5^3(5^(n-1)) = 5 ^(n+2).
But the hypothesis of the problem now yields that
5 ^(n+2) = 5^n
which is clearly impossible.
Conclusion: All of the statements are true
because the original statement can never hold!!

Answer 2

Did you transcribe the problem accurately? If we take the problem as it is written here, we arrive at the contradiction 0 = 2. Also, where do come up with "k" all of a sudden?

Now...if k = 3 and n = 5, then (5^(k-1) - 1)/8 = 5^n. 24/8 = 3 and 3125 = 5^5.

This will also hold if we pick any integer k and let n = k + 2.

Answer 3

Are you saying that 125 are the last three digits of the number equalling 5^n? If so, n must be odd, so (d). Otherwise I don't understand what you mean by "appended."