# How many numbers end in the four digits 1995 and become an integer number of times smaller when these digits are erased?

How many numbers end in the four digits 1995 and become an integer number of times smaller when these digits are erased?

I do not understand the question through but I think this is asking for all numbers of the form: 19951995 -> 1995

But since there is no restriction on the length so could we have a finite solution for this problem? If yes, how?

Thank you,

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If this is a homework problem, please remember to add the homework tag. And you should understand your question thoroughly and phrase it carefully when posting it here. –  Dimitrije Kostic Dec 18 '11 at 18:53
Note that $\frac{199519951995}{19951995}$ is not an integer. –  Ross Millikan Jan 17 '12 at 20:21

If $n$ is the number remaining after removing the last four digits, then we are given that $$\frac{10000n + 1995}{n}$$ is an integer. But this is equal to $$10000 + \frac{1995}{n}$$ So the answer is simply the number of divisors of $1995$.

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Let $n$ be your number. Then $10000$ divides $n-1995$ and $\frac{n-1995}{10000}$ is a divisor of $n$.

Thus

$$n= k \times \frac{n-1995}{10000} \,.$$

$$10000n = kn-1995k \,.$$

or

$$1995k =n (k -10000) \,.$$

Let $d =$ gcd $(k,n)$. Then $n=dn_1$ and $k=dk_1$. The equation becomes

$$1995k_1=n_1(k_1d-10000)$$

Since $n_1, k_1$ are relatively prime, $n_1$ is a divisor of $1995$, and a case by case analysis should solve the problem...

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