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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

2 Answers 2

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

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

This leads to

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

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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