Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,

share|cite|improve this question
    
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 $1995$.

share|cite|improve this answer

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...

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.