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I am having difficulties with a problem from my exercise sheet for a University exam.

The problem statement is as following:

When dividing a number by 12, 15 or 48 there will always be a remainder of 10. If the number is the least posible, how many divisors does the number have?

How would you solve that problem?

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If by least possible you mean least positive, the number is obviously $10$, which has $4$ positive divisors. Note that in general the numbers in question are the numbers that have remainder $10$ on division by $240$. – André Nicolas Dec 21 '11 at 19:21

For the specific problem, I would note that $10$ divided by any of these leaves a remainder of $10$. The next value would be $10+LCM(12,15,48)=250$. Then you can factor your favorite and use the fact that the number of divisors of $p^aq^b$ (p,q primes) is $(a+1)(b+1)$-have you seen that?

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Everything after "factor your favorite" I don't understand... could you please be more specific? – Kenny Meyer Dec 21 '11 at 19:37
@KennyM.: If you want a factor of $p^aq^b$, you can have anywhere from $0$ to $a$ factors of $p$, $(a+1)$ choices, and from $0$ to $b$ factors of $q$, $(b+1)$ choices. So the total number of choices is $(a+1)(b+1)$. You could try it by hand for $2^43^2=144$-you should find $5*3=15$. You might see – Ross Millikan Dec 21 '11 at 19:49

for this problem $250$ is the least as you can see $250 \; \% \;12=10$, $250 \; \% \; 15=10$, and $250 \; \% \; 48=10$. You can find the factors by expressing it as $p^a \cdot q^b\dots$ and the number of factors is $(a+1)(b+1)\dots$.

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Sorry, I still don't really understand why it is 250, if 250%48=15, and the problem statement says a remainder of 10. – Kenny Meyer Dec 21 '11 at 19:42
if the number can be represented as powers of prime factors (p^a*q^b) then number of factors is (a+1)*(b+1) is number of divisors see this for finding factors i corrected my mistake – harish.venkat Dec 21 '11 at 19:47
But $250\%48=10$ because $250-5*48=10$ – Ross Millikan Dec 21 '11 at 22:05

HINT $\rm\ 10\ $ is the least solution $> 0\: $ since any smaller one is $\equiv 10\pmod{48}$ so is $\:\le 10-48 < 0\:.$

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