1
$\begingroup$

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?

  • 0
    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 $240$.2011-12-21

3 Answers 3

2

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?

  • 1
    @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 http://mathforum.org/library/drmath/view/57151.html2011-12-21
1

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 $(a+1)(b+1)\dots$.

  • 0
    But $250\%48=10$ because $250-5*48=10$2011-12-21
0

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