Number analysis I was preparing for SAT, and saw this question (without answers) and was stumped.
If $2015 = 101a + 10b$, what is $a + b$?
I broke $2015$ into primes, one of which is $5$, and if you break $10$ into primes one of them is $5$.  However, $101$ is prime. So, I'm not sure how to get to $a + b$ since I can't factor anything out.
So, I solved for $a$ to get $a = \frac{2015}{101} -\frac{10b}{101}$ then added $b$ to both sides to get $a + b = \frac{2015 +91b}{101}$.
That's as far as I could get. Would you think this is what they are looking for?
 A: It is impossible to determine $a+b$ uniquely from the information that you’ve given, even assuming that $a$ and $b$ are required to be positive integers: 
$$2015=101\cdot5+10\cdot151=101\cdot15+10\cdot50\;,$$ so even with that limitation we can have $a+b=156$ or $a+b=65$. 
If negative integers are allowed, there are infinitely many solutions. For example, $$101\cdot1+10\cdot(-10)=1\;,$$
so $101\cdot2015+10\cdot(-20150)=2015$, giving $a+b=-18,135$.
A: Just adding on to Brian's answer, there is no unique value that can be assigned to $a+b$. However, it is possible to prove that the smallest positive value that $a+b$ can take is $65$.
$gcd(101,10) = 1$ so the equation clearly has solutions.
One solution can be found by the Extended Euclidean algorithm. You can use this to show that:
$101(1) + 10(-10) = 1$
so multiplying throughout by $2015$, you get:
$101(2015) + 10(-20150) = 2015$.
So you've found one possible set of values for $(a,b)$, i.e. $a_1= 2015, b_1 = -20150$
To find the full set of solutions, you note that if $a_1,b_1$ are solutions, then $a_1+\frac{10}{gcd(101,10)}k=a_1+10k$ and $b_1-\frac{101}{gcd(101,10)}k=b_1-101k$ are also solutions. This is because $101(a_1 + 10k) + 10(b_1-101k) = 101a_1 + 10b_1 + 1010k - 1010 k = 101a_1 + 10b_1$.
So the full solution set is $a = 2015 + 10k, b = -20150-101k$.
To simplify that a little, put $k = t-200$ to give $a = 10t + 15, b = 50-101t$.
So $a+b = 65-91t$, where $t$ is any integer. Clearly, the smallest positive value that $a+b$ can take is $65$.
