AOPS

1. (1)2、3、5、7、11、13、17、19、23、29、31、37、
41、43、47、53、59、61、67、71、73、79、83、89、97

(2) $10=3+7$, $26=7+19$, $30=7+23$.

2.C

Solution 1

Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $oxed{ extbf{(C)} 3}$.

Solution 2

Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $frac{a^2 + ab + b^2}{a^2 – 2ab + b^2} = frac{73}{3}$.

Set $x = a^2 + b^2$, and $y = ab$. Then $frac{x + y}{x – 2y} = frac{73}{3}$. Cross multiplying gives $3x + 3y = 73x – 146y$, and simplifying gives $frac{x}{y} = frac{149}{70}$. Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$, giving $a^2 + b^2 = 149$ and $ab = 70$. Since $a>b>0$, the only solution is $(a,b) = (10, 7)$, which can be seen upon squaring and summing the various factor pairs of $70$.

Thus, $a – b = oxed{ extbf{(C)} 3}$.

Remarks:

An alternate method of solving the system of equations involves solving the second equation for $a$, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of $u = b^2$. The four solutions correspond to $(pm10, pm7), (pm7, pm10).$

Also, we can solve for $a-b$ directly instead of solving for $a$ and $b$: $a^2-2ab+b^2=149-2(70)=9 implies a-b=3.$

Note that if you double $x$ and double $y$, you will get different (but not relatively prime) values for $a$ and $b$ that satisfy the original equation.

3. A

Solution 1

The sum of a number’s digits $mod{3}$ is congruent to the number $pmod{3}$. $74A52B1 mod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 mod{3}$ is also congruent to 0. $7+4+5+2+1 equiv 1 pmod{3}$, so $A+Bequiv 2 pmod{3}$. As we know, $326AB4Cequiv 0 pmod{3}$, so $3+2+6+A+B+4+C =15+A+B+Cequiv 0 pmod{3}$, and therefore $A+B+Cequiv 0 pmod{3}$. We can substitute 2 for $A+B$, so $2+Cequiv 0 pmod{3}$, and therefore $Cequiv 1pmod{3}$. This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $oxed{ extbf{(A) }1}$.

Solution 2

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1, so $oxed{ extbf{(A) }1}$ is your answer.

4. D