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Sample Question 13:
Step 1: Let's call the two primes \(a\) and \(b\). The two given conditions are then expressed as \(a-b=2\) and \(a^3-b^3=31106\).
Step 2: Using the formula for the difference of cubes, \(a^3-b^3=(a-b)(a^2+ab+b^2)\), and inserting the two given conditions, we can simplify this to \(a^2+ab+b^2=15553\).
Step 3: We substitute \(a\) and \(b\) with \(x+2\) and \(x\), where \(x+2\) is \(a\) and \(x\) is \(b\). For these values, \(a^2+ab+b^2\) transforms to \((x+2)^2+x(x+2)+x^2\).
Step 4: If we expand this, we get \(x^2+4x+4+x^2+2x+x^2\), which simplifies to \(3x^2+6x+4\). This is equal to \(15553\), so \(3x^2+6x=15549\).
Step 5: Now, notice that all three numbers are divisible by 3. Dividing the entire equation by 3, we get \(x^2+2x=5183\).
Step 6: We can turn the left side into a perfect square trinomial if we add 1 to both sides: \(x^2+2x+1=5184\). This is equivalent to \((x+1)^2=5184\).
Step 7: The square root of 5184 is 72, so the two numbers are 71 and 73.
Step 8: Since the next prime number is 79 and the sum of the digits is \(7+9=16\), the answer that fits these conditions is \(\textbf{(E)}\ 16\).
The positive difference between a pair of primes is equal to \(2\), and the positive difference between the cubes of the two primes is \(31106\). What is the sum of the digits of the least prime that is greater than those two primes?
\(\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16\)
Answer Keys
Question 13: ESolutions
Question 13Step 1: Let's call the two primes \(a\) and \(b\). The two given conditions are then expressed as \(a-b=2\) and \(a^3-b^3=31106\).
Step 2: Using the formula for the difference of cubes, \(a^3-b^3=(a-b)(a^2+ab+b^2)\), and inserting the two given conditions, we can simplify this to \(a^2+ab+b^2=15553\).
Step 3: We substitute \(a\) and \(b\) with \(x+2\) and \(x\), where \(x+2\) is \(a\) and \(x\) is \(b\). For these values, \(a^2+ab+b^2\) transforms to \((x+2)^2+x(x+2)+x^2\).
Step 4: If we expand this, we get \(x^2+4x+4+x^2+2x+x^2\), which simplifies to \(3x^2+6x+4\). This is equal to \(15553\), so \(3x^2+6x=15549\).
Step 5: Now, notice that all three numbers are divisible by 3. Dividing the entire equation by 3, we get \(x^2+2x=5183\).
Step 6: We can turn the left side into a perfect square trinomial if we add 1 to both sides: \(x^2+2x+1=5184\). This is equivalent to \((x+1)^2=5184\).
Step 7: The square root of 5184 is 72, so the two numbers are 71 and 73.
Step 8: Since the next prime number is 79 and the sum of the digits is \(7+9=16\), the answer that fits these conditions is \(\textbf{(E)}\ 16\).