Sample Question 21:

Four positive integers $a$, $b$, $c$, and $d$ have a product of $8!$ and satisfy:

What is $a-d$?

$\text{(A) }4 \qquad \text{(B) }6 \qquad \text{(C) }8 \qquad \text{(D) }10 \qquad \text{(E) }12$

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Answer Keys

Question 21: D

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Solutions

Question 21
Step 1: Apply Simon's Favorite Factoring Trick to rewrite the three given equations in the following form:

$(a+1)(b+1) = 525, \quad (b+1)(c+1) = 147, \quad (c+1)(d+1) = 105$

Step 2: Define new variables $e=a+1, f=b+1, g=c+1, h=d+1$, so we can express the above equations in terms of products of prime numbers as:

$ef = 3 \cdot 5 \cdot 5 \cdot 7, \quad fg = 3 \cdot 7 \cdot 7, \quad gh = 3 \cdot 5 \cdot 7$

Step 3: Notice that $7^2$ divides $fg$, but $7^2$ cannot divide either $f$ or $g$ as it would then also divide $ef$ or $gh$. Therefore, $7$ divides both $f$ and $g$.

Step 4: We have two possible combinations for $f$ and $g$: $(f,g) = (7,21)$ or $(f,g) = (21,7)$.

Step 5: For the first case $(f,g) = (7,21)$, we get $e=75, h=5$, and therefore, the original numbers are $(a,b,c,d) = (74,6,20,4)$. However, $a \cdot b \cdot c \cdot d$ is not equal to $8!$ and $a-d$ is not in the answer choices. Hence, this is not a valid solution.

Step 6: For the second case $(f,g) = (21,7)$, we get $e=25, h=15$, and therefore the original numbers are $(a,b,c,d) = (24,20,6,14)$.

Step 7: Thus, $a - d = 24 - 14 = 10$. Therefore, the answer to the problem is $(D)$ 10.