Question

Consider a seven-digit number $735x6y4$, divisible by $44$, where the two digits $x$ and $y$ are unknown. Consider the following two additional pieces of information: I. $x$ and $y$ are even numbers II. $x$ and $y$ are equal To determine the values of $x$ and $y$ UNIQUELY, which of the above pieces of information is/are MINIMALLY SUFFICIENT?

Accepted Answer

C. For $735x6y4$ to be divisible by $44$, it must be divisible by $4$ and $11$. Divisibility by $4$: $y4$ must be divisible by $4 \implies y \in \{0, 2, 4, 6, 8\}$. Divisibility by $11$: $(7 + 5 + 6 + 4) - (3 + x + y) = 22 - (3 + x + y) = 19 - (x + y)$ must be divisible by $11$. Since $x, y \in [0, 9]$, $x+y = 8$ or $x+y = 19$ (impossible as max is 18). So $x + y = 8$. Condition I ($x, y$ are even): Pairs for $(x, y)$ can be $(8,0), (6,2), (4,4), (2,6), (0,8)$. Multiple solutions. Condition II ($x = y$): $x + x = 8 \implies x = 4, y = 4$. Unique solution! Therefore, Condition II alone is minimally sufficient.

Divisibility of seven-digit number

Explanation

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