Track / Stage 3 / 46 of 260 #46 of 2000

Problem 46

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Find the answer

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ ?

$\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

Your answer

Official solution

The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ :

If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most $12$ . Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$ . We can construct a working scenario where $x=8$ :

So, our answer is $\boxed{\textbf{(D) } 8}$ .
~ihatemath123

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.