Try Find \( k \) if the number \( 25348937 k 4 \) is divisible by 4. it! \( \begin{array}{llllll}\text { (4) } 3 & \text { B } 5 & \text { C } 6 & \text { D } 7\end{array} \)
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To determine the value of \( k \) that makes the number \( 25348937 k 4 \) divisible by 4, we only need to focus on the last two digits of the number. A number is divisible by 4 if the number formed by its last two digits is divisible by 4. In this case, the last two digits are \( k4 \). Now we can examine the possible values of \( k \) from the provided options \( (4), 3, 5, 6, 7 \): - If \( k = 0 \), \( 04 \div 4 = 1 \) (divisible) - If \( k = 1 \), \( 14 \div 4 = 3.5 \) (not divisible) - If \( k = 2 \), \( 24 \div 4 = 6 \) (divisible) - If \( k = 3 \), \( 34 \div 4 = 8.5 \) (not divisible) - If \( k = 4 \), \( 44 \div 4 = 11 \) (divisible) - If \( k = 5 \), \( 54 \div 4 = 13.5 \) (not divisible) - If \( k = 6 \), \( 64 \div 4 = 16 \) (divisible) - If \( k = 7 \), \( 74 \div 4 = 18.5 \) (not divisible) - If \( k = 8 \), \( 84 \div 4 = 21 \) (divisible) - If \( k = 9 \), \( 94 \div 4 = 23.5 \) (not divisible) From the options (4), 3, 5, 6, 7, the viable candidates that satisfy our divisibility rule are \( k = 4 \) and \( k = 6 \). So the complete solution is \( k \) can be \( 4 \) or \( 6 \).
