Question

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Explanation:

Step1: Analyze the dividend's tens place

The number is 114, so the tens digit is 1 (wait, no, 114 has 1 ten? Wait, no, 114 is 1 hundred, 1 ten, and 4 ones? Wait, no, 114 is 1100 + 110 + 4*1? Wait, no, when dividing 114 by 2, we start with the hundreds and tens? Wait, the problem says "Divide the tens by 2". Wait, maybe the number is 11 tens? Wait, no, the division is \(

$$\begin{array}{r} 2\enclose{longdiv}{114} \end{array}$$

\), and we are looking at the tens place. The first digit of the dividend is 1 (hundreds place), then 1 (tens place). Wait, maybe it's 11 tens? Wait, 114: the tens digit is 1, but maybe we consider the first two digits, 11? Wait, 11 tens? Wait, no, 114 is 11 tens and 4 ones? Wait, 11 tens is 110, plus 4 is 114. So we have 11 tens, and we divide by 2. How many tens in each group? 11 divided by 2 is 5 with a remainder of 1. Wait, but the options are 3,4,5,6. Wait, 11 divided by 2: 25=10, remainder 1. So 5 tens can go into each group, with 1 ten left over? Wait, but the options for the first box are 3,4,5,6. Let's check: 25=10, which is less than 11. 2*6=12, which is more than 11. So the maximum number of tens per group is 5, with 1 ten left over. Wait, but the problem says "Divide the tens by 2". Wait, maybe the number of tens is 11? Wait, 11 divided by 2: quotient 5, remainder 1. So 5 tens can go in each group, with 1 ten left over. Wait, but the options for the first box are 3,4,5,6. So 5 is an option. So:

Step1: Divide 11 (tens) by 2

We have 11 tens. Dividing 11 by 2: \( 11 \div 2 = 5 \) with a remainder of \( 11 - 2 \times 5 = 1 \). So 5 tens can go into each group, and 1 ten is left over.

Answer:

5 tens can go in each group, with 1 ten left over. (For the first box, select 5 tens; for the remainder, 1 ten)