Question

there are no more digits in the dividend to bring down. how can you continue dividing? write a 0 in the hundredths place of the dividend. how many times does 2 go into 10? times, with a remainder of

Explanation:

Step1: Determine how many times 2 fits into 10

We know that \(2\times5 = 10\), so we need to find the number of times 2 goes into 10.

Step2: Calculate the remainder

We use the division formula: \(Dividend = Divisor\times Quotient+Remainder\). Here, dividend is 10, divisor is 2, and we found quotient is 5. So \(10=2\times5 + Remainder\), which means \(Remainder = 10 - 2\times5=0\). Wait, but maybe I misread. Wait, the problem is "How many times does 2 go into 10?". Wait, 2 times 5 is 10, so 5 times, with remainder 0. Wait, maybe the initial thought was wrong. Wait, let's re - calculate. \(2\times5 = 10\), so when we divide 10 by 2, the quotient is 5 and the remainder is 0.

Wait, maybe there was a typo in my first step. Let's do it properly. The division of 10 by 2: we know that multiplication table, 2*5 = 10. So the number of times 2 goes into 10 is 5, and the remainder is \(10-(2\times5)=0\).

Wait, maybe the original problem had a different context, but based on the question "How many times does 2 go into 10?".

Step1: Find the quotient

We want to find \(n\) such that \(2\times n\leq10\) and \(n\) is the largest integer possible. We know that \(2\times5 = 10\), so \(n = 5\).

Step2: Find the remainder

Using the formula \(Remainder=Dividend - Divisor\times Quotient\), substituting dividend = 10, divisor = 2, quotient = 5. We get \(Remainder = 10-2\times5=0\).

Answer:

5 times, with a remainder of 0