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Sagot :
Certainly! Let's perform the long division of [tex]\( 6 \)[/tex] by [tex]\( 27 \)[/tex].
1. Setup: We start by setting up 6 as the dividend and 27 as the divisor.
2. Determine the quotient: We observe how many times 27 can fit into 6.
- 27 is larger than 6, so it cannot fit even once.
- Therefore, the quotient is [tex]\( 0 \)[/tex].
3. Calculate the remainder: Since 27 cannot fit into 6, we have used none of the 6, leaving the entire 6 still present.
- This means the remainder is [tex]\( 6 \)[/tex].
Thus, when [tex]\( 6 \)[/tex] is divided by [tex]\( 27 \)[/tex], the quotient is [tex]\( 0 \)[/tex] and the remainder is [tex]\( 6 \)[/tex].
So, the final answer is:
- Quotient: [tex]\( 0 \)[/tex]
- Remainder: [tex]\( 6 \)[/tex]
1. Setup: We start by setting up 6 as the dividend and 27 as the divisor.
2. Determine the quotient: We observe how many times 27 can fit into 6.
- 27 is larger than 6, so it cannot fit even once.
- Therefore, the quotient is [tex]\( 0 \)[/tex].
3. Calculate the remainder: Since 27 cannot fit into 6, we have used none of the 6, leaving the entire 6 still present.
- This means the remainder is [tex]\( 6 \)[/tex].
Thus, when [tex]\( 6 \)[/tex] is divided by [tex]\( 27 \)[/tex], the quotient is [tex]\( 0 \)[/tex] and the remainder is [tex]\( 6 \)[/tex].
So, the final answer is:
- Quotient: [tex]\( 0 \)[/tex]
- Remainder: [tex]\( 6 \)[/tex]
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