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Original Question
create up to 5 or more smaller friendly equations to solve 6841 divided by 30 no decimals, no fractions, only division.Asked by Everything is Nothing
create up to 5 or more smaller friendly equations to solve 543 divided by 15
no decimals, no fractions, only division,
no decimals, no fractions, only division,
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Answered by
GPT-4o mini
Sure! Here are five smaller friendly equations that help in solving \(543 \div 15\) using only division:
1. **Break down 543:**
\(540 \div 15\) (easier number to work with)
\(540 \div 15 = 36\)
2. **Leftover from 540:**
\(3 \div 15\)
Since \(3 < 15\), this equals \(0\).
Combining these gives us:
\[
543 \div 15 = 36 + 0 = 36
\]
3. **Using multiples of 15:**
\(15 \div 15 = 1\)
\(30 \div 15 = 2\)
\(45 \div 15 = 3\)
\(60 \div 15 = 4\)
\(540 \div 15 = 36\) (this leads us back to the earlier equation)
4. **Another approach with adjustments:**
\(525 \div 15 = 35\)
Next, consider the remaining:
\(18 \div 15 = 1\) (18 is \(543 - 525\))
5. **Re-checking larger parts:**
\(150 \div 15 = 10\)
\(300 \div 15 = 20\)
Then, adding those gives
\(10 + 20 + 6 \div 15 = 36\) when calculated incrementally.
Combining these methods confirms the final solution,
\[
543 \div 15 = 36
\]
1. **Break down 543:**
\(540 \div 15\) (easier number to work with)
\(540 \div 15 = 36\)
2. **Leftover from 540:**
\(3 \div 15\)
Since \(3 < 15\), this equals \(0\).
Combining these gives us:
\[
543 \div 15 = 36 + 0 = 36
\]
3. **Using multiples of 15:**
\(15 \div 15 = 1\)
\(30 \div 15 = 2\)
\(45 \div 15 = 3\)
\(60 \div 15 = 4\)
\(540 \div 15 = 36\) (this leads us back to the earlier equation)
4. **Another approach with adjustments:**
\(525 \div 15 = 35\)
Next, consider the remaining:
\(18 \div 15 = 1\) (18 is \(543 - 525\))
5. **Re-checking larger parts:**
\(150 \div 15 = 10\)
\(300 \div 15 = 20\)
Then, adding those gives
\(10 + 20 + 6 \div 15 = 36\) when calculated incrementally.
Combining these methods confirms the final solution,
\[
543 \div 15 = 36
\]
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