Long Division Problem: Divide 325 by 14 Step-by-Step

Long Division with Two-Digit Divisors

14325

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:04 Let's solve this problem together.
00:07 First, divide the leftmost digit of the dividend.
00:11 Since 3 is less than 14, add the next digit to it before dividing.
00:18 Write the result above, without the remainder. Be precise!
00:22 Next, multiply what you wrote by the divisor.
00:26 Subtract this product from your current number.
00:31 Bring the next digit down and repeat.
00:35 Now, divide again.
00:40 Write the new result above. Position is key!
00:45 Multiply the result and subtract once more.
00:53 We ended with a remainder of 3.
00:58 And that's how we solved the problem. Great job!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

14325

2

Step-by-step solution

To solve this problem, we'll conduct a long division of 325 by 14 step by step:

  • Step 1: Determine how many times 14 fits into the first one or two digits of 325.
  • Step 2: The first part of 325 we consider is 32 (as 14 is larger than 3). 14 fits into 32 two times because 14×2=2814 \times 2 = 28 and 14×3=4214 \times 3 = 42, which is too large. Write 2 as part of the quotient.
  • Step 3: Subtract 2828 from 3232 to get 44.
  • Step 4: Bring down the next digit of the dividend, which is 5, forming 45.
  • Step 5: Determine how many times 14 fits into 45. It fits three times, since 14×3=4214 \times 3 = 42 and 14×4=5614 \times 4 = 56 is too large. Write 3 as part of the quotient.
  • Step 6: Subtract 4242 from 4545 to get 33.
  • Step 7: There are no more digits to bring down from the dividend, so the remainder is 3.

The quotient of 325÷14325 \div 14 is 2323 with a remainder of 33.

Therefore, the solution to the problem is 23 23 with a remainder of 33 .

3

Final Answer

23 23 with a remainder of 3

Key Points to Remember

Essential concepts to master this topic
  • Setup: Start with largest digits when divisor exceeds first digit
  • Technique: Check 14×2=2814 \times 2 = 28 fits into 32, then 14×3=4214 \times 3 = 42 fits into 45
  • Check: Verify 23×14+3=322+3=32523 \times 14 + 3 = 322 + 3 = 325

Common Mistakes

Avoid these frequent errors
  • Starting division with wrong digit groupings
    Don't try dividing 14 into just the 3 = impossible division! This leads to confusion and wrong quotient placement. Always group enough digits so the divisor fits at least once.

Practice Quiz

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216

FAQ

Everything you need to know about this question

Why can't I divide 14 into 3?

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Since 14 is larger than 3, it can't fit even once! That's why we look at the first two digits (32) instead. Always use enough digits so your divisor fits at least once.

How do I know which multiple of 14 to use?

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Try multiples until you find the largest one that doesn't exceed your number. For 32: 14×2=2814 \times 2 = 28 fits, but 14×3=4214 \times 3 = 42 is too big!

What if I make an error in my multiplication?

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Always double-check your multiplication tables! Write them out if needed: 14×1=1414 \times 1 = 14, 14×2=2814 \times 2 = 28, 14×3=4214 \times 3 = 42, etc.

How do I verify my final answer is correct?

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Multiply your quotient by the divisor, then add the remainder: 23×14+323 \times 14 + 3. If this equals your original dividend (325), you're right!

What does 'remainder 3' actually mean?

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The remainder is what's left over after division. It means 325 contains exactly 23 groups of 14, plus 3 extra that can't form another complete group.

Can the remainder ever be bigger than the divisor?

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Never! If your remainder equals or exceeds the divisor, you can divide at least one more time. The remainder must always be smaller than what you're dividing by.

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