Fundamental Theorem of Arithmetic — Prime Factorisation of 3825

easy CBSE NCERT Class 10 4 min read

Question

Find the prime factorisation of 3825 using the Fundamental Theorem of Arithmetic. Express your answer as a product of prime powers.


Solution — Step by Step

We check divisibility from the smallest prime — 2. Since 3825 is odd, 2 doesn’t divide it.

Try 3: sum of digits = 3 + 8 + 2 + 5 = 18, which is divisible by 3. So 3 divides 3825.

3825÷3=12753825 \div 3 = 1275

Check 1275 for divisibility by 3: digit sum = 1 + 2 + 7 + 5 = 15. Yes, divisible again.

1275÷3=4251275 \div 3 = 425

Now check 425: digit sum = 4 + 2 + 5 = 11. Not divisible by 3. Move to the next prime.

425 ends in 5, so it’s divisible by 5. This is the quickest check you’ll ever do.

425÷5=85425 \div 5 = 85 85÷5=1785 \div 5 = 17

We need to verify 17 is prime. Check divisibility by all primes up to 174.1\sqrt{17} \approx 4.1 — that means only 2 and 3.

17 is odd, and 1 + 7 = 8 (not divisible by 3). So 17 is prime. We stop here.

Collecting everything:

3825=3×3×5×5×17=32×52×173825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17
3825=32×52×173825 = 3^2 \times 5^2 \times 17

Why This Works

The Fundamental Theorem of Arithmetic guarantees that every composite number greater than 1 can be expressed as a product of primes in exactly one way (ignoring the order of factors). This uniqueness is the key — there is no other set of primes that multiply to give 3825.

The method we used is called the factor tree or successive division method. We always start with the smallest prime and keep dividing until we reach 1. The reason we go in order — 2, 3, 5, 7, 11, … — is so we never accidentally skip a prime factor.

The divisibility rule shortcut for 3 (sum of digits) and 5 (units digit) are your best friends here. They turn what looks like guesswork into a systematic process you can finish in under two minutes.


Alternative Method

We can draw a factor tree instead of doing successive division.

Start by splitting 3825 into any two factors — say, 25 and 153:

  • 25=5×525 = 5 \times 5
  • 153=3×51=3×3×17153 = 3 \times 51 = 3 \times 3 \times 17

Put it together: 3825=52×32×173825 = 5^2 \times 3^2 \times 17

Same answer, different path. This is exactly what the Fundamental Theorem promises — no matter how you factor, you always end up with the same primes.

In board exams, the successive division method (ladder method) is safer than the factor tree. It’s harder to miss a prime when you go step by step in a single column. The factor tree looks neat but it’s easy to stop early by accident.


Common Mistake

Stopping at 3825=25×1533825 = 25 \times 153 and leaving it there.

Many students write this as the final answer. But 25 and 153 are composite numbers, not primes. The Fundamental Theorem requires the final expression to have only prime bases. Always keep factoring every composite you see until every number in the product is prime.

A related slip: writing 3825=9×52×173825 = 9 \times 5^2 \times 17 and forgetting to break 9 into 323^2. The answer must use prime numbers, not prime powers hidden inside composites. Check each base — if it’s not prime, split it further.

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