NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Finding the HCF Directly from the Prime Factorisations
Book page 53–54 Updated on2026-09-19
Q1.
How do we directly find the HCF without listing all the factors?
Answer
Build the HCF prime by prime: keep only the primes that occur in every number, and take each one the fewest number of times it occurs.
Step 1 — write the prime factorisation of each number
Step 2 — mark the primes that appear in all of them
Step 3 — for each such prime, count how many times it occurs in each number and take the minimum
Step 4 — multiply what you have collected
Why it happens: the HCF is the largest subpart common to both factorisations. A prime missing from one number can never appear in a common subpart. And a prime that occurs twice in one number but three times in the other can be used at most twice — the smaller count is the limit. Taking the most that is allowed of each shared prime gives the largest common subpart.
Q2.
Example 4: Find the HCF of 30 and 72. [How many 2s will it contain? How many 3s will it contain?]
Answer
HCF(30, 72) = 6.
30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
Primes in both: 2 and 3 (5 is only in 30, so it is out)
How many 2s? 30 has one 2, 72 has three 2s. The minimum is one.
How many 3s? 30 has one 3, 72 has two 3s. The minimum is one.
Why it happens: a common factor may not ask for more 2s than 30 can supply, and 30 supplies only one. The same for the 3s. So one 2 and one 3 is the most any common factor can hold, and 2 × 3 = 6 is the highest.
Q3.
Example 5: Find the HCF of 225 and 750.
Answer
HCF(225, 750) = 75.
225 = 3 × 3 × 5 × 5
750 = 2 × 3 × 5 × 5 × 5
Primes in both: 3 and 5 (2 is only in 750)
How many 3s? 225 has two, 750 has one. The minimum is one.
How many 5s? 225 has two, 750 has three. The minimum is two.
Tip: the same recipe works for three or more numbers — take the primes common to all of them, each the minimum number of times across all the factorisations.