NCERT Solutions for Class 7th Maths Chapter 3 Try This — Property Involving both the HCF and the LCM (2026-27)

Q1.

Thus, our observations seem to suggest the following: HCF × LCM = Product of the two numbers. Why does this happen? Can you give an explanation or proof? [Hint: Consider the prime factorisation of the given numbers. Among their prime factors, some are common to both factorisations, and the rest occur in only one of them. Between the HCF and the LCM, see how the common and non-common prime factors get distributed. In the product, observe how these two kinds of prime factors occur. Compare them.]

Answer

Follow one prime at a time. That is enough to prove the whole statement.

Take any prime p. Suppose it occurs i times in the first number and j times in the second.

In the HCF, p occurs the smaller of i and j times
In the LCM, p occurs the larger of i and j times
So in HCF × LCM, p occurs (smaller) + (larger) = i + j times

In the product of the two numbers, p occurs i + j times as well
(the two lists of factors are simply written side by side)

Every prime occurs the same number of times on both sides, so the two numbers are equal.

HCF × LCM = product of the two numbers

A worked case, 60 and 90:

Prime	In 60 = 2×2×3×5	In 90 = 2×3×3×5	In HCF (min)	In LCM (max)	min + max
2	2	1	1	2	3
3	1	2	1	2	3
5	1	1	1	1	2

HCF = 2 × 3 × 5 = 30, LCM = 2 × 2 × 3 × 3 × 5 = 180
HCF × LCM = 30 × 180 = 5400
60 × 90 = 5400 ✓

Why it happens: the key is that "smallest + largest = first + second" for any two numbers. Whichever of i and j is bigger, the pair (min, max) is just the pair (i, j) rearranged. So the HCF and the LCM between them use up each prime exactly as many times as the plain product does — the HCF takes the shared copies, the LCM takes everything else, and nothing is lost or double-counted.

Q2.

Explore whether this property holds when 3 numbers are considered.

Answer

It does not hold in general for three numbers.

Numbers	HCF	LCM	HCF × LCM	Product	Equal?
2, 3, 5	1	30	30	30	yes
2, 4, 8	2	8	16	64	no
4, 6, 9	1	36	36	216	no
6, 10, 15	1	30	30	900	no

It happens to work when the three numbers are pairwise co-prime, as in 2, 3, 5 — but that is a special case, not a rule about three numbers.

Why it fails: run the same prime-counting argument with three numbers. For a prime p occurring i, j and k times, the HCF takes the smallest and the LCM takes the largest, so together they use (min + max) copies. But the product uses i + j + k copies. With two numbers those two counts always match; with three they match only when the middle count is 0 — which is why 4, 6, 9 fails (the prime 2 occurs 2, 1 and 0 times: min + max = 0 + 2 = 2, while the product needs 3).

Tip: for three numbers the correct statement is different. For 2, 4 and 8, HCF = 2 and LCM = 8, and 2 × 8 = 16, nowhere near 64. Do not carry the two-number rule across without checking.