Find the LCM Using Prime Factorization

Factor every input completely into primes

For 24, divide repeatedly by two: 24 = 2×12 = 2×2×6 = 2×2×2×3 = 2^3×3. For 90, the factorization is 2×3×3×5 = 2×3^2×5. A factor tree can begin with different factor pairs and still reach the same prime factors. Stop each branch only when its remaining factor is prime; leaving a composite factor such as six in the final list makes the exponent comparison harder to perform correctly.

Choose the highest exponent of every prime present

The primes appearing in either 24 or 90 are two, three and five. The greatest power of two is 2^3, the greatest power of three is 3^2, and five appears to the first power. Multiply these selected powers to obtain LCM = 2^3×3^2×5 = 360. A missing prime in one factorization has exponent zero for comparison purposes. It still must be included if another input requires it, because the final multiple must be divisible by every input.

Explain why the selected powers are necessary

Any common multiple of 24 and 90 must contain at least three factors of two to be divisible by 24, at least two factors of three to be divisible by 90, and at least one factor of five. The product 360 meets those requirements exactly. Removing one selected prime factor would make the product fail at least one divisibility requirement. Adding extra prime factors gives a larger common multiple, which is valid as a multiple but is no longer the least one.

Check the completed result by division

The quotients 360/24 = 15 and 360/90 = 4 are whole numbers. This confirms that the proposed value is a common multiple. The prime-power argument establishes that it is the least one; divisibility alone would also accept 720. In the local form, enter 24 and then 90 in separate runs to verify the two factorizations. The form lists repeated prime factors, so count their repetitions to obtain the exponents before building the LCM manually.

Work a three-number example

For 18, 40 and 60, the factorizations are 2×3^2, 2^3×5, and 2^2×3×5. The highest powers across all three are 2^3, 3^2 and 5. Their product is again 360. It is divisible by 18, 40 and 60, with quotients 20, 9 and 6. Comparing only the first and last numbers would miss the third factor of two required by 40 and would incorrectly produce 180.

Contrast the GCF exponent rule

For the greatest common factor, use only primes present in every input and select their lowest exponents. For 24 and 90, that gives 2^1×3^1 = 6. Using the smallest powers for the LCM would therefore calculate a different quantity. A useful memory aid is to connect the rules to divisibility: a common factor must fit inside every input, while a common multiple must contain everything needed by each input. The reasoning is more reliable than memorizing largest and smallest without their purpose.

Account for one and repeated inputs

The number one has no prime factors and imposes no additional prime-power requirement, so LCM(1,n) = n for positive n. Repeating an input also adds no new requirement: LCM(24,24,90) remains 360. Do not add exponents from different input rows when the aim is an LCM; that would often form the full product rather than the least multiple. Add exponents only when multiplying factors within a single factorization, and take maxima when comparing the completed input factorizations.

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