If a-23- 3- b-2 - 3 - 5- c-3n- 5 and LCMa- b- c-23- 32- 5- then n-a 1b 2c 3d 4

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Standard X

Mathematics

Method of Finding LCM

If a=23× 3, b...

Question

If a = 2³ ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3ⁿ ✕ 5 and LCM (a, b, c) = 2³ ✕ 3² ✕ 5, then n =

(a) 1
(b) 2
(c) 3
(d) 4

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Solution

LCM ……(I)

We have to find the value for n

Also

We know that the while evaluating LCM, we take greater exponent of the prime numbers in the factorization of the number.

Therefore, by applying this rule and taking we get the LCM as

LCM ……(II)

On comparing (I) and (II) sides, we get:

Hence the correct choice is (b).

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If $a = 2^{3} \times 3$ , $b = 2 \times 3 \times 5$ , $c = 3^{n} \times 5$ and LCM ( $a, b, c$ ) $= 2^{3} \times 3^{2} \times 5$ , then $n =$ ? (Here, $n$ is a natural number)

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If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n = (a) 1 (b) 2 (c) 3 (d) 4

If a = 2³ ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3ⁿ ✕ 5 and LCM (a, b, c) = 2³ ✕ 3² ✕ 5, then n =

(a) 1
(b) 2
(c) 3
(d) 4