(B) The $LCM$ of two numbers must always be divisible by their $HCF$.Given $HCF(a, b) = 25$.We check the divisibility of the given options by $25$:$1$

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(B) The $LCM$ of two numbers must always be divisible by their $HCF$.Given $HCF(a, b) = 25$.We check the divisibility of the given options by $25$:$1$. $50 / 25 = 2$ (Divisible)$2$. $105 / 25 = 4.2$ (Not divisible)$3$. $100 / 25 = 4$ (Divisible)Since $105$ is not divisible by $25$,it cannot be the $LCM$ of two numbers whose $HCF$ is $25$.

Class 10 Mathematics Real Numbers HCF and LCM by Prime Factorisation

Medium

If the $HCF(a, b) = 25$,then which of the following cannot be the $LCM(a, b)$? (Options: $50, 105, 100$)

A
$50$
B
$105$
C
$100$
D
None of these

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Class 10

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Real Numbers

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HCF and LCM by Prime Factorisation

What is the largest number that divides $626$,$3127$,and $15628$ and leaves remainders $1$,$2$,and $3$ respectively?

Medium

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By using the fact $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$,find $\text{l.c.m.}(306, 657)$.

Medium

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If the $HCF(a, b) = 16$, then which of the following cannot be the $LCM(a, b)$? (Options: $32, 72, 64$)

Medium

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If the $HCF(a, b) = 25$,then which of the following cannot be the $LCM(a, b)$?

Medium

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$L.C.M.(115, 25) = \ldots$

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If the $HCF(a, b) = 25$,then which of the following cannot be the $LCM(a, b)$? (Options: $50, 105, 100$)

Similar Questions

What is the largest number that divides $626$,$3127$,and $15628$ and leaves remainders $1$,$2$,and $3$ respectively?

By using the fact $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$,find $\text{l.c.m.}(306, 657)$.

If the $HCF(a, b) = 16$, then which of the following cannot be the $LCM(a, b)$? (Options: $32, 72, 64$)

If the $HCF(a, b) = 25$,then which of the following cannot be the $LCM(a, b)$?

$L.C.M.(115, 25) = \ldots$

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