The HCF of two numbers is 23 and their LCM is 1449 if one of the number is 161, find the other

Ans: It is given that the HCF of the two numbers is 23 and their LCM is 1449 We have to find the other number if one of the numbers is 161 . Let the other number be a. We know that, LCM $\times$ HCF $=$ Product of two numbers. Therefore, $$ \begin{array}{l} 1449 \times 23=a \times 161 \\ \Rightarrow a=\frac{1449 \times 23}{161} \\ \therefore a=207 \end{array} $$ $15 .$ Show that every positive odd integer is of the form $(4 \mathrm{q}+1)$ or $(4 \mathrm{q}+3)$ for some integer $q$. $\Rightarrow \mathrm{a}=\frac{1449 \times 23}{161}$

$\therefore a=207$

Text Solution

`205``208``207``209`

Answer : C

Solution : For two numbers `a` and `b,` we know that <br> ` ( axx b)` = ( HCF (a ,b) ` xx ` ( LCM (a,b)) . <br> Here a '= 161 , HCF = w23 and LCM = 1449. <br> And we have to find b. <br> ` ( 161 xxb) = ( 23xx 1449) Rightarrow b = ((23 xx 1449))/161 = 207` <br> Hence , the other number is `207. `

The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Let the two numbers be a and b.Let the value of a be 161.Given: HCF = 23 and LCM = 1449We know, a × b = HCF × LCM                ⇒ 161 × b = 23 × 1449                ⇒ ∴ b =`( 23 ×1449)/161 = 33327/161 = 207`

Hence, the other number b is 207. 

Concept: Euclid’s Division Lemma

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The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find
the other.

HCF of two numbers = 145LCM of two numbers = 2175Let one of the two numbers be 725 and other be x.Using the formula, product of two numbers = HCF × LCMwe conclude that725 × x = 145 × 2175x = `(145 ×2175)/ 725`= 435

Hence, the other number is 435.

Concept: Euclid’s Division Lemma

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The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.