Question

the hcf and LCM of number are 33 and 264 respectively . the first number is completely divisble by 2 and gives quotient 33.find the other number​

Answer 1

Answer:

first number is 66 second number is 132

Step-by-step explanation:

product of 2 numbers=product of hcf and lcm of the 2 numbers

first lets find the product of lcm and hcf

264×33=8712

now let the first number be y

it is divisible by 2 and gives quotient 33

y/2=33

y=33×2

y=66

now let the second number be z

according to the formula

product of 2 numbers=product of hcf and lcm of the 2 numbers

66×z=264×33

66×z=8712

z=8712/66

z=132

Answer 2

Given that, HCF and LCM of the Number are 33 and 264 respectively.

Need to find: The other Number.

❍ Let us consider that the required first number be x and other number be y respectively.

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$⠀⠀

• The first Number is completely divisible by 2 and the Quotient is 33. Remainder will be 0.

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E U C L I D'S ⠀D I V I S I O N⠀ L E M M A :

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$\bf{\dag}\;\boxed{\sf{a = b(q) + r}}$

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$:\implies\sf a = b(q) + r\\\\\\:\implies\sf x = 2 \times (33) + 0 \\\\\\:\implies\sf x = 66 + 0\\\\\\:\implies{\underline{\boxed{\frak{\pink{x = 66}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence, \; required\: first\; number\;is\;\bf{66 }.}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀⠀⠀⠀

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$\bf{\dag}\;\boxed{\sf{HCF \times LCM = Product\;of \; two\;numbers}}$

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$:\implies\sf 33 \times 264 = 66 \times y \\\\\\:\implies\sf y = \dfrac{33 \times 264}{66}\\\\\\:\implies\sf y = \cancel\dfrac{8712}{66}\\\\\\:\implies{\underline{\boxed{\frak{\pink{y = 132}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence,\;the\;other\; required\; number\; is\; \bf{132 }.}}}$