Tag: fundamental theorem of arithmetic

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Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

HCF of the two numbers =

  1. Product of numbers + their LCM

  2. Product of numbers - their LCM
  3. Product of numbers $\times$ their LCM
  4. Product of numbers $\div$ their LCM

  5. Answer required
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The product of highest common factor $(H.C.F.)$ and lowest common multiple $(L.C.M.)$ of two numbers is equal to the product of two numbers.

If two numbers are $a$ and $b$ then
$HCF\ \times LCM=a\times b$ 

Hence,
$HCF=\dfrac{ab}{LCM}$

For example: 
Let $a=10\Rightarrow 2\times 5$ and $b=15\Rightarrow 3\times 5$
So, the $LCM$ of both numbers $=2\times 3\times 5\Rightarrow 30$
Then 
$HCF=\dfrac{10\times 15}{30}\Rightarrow 5$

Hence,
$HCF=\dfrac{Product\ of\ numbers}{Their\ LCM}.$

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

What is the HCF of $4x^{3} + 3x^{2}y - 9xy^{2} + 2y^{3}$ and $x^{2} + xy - 2y^{2}$?

  1. $x - 2y$
  2. $x - y$
  3. $(x + 2y)(x - y)$
  4. $(x - 2y)(x - y)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $f(x, y) = 4x^{2} + 3x^{2}y - 9xy^{2} + 2y^{3}$
and $g(x, y) = x^{2} + xy - 2y^{2}$
Clearly $(x - y) = 0$ or $x = y$ satisfy both equation, since for $x = y$ makes both equations zero. 

Hence, $(x - y)$ is the H.C.F. 
Similarly $(x + 2y) = 0$ i.e., $x = -2y$ also makes both equations zero. 
Thus, $(x + 2y)(x - y)$ is the H.C.F.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The smaller value of n for which $x^{2} - 2x - 3$ and $x^{3} - 2x^{2} - nx - 3$ have an H.C.F. involving $x$ is

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the two polynomials to have an HCF involving x, they must share a common root. x^2 - 2x - 3 factors to (x-3)(x+1). Testing x=3 in the second polynomial: 27 - 18 - 3n - 3 = 0 => 6 - 3n = 0 => n=2.

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

The number of possible pairs of number, whose product is 5400 and the HCF is 30 is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Given\quad that\quad product\quad of\quad the\quad number\quad is\quad 5400=30\times 3\times 2\times 30.\ \therefore \quad Possible\quad pairs\quad as\quad per\quad the\quad requirment\quad are-\ (1)\quad 30\times (3\times 2\times 30)=30\times 180\ (2)\quad (30\times 3)\times (2\times 30)=90\times 60\ \therefore \quad Total\quad number\quad of\quad pairs=2\quad \quad (Ans) $

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

If HCF of numbers $408$ and $1032$ can be expressed in the form of $1032x -408 \times 5$, then find the value of $x$.

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$408=2\times2\times2\times3\times17$

$1032=2\times2\times2\times3\times43$

Hence, $HCF=2\times2\times2\times3=24$

Now, $1032x-408\times5=24\Rightarrow 1032x=2064\Rightarrow x=2$

Multiple choice maths real number real numbers on number line fundamental theorem of arithmetic common factors and hcf

Find the LCM and HCF of the following integers by the prime factorization mass

  1. 12, 15 and 21

  2. 17, 23, and 29

  3. 8, 9 and 25

  4. 72 and 108

  5. 72 and 108

  6. 306 and 657
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The question asks for the LCM and HCF of integers but provides a list of sets. Option A (12, 15, 21) is a valid set of integers for which LCM and HCF can be calculated.