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3 years ago

Two integers A and B have product of 24 what is the least possible sum of A and B

Mathematics

2 answers:

jeka943 years ago

3 0

The answer would be -25.

Before we get to the solution, we need to remind ourselves that integers are whole numbers, but they include even negative numbers.

First we need to look at least two factors of 24.

Positive:

24 x 1 = 24

12 x 2 = 24

8 x 3 = 24

6 x 4 = 24

Negative:

-24 x -1 = 24

- 12 x -2 = 24

-8 x - 3 = 24

-6 x -4 = 24

Now lets add these factors and see the sum of each of these factors:

24 + 1 =25

12 + 2 = 14

8 + 3 = 11

6 + 4 = 10

-24 + -1 = -25

- 12 + -2 = -14

-8 + - 3 = -11

-6 + -4 = -10

aleksley [76]3 years ago

3 0

The least possible sum is 10 , if the integers are 6 and 4 .

The greatest possible sum is 25 , if the integers are 1 and 24 .

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3 years ago

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2 years ago

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quester [9]

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Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

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$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

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Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

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