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3 years ago

How many positive integers between 1000 and 9999 inclusive

Mathematics

1 answer:

bekas [8.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

first number is 1000-1+9=1008

9)1000(1

-------

-----

----

last number is 9999

9| 9999

---------

1111 |0

--------

9999=1008+(n-1)9

9999-1008=(n-1)9

n-1=8991/9=999

n=999+1=1000

first digit=1000

last digit=9999-1=9998

2 |9999

---------

|4999|1

9998=1000+(n-1)2

(n-1)2=9998-1000=8998

n-1=4499

n=4499=1=5000

c.not sure

total numbers=9000

9999=1000+(n-1)1

9999-1000=n-1

n=8999+1=9000

numbers divisible by 3=3000

first number=1002

last number=9999

9999=1002+(n-1)3

(n-1)3=9999-1002=8997

n-1=2999

n=2999+1=3000

numbers not divisible by 3=9000-3000=6000

numbers divisible by 5=1800

first number=1000

last number=9995

9995=1000+(n-1)5

(n-1)5=9995-1000=8995

n-1=1799

n=1799+1=1800

numbers divisible by 7=1286

7 | 1000

---------

| 142-6

1000-6+7=1001

7 | 9999

|---------

1428-3

9999-3=9996

first digit=1001

last digit=9996

9996=1001+(n-1)7

(n-1)7=9996-1001=8995

n-1=1285

n=1285+1=1286

numbers divisible by 35=257

first digit=1015

35 ) 1000 ( 28

----

300

280

------

---

1000-20+35=1015

35)9999(285

----

299

280

-----

199

175

----

last digit=9999-24=9975

9975=1015+(n-1)35

(n-1)35=9975-1015=8960

n-1=8960/35=256

n=257

reqd. numbers=1800+1286-257=3019

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The equation $f(x)=g(x)$ is

$|x-4|-11=-\sqrt{5x}$

Some observations:

$\sqrt{5x}$ is defined only as long as $5x\ge0$ , or $x\ge0$
wherever $\sqrt{5x}$ is defined, its value must be non-negative, so that $-\sqrt{5x}$ is never positive
by the definition of absolute value, we have $|x-4|=x-4$ if $x\ge4$ , and $|x-4|=-(x-4)=4-x$ if $x$ . Then

$|x-4|-11=\begin{cases}x-15&\text{for }x\ge4\\-x-7&\text{for }x$

If $x$ , the equation becomes

$-x-7=-\sqrt{5x}\implies x+7=\sqrt{5x}$

Taking the square of both sides gives

$(x+7)^2=\left(\sqrt{5x}\right)^2\implies x^2+14x+49=5x\implies x^2+9x+49=0$

but since the discriminant is $9^2-4\cdot1\cdot49$ , there are no real solutions.

If $x\ge4$ , then

$x-15=-\sqrt{5x}$

Taking squares gives

$(x-15)^2=\left(-\sqrt{5x}\right)^2\implies x^2-30x+225=5x$

and solving by the quadratic formula gives two potential solutions,

$x=\dfrac{35\pm5\sqrt{13}}2$

which have approximate values of 8.49 and 26.51.

We know for any value of $x$ that $g(x)\le0$ . We have $f(8.49)\approx-6.51$ and $f(26.51)\approx11.51$ , so only the first solution 8.49 is valid.