Free Practice Questions for Number-system in Maths

When a number is divided by 156 it gives remainder 29.

It means the number is in the form of 156a+29.

When the same number is divided by 13,

 the remainder will be $=29÷13=3$

We know that when $(a^n-b^n)$

1. It is always divisible by $a-b$.

2. When $n$ is even it is also divisible by $a+b$.

3. When $n$ is odd it is not divisible by $a+b$.

Hence it is divisible by 59+57 = 116

ATQ,

Only Red and yellow color bought a customers $=$ 30
Only Yellow and maroon colors bought a customers $=20$
Only Red and maroon colors bought a customers $=25$
All three bought color $=29$
Only one color of 2 meters of cloth bought
$=(30-29)+(29-20)+(29-25)$
$=1+9+4$
$=14$

We know that
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
According to question
Divisor $=5\times$ Quotient $=9\times$ Remainder
So,
Divisor $=9\times 25=225$
Quotient $=\frac{225}{5}=45$
Hence,
Dividend $=225\times 45+25$

$\Rightarrow 10,125+25=10,150$

a – b – c = 0, then $a^3-b^3-c^3=3abc$

99 - 63 - 36 = 0
${99}^3-{63}^3-{36}^3=3\times 99\times 63\times 36$
$=3\times 3^2\times 11\times 3^2\times 7\times 3^2\times 2^2$
$=2^2\times 3^7\times 7^1\times {11}^1$
Number of factors = (power+1) 

= (2+1) + (7+1) + (1+1) + (1+1)

= 3 × 8 × 2 × 2 = 96

For a number to be divisible by 7,11 and 13, it must be divisible by $7\times 11\times 13$ i.e. $1001.$
Now, we know that a 6 - Digit number divisible by 1001 is of the from xyz xyz.
$\Rightarrow$ If $5\mathrm{a}4\mathrm{b}6\mathrm{c}$ to be divisible by $1001.$
$b=5,a=6,c=4$
So, $3a-5b+7c=18-25+28$
$=21$

Given, 

$789x6378y$ is divisible by 72 i.e $9\times 8$

Divisibility Rule of 8

If the last three digits of a number are divisible by 8, then the number is completely divisible by 8.

For 8, y = 4

Divisibility Rule of 9

If the sum of digits of the number is divisible by 9, then the number itself is divisible by 9.

For 9, x = 2

so, x+y = 4 + 2 = 6

Unit digit of (433)²⁴ – (377)³⁸ + (166)⁵⁴ = 1 – 9 + 6 = -2 

Unit digit can not be negative so unit digit = 10 – 2 = 8

If the sum of the digits is a multiple of 9 , then the number is divisible by 9 .

$X+8+4+Y$

Minimum value of X and Y is 2 and 4.

$2+8+4+4=18$ is divisible by 9 .

$X+Y=2+4=6$

If the last two digits of a number are divisible by 4, the number is divisible by 4.

Clearly, 4864 is divisible by 4 but the sum of these numbers is 22, so it is not divisible by 3. 

So 9P2 will be the number that is divisible by 3.

By hit and trial method, we put p =1

Therefore

The number 912 is divisible by 3

In other words

To divide a whole number by 12

put p = 1

$a^n-b^n$ is divisible by $a-b$ for all $n$.

$a^n-b^n$ is divisible by $a+b$ if $n$ is Even.

$a^n+b^n$ is divisible by $a+b$ if $n$ is Odd.

$(49{)}^{15}-(1{)}^{15}$, here $\mathrm{n}=15$ i.e. odd

So it will be divisible by $49-1=\mathbf{4}\mathbf{8}$

$(49{)}^{15}$ can also be written as ${\left(7^2\right)}^{15}=(7{)}^{30}$

$\therefore (49{)}^{15}-(1{)}^{15}=(7{)}^{30}-(1{)}^{30}$

So it will also be divisible by $7+1=\mathbf{8}$

48 is multiple of 8.

$88=8\times 11$
So, $7\mathrm{A}6245\mathrm{B}$ is divisible by both 8 and 11 .
Divisibility of 8 : If the last three digits of any number is divisible by 8 , then the number will be divisible by 8 .
$45\mathrm{B}$ is divisible by 8 , So $\mathrm{B}=6$
So, Number $=7\mathrm{A}62456$
Divisibility of 11 : If the difference between the sum of digits at even places and that of those at odd places is 0 or a multiple of 11 , then the number is divisible by 11 .

$(7+6+4+6)-(A+2+5)=23-7-A=16-A$

$16-A=0or11$

16-A = 11

$A=5$

Hence,  $B^2-A^2=36-25=11$

Prime numbers between 70 and $100=71,73,79,83,89,$ and $97$

Average $=\frac{71+73+79+83+89+97}{6}=\frac{492}{6}=82$

LCM of $6,9,12$ and 15 $=180$

Number $=180k+3$, which is a multiple of 7

when $k=5$

$175k+5k+3$

$=175\times 5+5\times 5+3$

$=875+25+3$

$=903$

Hence, the number $=903$

$\frac{330\times 613\times 860}{17}=\frac{7\times 1\times 10}{17}=\frac{70}{17}$

Remainder $=2$

Average of Arithmetic progression $=\frac{a+l}{2}$, where $\mathrm{a}$ is the first term and {l} is the last term.
$\frac{a+l}{2}=40$
$a+l=80$
$l=a+(n-1)d$, where $\mathrm{d}=1$, as the numbers are consecutive.
$l=a+10$
So, $a+a+10=80$
$\begin{array}{rl}& 2a=70\\ & a=35\end{array}$

156 - 6 = 150

181 - 6 = 175

331 - 6 = 325

HCF of 150, 175 and 325 = 25

$56=7\times 8$
So, the number must be divisible by both 7 and 8 .
Divisibility rule of 8 : If the last three digits of any number is divisible by 8 , then the number will also be divisible by 8 .

Divisibility rule of 7 : To check the divisibility of 7 , we multiply the unit digit by 2 and subtract it from the rest, and this process continues till the number
reduces to a smaller number. If the number so formed is 0 or multiple of 7 , then the number is divisible by 7 else it is not divisible by 7 .
$2y4$ is divisible by 8 , So $y=2$ or 6
As we need the maximum value of $y$, So $y=6$
So, Now $x894264$ is divisible by 7
$x89426-8=x89418$
$x8941-16=x8925$
$x892-10=x882$
$x88-4=x84$
$x8-8=x$
$x$ can be 0 and 7 both, but since the number is a 7 digit number so the value of $x$ can not be 0 , as it will make the number a six digit number.
Hence $x=7$
$x^2-y=7^2-6=49-6=43$

Total numbers from 600 to $750=(750-600)+1=151$

Total numbers from 1 to 750 that are divisible by 3 or 7 $=\frac{750}{3}+\frac{750}{7}-\frac{750}{21}=250+107-35=322$
Total numbers from 1 to 599 that are divisible by 3 or 7 $=\frac{599}{3}+\frac{599}{7}-\frac{599}{21}=199+85-28=256$
Numbers from 600 to 750 that are divisible by 3 or 7 $=322-256=66$
So, numbers that are divisible neither by 3 nor 7 $=151-66=85$

Total numbers from 600 to $750=(750-600)+1=151$

Total numbers From 1 to 750 that are divisible by 3 or 7 $=\frac{750}{3}+\frac{750}{7}-\frac{750}{21}=250+107-35=322$
Total numbers From 1 to 599 that are divisible by 3 or 7 $=\frac{599}{3}+\frac{599}{7}-\frac{599}{21}=199+85-28=256$
Numbers From 600 to 750 that are divisible by 3 or 7 $=322-256=66$
So, Numbers that are divisible neither by 3 nor 7 $=151-66=85$