Tag: large numbers

Questions Related to large numbers

The digit in the units place of the number $2015!+3^{7886}$ is

maths numbers and place value face value of digit large numbers general form of number

  1. $1$

  2. $3$

  3. $7$

  4. $9$

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Correct Option: D
Explanation:

$2015! + 3^{7886}$

$2015! \Rightarrow$ Unit digit $\Rightarrow \underline { 0 }$

$3^{7886}=3^{4n+2}$, So unit digit $\Rightarrow 9$

$\therefore 0+9 \Rightarrow \underline{9}$.

$\therefore$ Digit at unit place $\Rightarrow 9$ 

The sum of face values of 4 and 8 in any number having 4 and 8 as digits, is:

maths numbers and place value face value of digit large numbers general form of number

  1. 12

  2. 32

  3. 6

  4. 48

  5. None of these

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Correct Option: A

The last two digits of the number $3^{400}$ are ...........

maths numbers and place value face value of digit large numbers general form of number

  1. $11$

  2. $91$

  3. $10$

  4. $01$

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Correct Option: D

A man has 480 rupees in the denominations of one-rupee five rupee notes and ten-rupee notes The number of notes of each denomination is equal. What is the total number of notes that he has?

maths numbers and place value face value of digit large numbers general form of number

  1. 90

  2. 75

  3. 45

  4. 60

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Correct Option: A
Explanation:

Let the numbers Of all kind is $x$

Then amount of one rupee note=$1\times x=x$
And  amount of Five rupee note=$5\times x=5x$
And  amount of Ten  rupee note=$10\times x=10x$

Then $x+5x+10x=480$
Or $16x=480$  or $x=30$

Then total numbers of note =$3\times 30=90$

Face value of '$3$' in $31005660$ is:

maths numbers and place value face value of digit large numbers general form of number

  1. $3$ crores

  2. $30$ lakhs

  3. $3$

  4. $0$

Show answer
Correct Option: C
Explanation:

Face value of a digit is the value of the digit in the number. 

So, the face value of $3$ in $31005660$ is $3$.
Hence, the answer is $3$.

How many times does the digit $1$ appear in numbers from $1$ to $100$?

maths numbers and place value face value of digit large numbers general form of number

  1. $18$

  2. $19$

  3. $20$

  4. $21$

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Correct Option: D
Explanation:

$\Rightarrow$  1–10 = 2 times

$\Rightarrow$  11–20 = 10 times
$\Rightarrow$  21–30 = 1 time
$\Rightarrow$  31–40 = 1 time
$\Rightarrow$  41–50 = 1 time
$\Rightarrow$  51–60 = 1 time
$\Rightarrow$  61–70 = 1 time
$\Rightarrow$  71–80 = 1 time
$\Rightarrow$  81–90 = 1 time
$\Rightarrow$  91–100 = 2 times
$\Rightarrow$  $Total = 2+10+1+1+1+1+1+1+1+2=21$
$\therefore$   The digit 1 appear in number from 1 to 100 is $21$.

What is the sum of all integers between $50$ and $350$ which have $1$ as the units digit?

maths numbers and place value face value of digit large numbers general form of number

  1. $5880$

  2. $5985$

  3. $6230$

  4. $6800$

Show answer
Correct Option: A
Explanation:

The sequence $51+61+71...+341$ is an arithmetic progression.

$\Rightarrow$  To find the sum of n terms of an AP we use the formula.
$\Rightarrow$  Here, $n=30,\,a=51$ and $d=10$.
$\therefore$   $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\therefore$   $S _n=\dfrac{30}{2}[2\times 51+(30-1)10]$

$\therefore$   $S _n=15[102+290]$
$\therefore$   $S _n=15\times 392$
$\therefore$   $S _n=5880$

A number consists of two digits whose sum is $11$. If $27$ is added to the number, then the digits change their places. What is the number?

maths numbers and place value face value of digit large numbers general form of number

  1. $47$

  2. $65$

  3. $83$

  4. $92$

Show answer
Correct Option: A
Explanation:

Let the ten's digit be $x$. Then, unit's digit $= \left(11 - x\right)$.
So, number $= 10x + \left(11 - x\right) = 9x + 11$.
Therefore $\left(9x + 11\right) + 27 = 10 \left(11 - x\right) + x  \Leftrightarrow  9x + 38 = 110 - 9x \Leftrightarrow   18x = 72 \Leftrightarrow   x = 4$.
Thus, ten's digit $= 4$ and unit's digit $= 7$.
Hence, required number $= 47$.

For $Z _1=\displaystyle \sqrt[6]{\frac{1-i}{1+i\sqrt{3}}}; Z _2=\sqrt[6]{\frac{1-i}{\sqrt{3}+i}}; Z _3=\sqrt[6]{\frac{1+i}{\sqrt{3}-i}}$ which of the following holds good?

maths numbers and place value face value of digit large numbers general form of number

  1. $\displaystyle\sum|Z _1|^2=\frac{3}{2}$

  2. $\displaystyle|Z _1|^4+|Z _2|^4=|Z _3|^{-8}$

  3. $\displaystyle\sum|Z _1|^3+|Z _2|^3=|Z _3|^{-6}$

  4. $|Z _1|^4+|Z _2|^4=|Z _3|^8$

Show answer
Correct Option: B
Explanation:
$z _1 =\sqrt {\dfrac {1-i}{1+i\sqrt 3}}, z _2=\sqrt {\dfrac {1-i}{\sqrt 3 +i}}, z _3=\sqrt {\dfrac {1+i}{\sqrt 3-1}}$
$z _1 =\sqrt [6]{\dfrac {1-i}{1+\sqrt 3}}=\sqrt [6]{\dfrac {(1-i)(1-i\sqrt 3)}{1+3}}=\sqrt [6]{\dfrac {1(1-\sqrt 3)-i(1+\sqrt 3)}{4}}$
$|z _1|^2 =z _1 \bar {z} _1 =\sqrt [6]{\dfrac {(1-\sqrt 3)}{4}}\times \sqrt [6]{\dfrac {(1-\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _1|^2 =\sqrt [6]{\dfrac {(1-\sqrt 3)^2 +(1+\sqrt 3)^2}{16}}$
$|z _1|^2 =\sqrt [6]{\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _2 =\sqrt [6]{\dfrac {1-i}{\sqrt 3+i}}=\sqrt [6]{\dfrac {(1-i) (\sqrt 3 -i)}{(3+1)}}=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$|z _2|^2 =z _2 \bar {z} _2=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _2|^2 =\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _3 =z _3 \bar {z} _3 =\sqrt [6]{\dfrac {(\sqrt 3-1)+(1+\sqrt 3)}{4}\times \dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$=\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2)1/6}$
$|z _1|^4 =\dfrac {1}{2^{2/6}}\quad |z _2|^4 =\dfrac {1}{2^{2/6}}$
$|z _3|^8 =\dfrac {1}{2^{4/6}}\ \Rightarrow \ |z _3|^{-8}=2^{4/6}$
$\Rightarrow \ |z _1|^4 +|z _2|^4 =\dfrac {1}{2^{2/6}}+\dfrac {1}{2^{2/6}}=\dfrac {2}{2^{2/6}}=2^{4/6}$
$=|z _3|^{-8}$
so, $\boxed {|z _1|^4 +|z _2|^4 =|z _3|^{-8}}$ as
so, option $(B)$ is right.