Quantitative Aptitude

Number System and Digits

380 Questions

Number system and digits questions test the ability to manipulate numbers, identify significant digits, and form specific values. These problems often require finding missing digits or determining the properties of large sums. They form a vital component of the quantitative aptitude section.

Number formationMissing digitsSignificant digitsLargest and smallest numbersDigit sum properties

Number System and Digits Questions

Multiple choice maths hcf-lcm introduction to multiples multiples lcm

The greatest number with four digits which when divided by $3, 5, 7, 9$ leaves the remainders $1, 3, 5, 7$ respectively, is _______.

  1. $9763$
  2. $9673$
  3. $9367$
  4. $9969$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since on dividing by $3$ the remainder is $1$, the sum of digits of the number must add upto a number, dividing which by $3$ we get remainder $1$

Since on dividing by $5$ remainder is $3$, unit place digit has to be either $3$ or $8$
Since on dividing by $9$ remainder is $7$, sum of digits should give remainder $7$ when divided by $9$ 
Only options (A), (B) satisfy these criteria
Since (A) is bigger, we divide it by $7$ and find remainder which turns out to be $5$. 

Multiple choice maths combinatorics and mathematical induction fundamental principle of addition fundamental principles of counting principles of counting

The number of all three digit even number such that if $3$ is one of the digits, then next digit is $5$, is 

  1. $359$
  2. $360$
  3. $365$
  4. $380$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We need even numbers so the last digit must be $0, 2, 4, 6, 8$. So $5$ possibilities
Now the ten's place can be any number between $0$ to $9$ except $3$ because in that case, the $3$ is to be followed by $5$ in the units place then it won't be an even no., so there are $9$ possibilities.
Now in hundred's place can be anything between $1$ to $9$ except $3$ because if $3$ is present then the next digit must be $5$ so there are $8$ possibilities.
Now if there is $3$ in hundred's place then $5$ will be in ten's place and the numbers must be $350, 352, 354, 356, 358$ which will be treated seperately. So $5$ possibilities
Therefore, 
Total no. of even numbers $= 8 \times 9 \times 5 + 5 = 365$
Multiple choice maths concepts of seven and eight digit numbers comparison of numbers comparing numbers operations on rational numbers indian place value chart largest and smallest numbers writing and expanding numbers

Smallest $6$-digit number that can be formed using $9,2,6,0,3,1$ (using each digit only once) is _________ .

  1. $012369$
  2. $102369$
  3. $106239$
  4. $103269$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the smallest digit start arranging the numbers in ascending order.

However, $0$ cannot be the first or else the number would become $5$ digit.
Therefore, $102369$ is the correct answer.

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

What is the unit digit in ${({6374}^{1793}\times {625}^{317}\times{341}^{491})}$?

  1. $0$
  2. $2$
  3. $3$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Unit digit in ${6374}^{1993}=$ Unit digit in ${(4)}^{1793}$
=Unit digit in $[{({4}^{2})}^{896}\times 4]$
=Unit digit in $(6\times 4)=4$
Unit digit in ${(625)}^{317}=$ Unit digit in ${(5)}^{317}=5$
Unit digit in ${(341)}^{491}=$ Unit digit in ${(1)}^{491}=1$
Required digit$=$ Unit digit in $(4\times 5\times1)=0$

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

The smallest odd number formed by using the digits $1,0,3,4$ and $5$ is

  1. $10345$
  2. $10453$
  3. $10543$
  4. $10534$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The smallest odd number using digits $1,0,3,4,5$


$\rightarrow $ We have five digits and we have to make smallest five digit odd numbers.


$\rightarrow$ So, the number cannot start with $0$

$\rightarrow$ For the smallest it should be start with $1$

$\rightarrow$ and second space should be $0$

    $1\\ \overline { 1st } $  $0\\ \overline { 2nd }$  $\;\\ \overline { 3rd } $  $\;\\ \overline { 4rt } $  $\;\\ \overline { 5th } $

$\rightarrow$ Now two space are filled and $3$ are left.

$\rightarrow$ For smallest third place for should be $3$ 

          $\underline { 1 } \underline { 0 } \underline { 3 } \underline {  } \underline {  } $

$\rightarrow $ Now two places are left for and no. should be odd so, last digit should be $5$

So, the number $=10345.$

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

  1. $192$
  2. $375$
  3. $400$
  4. $720$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

we have the number $0,1,2,3,5,7$

Now the digit should be odd and hence last digit should be 
filled with an odd number 
$ \Rightarrow $ Number of way to filled last number $= 4$ $({\text{i}}{\text{.e}}{\text{. }}1,3,5,7)$
$ \Rightarrow $ Number of way to filled third digit  $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled second digit $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled first digit $= 5$ $({\text{i}}{\text{.e}.}1,2,3,5,7)$
$ \Rightarrow $ Total 4digit number $=4\times6\times6\times5$
$= 720$
hence,
Opton $D$ is correct answer.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

The difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$. What is the difference between the two digits of number.

  1. $3$
  2. $2$
  3. $1$

  4. Cannot be determined

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the unit's digit be $y$ and ten's digit be $x$.

Then, the number $= 10x + y$. When we interchange the digits, the number will be $10y + x$.

Now, it is given that the difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$, therefore, we have:

$(10x+y)−(10y+x)=9\\ \Rightarrow 9x-9y=9\\ \Rightarrow 9(x-y)=9\\ \Rightarrow x-y=\frac { 9 }{ 9 } \\ \Rightarrow x-y=1$

Hence, the difference between the two digits of number is $1$.
Multiple choice maths decimal fractions rounding decimals non-terminating recurring decimals in rational numbers rounding of decimals estimation and rounding off estimations, bounds and rounding off

While rounding off, if the digit to be dropped is less than $5$, then the preceding digit:

  1. increases by $1$

  2. remains unchanged
  3. decreases by $1$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
When rounding, you examine the digit following (i.e., to the right of) the digit that is to be the last digit in the rounded off number. The digit you are examining is the first digit to be dropped.

  1. If that first digit to be dropped is less than $5$ (that is, $1, 2, 3 $ or $4$), drop it, and also drop all the digits to the right of it.
  2. If that first digit to be dropped is more than $5$ (that is, $6, 7, 8$ or $9$), increase by 1 the number to be rounded, that is, the preceeding digit (to the digit being dropped).
Thus, in our case, since the digit to be dropped is less than $5$, we make no change to the preceding digit. 
Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

Find the unit digit of ${3^{46}} + 125 \times 436 + 256 \times {7^{345}}$

  1. $1$
  2. $3$
  3. $7$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Find the unit digit of each term: 3^46 has a cycle of 4 for powers (46 mod 4 = 2, so 3^2 ends in 9). The middle term ends in 0 because 125 * 436 is an even number times 5. The last term ends in 6 * 7^345 (345 mod 4 = 1, so 7^1 ends in 7, and 6 * 7 ends in 2). Summing the unit digits: 9 + 0 + 2 = 11, which ends in 1.

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

Find the last two digits of $3^{1997}$.

  1. $67$
  2. $63$
  3. $80$
  4. $56$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is same as asking what is remainder when $3^{1997}\div 100$
$3^{4}\equiv 81  mod  100$
$3^{8}\equiv 61  mod  100$
$3^{12}\equiv 41  mod  100$
$3^{16}\equiv 21  mod  100$
$3^{20}\equiv 1  mod  100$


Now, $3^{40}, 3^{60}, 3^{80}, 3^{100}, ...., 3^{1980}$ all are $\equiv 1  mod  100$

We know $3^{16}\equiv 21  mod  100$

$3^{17}\equiv 21\times 3  mod  100$

$3^{17}\equiv 63  mod  100$

$\therefore 3^{1997}\equiv 3^{1980}\times 3^{17}$

since, $3^{1980}\equiv 1  mod  100$

and $3^{17}\equiv 63  mod  100$

$\therefore 3^{1997}\equiv 63  mod  100$

$\therefore $ Last two digit is 63

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

A two digit number in such that the product of its digits is $8$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.

  1. 18

  2. 72

  3. 27

  4. 81
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the digit at unit's place $= x$
$\displaystyle \therefore $ Digit at ten's place $\displaystyle \frac { 8 }{ x } $ and the number is $\displaystyle \left( \frac { 80 }{ x } +x \right) $
New number on interchanging the places of digits $\displaystyle =10x+\frac { 8 }{ x } $
$\displaystyle \therefore $ According to given condition
$\displaystyle \frac { 80 }{ x } +x-63=10x+\frac { 8 }{ x } $
$\displaystyle 80+{ x }^{ 2 }-63x=10{ x }^{ 2 }+8$
$\displaystyle { 9x }^{ 2 }+63x-72=0$
$\displaystyle { x }^{ 2 }-7x-8=0$
$\displaystyle { x }^{ 2 }+8x-x-8=0$
$\displaystyle x\left( x+8 \right) -1\left( x+8 \right) =0$
$\displaystyle \left( x+8 \right) \left( x-1 \right) =0$
$\displaystyle i.e.\quad x=-8$  and $ x=1$
Rejecting $\displaystyle x=-8$ and putting $\displaystyle x=1$ the required no. is $\displaystyle \left( \frac { 80 }{ 1 } +1 \right) =81$.

Multiple choice maths square and square root scientific notation use of exponents power of 10

The digit in the ten's place of a two-digit number is three times that in the one's places if the digits are reversed the new number will be 36 less than the original number Find the number 

  1. 64

  2. 52

  3. 62

  4. 42
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the digits be $ x $ and $ y $
Given, "The digit in the ten's place of a two-digit number is three times that in the one's places "
$ => x = 3y $ 

Now, when the digits are reversed, the number will be $ 10y + x $
Also,  if the digits are reversed the new number will be $ 36 $ less than the original number. $ => 10y + x = (10x + y) - 36 $
$ => 9x -9y = 36 $

Putting $ x = 3y $ in this,
$ 9(3y) -9y = 36 $
$ => 27y - 9y = 36 $
$ 18y = 36 => y = 2 $

So, $ x = 3y = 6 $
Hence, the number is $ 62 $