The aim of this topic is to help you solve the various kinds of problems on numbers. The first lesson in this topic is an introduction to problems on numbers where we have shared some important points and formulae about numbers . The exercises contain solved question and answers on Numbers. When solving exercises, we recommend you to try solving the problems on your own, then view the solution.
By the end of this post, you will learn to solve the following types of problems:
- Can you find the remainder of an + bn by a number that is equal to a+b?
- A number when successively divided by a,b and c leaves remainders x, y and z respectively. Can you find the remainders when the same number is divided by c, b and a successively?
Introduction to Numbers
Important Formulas
- (a + b)(a – b) = (a2 – b2)
- (a + b)2 = (a2 + b2 + 2ab)
- (a – b)2 = (a2 + b2 – 2ab)
- (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
- (a3 + b3) = (a + b)(a2 – ab + b2)
- (a3 – b3) = (a – b)(a2 + ab + b2)
- (a3 + b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
- When a + b + c = 0, then a3 + b3 + c3 = 3abc
Number Series
- 1 + 2 + 3 + 4 + 5 + … + n = n(n + 1)/2
- (12 + 22 + 32 + ….. + n2) = n ( n + 1 ) (2n + 1) / 6
- (13 + 23 + 33 + ….. + n3) = (n(n + 1)/ 2)2
- Sum of first n odd numbers = n2
- Sum of first n even numbers = n (n + 1)
Divisibility Rules
| Divisibility Rule | Example | |
|---|---|---|
| 2 | Number must be even | 124 is even, hence it is divisible by 2. 421 is odd, hence it is not divisible by 2 |
| 3 | Sum of the digits must be divisible by 3 | Consider 123. The sum of its digits is 1+2+3 = 6. 6 is divisible by 3. Hence 123 is divisible by 3 |
| 4 | The last two digits must be divisible by 4 | In 123496, the last two digits 96 is divisible by 4, hence 123496 is divisible by 4. |
| 5 | The last digit is either 5 or 0, then the number is divisible by 5 | 947245. Here the last digit is 5. Hence 947245 is divisible by 5. |
| 6 | The number must be divisible by both 2 and 3 | 96312. The number is even hence its divisible by 2 and the sum of its digits is 21, which is divisible by 3. Hence 96312 is divisible by 6 |
| 7 | Multiply the last digit by 2, then subtract it from the rest of the number. If the result is divisible by 7, then the number is divisible by 7 | 672. 1. Multiplying the last digit 2 by 2 gives 4. 2. Subtract 4 from the rest of the number i.e 67 - 4 = 63. Since 63 is divisible by 7. 672 is divisible by 7 |
| 8 | The last three digits must be divisible by 8 | 26032. Since the last three digits 032 is divisible by 8. The number 26032 is divisible by 8 |
| 9 | Sum of the digits must be divisible by 9 | 11106. The sum of the digits is 1+1+1+0+6 = 9, which is divisible by 9. Hence 11106 is divisible by 9. |
| 10 | If the last digit is 0, the number is divisible by 10 | 123450. Since the number ends in 0. It is divisible by 10 |
| 11 | Subtract the sum of the digits in even numbered places with the sum of the digits in odd numbered places. The result obtained must be divisible by 11 | 107415. Sum of the digits in even places is 1+7+1=9 Sum of the digits in odd places is 0+4+5 = 9 Difference in the even and odd places sum is 9-9=0, which is divisible by 11. Hence 107415 is divisible by 11 |
| 12 | Number must be divisible by both 3 and 4 | 132. Sum of the digits is 1+3+2 = 6, which is divisible by 3. Hence the number is divisible by 3. Last two digits 32 is divisible by 4. Hence the number is divisible by 4. Since the number is divisible by 3 and 4. The number is divisible by 12. |
| 13 | Multiply the last digit by 9, then subtract it from the rest of the number. If the result obtained is divisible by 13, then the number is divisible by 13 | 247. Multiply 7 by 9, which gives 63. Subtract 63 from 24, which gives -39. -39 is divisible by 13. Hence 247 is divisible by 13. |
Exercises
Problems on Numbers - Exercise - Blog
20 questions
- The difference between a two digit number and number formed by reversing its digits is 27. If the sum of the digits of the number is 9. Find the number.
- 81
- 72
- 63
- 54
Show the answer
Answer: 63
Let the digit in tens place of the number be X and digit in the units place of the number be Y. Then the number will be 10X + Y and the number obtained by reversing its digits will be 10Y + X.
From the problem,
10X + Y - (10Y+X) = 27
9X - 9Y = 27
X - Y = 3 _____ 1
Its given that sum of the digits of the number is 9.
i.e X + Y = 9 _____ 2
Solving 1 and 2 above
X = 6 and Y = 3.
Hence, the number is 63.
- The average of a two digit number and a number obtained by interchanging it digits is 33. If the product of the digits of the number is 8. Find the number.
- 81
- 42
- 80
- 26
Show the answer
Answer: 42
Let the digit in tens place of the number be X and digit in the units place of the number be Y. Then the number will be 10X + Y and the number obtained by reversing its digits will be 10Y + X.
From the problem,
$\frac{10\text{X} + Y + 10Y + \text{X}}{2}$ = 33
$\frac{11\text{X} + 11Y}{2}$ = 33
X + Y = 6 _____ 1
Given, X x Y = 8 _____ 2
Using 1 and 2,
(X-Y)$^2$ = (X+Y)$^2$ - 4XY
= 36 - 32
= 4
X - Y = 2 _____ 3
Solving for X and Y using 1 and 3,
X = 4 and Y = 2
Hence the number is 42.
- A two digit number is such that the product of its digits is 12 and when 9 is subtracted from the number the digits interchange their position. Find the number.
- 62
- 34
- 26
- 43
Show the answer
Answer: 43
Let the digit in tens place of the number be X and digit in the units place of the number be Y. Then the number will be 10X + Y and the number obtained by reversing its digits will be 10Y + X.
From the problem,
X x Y = 12 _____ 1
and
10X + Y - 9 = 10Y + X
=> X - Y = 1 _____ 2
Using 1 and 2,
(X + Y)$^2$ = (X - Y)$^2$ + 4XY
= 1 + 4 * 12
= 49
X + Y = 7 _____ 3
Solving 2 and 3,
X = 4 and Y = 3
Hence the number is 43.
- The sum of a rational number and its reciprocal is 25/12. Find the number
- $\frac{4}{3}$
- $\frac{1}{4}$
- $\frac{3}{2}$
- $\frac{2}{3}$
Show the answer
Answer: $\frac{4}{3}$
Let the number be x. Then,
x + 1/x = 25/12
12$(\text{x}^2 + 1)$*12 = 25x
12x$^2$ - 25x + 12 = 0
12x$^2$ - 16x - 9x + 12 = 0
4x(3x - 4) - 3(3x - 4) = 0
x = $\frac{3}{4}$ or x = $\frac{4}{3}$
- If the sum of squares of three consecutive numbers is 83 and sum of product of these numbers taken 2 at a time is 71. Find the smallest of the numbers.
- 4
- 5
- 7
- 9
Show the answer
Answer: 4
Let the 3 numbers be a, b and c. Then, given that
a$^2$ + b$^2$ + c$^2$ = 83 &
ab + bc + ca = 71
We have,
(a + b + c)$^2$ = a$^2$ + b$^2$ + c$^2$ + 2(ab + bc + ca)
=> 83 + 2 x 71
=> 225
a + b + c = 15
Since a, b and c are consecutive numbers, if a = x, b = x + 1, c = x + 2.
x + x + 1 + x + 2 = 15
3x = 12
x = 4
Hence the smallest of the numbers is 4.
- 247 x 253 = ?
- 62491
- 64219
- 62419
- 64129
Show the answer
Answer: 62491
247 x 253 = (250 - 3)(250 + 3)
=> 250$^2$ - 3$^2$
=> 62491
- $167^3 + 145^3 - 312^3$ = ?
- 22565240
- 22665240
- 22665220
- 22655240
Show the answer
Answer: 22665240
Let a=167, b=145, c = -312
Then, a + b + c = 167 + 145 - 312 = 0
We know that, if a + b + c = 0, then a$^3 + b$^3 - c$^3$ = 3abc
∴ $167^3 + 145^3 - 312^3$ = 3 x 167 x 145 x (-312)
= 22665240
- $65^3 + 35^3$ = ?
- 3175
- 31750
- 317500
- 3175000
Show the answer
Answer: 317500
We have,
a$^3$ + b$^3$ = (a + b)(a$^2$ - ab + b$^2$)
$65^3 + 35^3$ = (100) ($65^2$ - 65*35 + 35^2)
= 317500
- A number when divided by 77, gives 31 as remainder. What is the remainder when it is divided by 11?
- 0
- 30m
- 6
- 9
Show the answer
Answer: 9
Let N be the number. Then,
N = 77k + 31, where k = 0,1,2...
= 11(7k) + 22 + 9
= 11(7k +2) + 9
Hence, N divided by 11 leaves 9 as the remainder.
- A number when successively divided by 5, 6, and 7 leaves remainders 1, 5 and 4 respectively. Find the remainders when the same number is divided by 7, 6 and 5 successfully?
- 20, 6 and 3 respectively
- 6, 20 and 3 respectively
- 3, 3 and 20 respectively
- 20, 3 and 3 respectively
Show the answer
Answer: 20, 3 and 3 respectively
Let N be the number, then from the problem successive division by 5, 6 and 7 is given in the following steps
- N divided by 5 leaves remainder 1 and Quotient Q.
- then Q divided by 6 leaves remainder 5 and Quotient Q1.
- then Q1 divided by 7 leaves remainder 4 and Quotient Q2.Now, lets assume the Quotient Q2 to be 0. Then, quotient Q1 must be 4. So the number Q must be must be 6*4 + 5 = 29. This must be the quotient (Q) after division by 5. Original number must be 29*5 + 1 = 146
When you divide 146 by 7, remainder is 6, quotient is 20
When you divide 20 by 6, remainder is 2, quotient is 3
When you divide 3 by 5, remainder is 3
- If the number 523P741 is exactly divisible by 9. Find the smallest value of P.
- 3
- 5
- 7
- 9
Show the answer
Answer: 5
If the number is divisible by 9, then the sum of the digits must be divisible by 9. Hence,
5+2+3+P+7+4+1 => 22 + P must be divisible by 9.
Hence the smallest value of P will be 5
- What is the highest four digit number that is divisible by 46?
- 9996
- 9954
- 9976
- 9982
Show the answer
Answer: 9982
The highest four digit number is 9999.
9999 when divided by 46 leaves 17 as remainder.
Hence, the required number is 9999 - 17 = 9982
- What is the smallest value that should be added to 2376 so that the resulting sum is perfectly divisible by 74.
- 66
- 45
- 24
- 36
Show the answer
Answer: 66
2376 when divided by 74 leaves 8 as remainder.
Hence the smallest that should be added to 2376 to make it exactly divisible by 2376 is 74-8 = 66
- Find the number of 4 digit numbers that end in 75 and are divisible by 75.
- 100
- 70
- 50
- 30
Show the answer
Answer: 30
Four digit numbers ending in 75 can be written as 100k + 75, where k is a number from 10 to 99
Since 100 = 4*25, then 100k = 4*25k will be divisible by 75 if k is a multiple of 3. The number of multiples of 3 from 10 to 99 is $\frac{99}{3}$ - $\frac{10}{3}$ = 33 - 3 = 30
Hence there are 30 four digit numbers that end in 75 and are divisible by 75.
- A number when successfully divided by 3 and 5 leaves remainders 2 and 1 respectively. Find the remainder when the number is divided by 15.
- 0
- 1
- 3
- 5
Show the answer
Answer: 5
Let N be the number.
Since N leaves a remainder 2 when divided by 3
N = 3M + 2, where M is the quotient.
In the next successive division, quotient M is divided by 5, which leaves a remainder 1. Hence
M = 5L + 1, where L is the quotient obtained when M is divided by 5.
∴ N = 3(5L + 1) + 2
N = 15L + 5
Now, its evident that when N is divided by 15, it leaves a remainder 5.
- What is the remainder when 71$^{22}$ - 1 is divided by 72?
- 0
- 23
- 32
- 71
Show the answer
Answer: 0
71$^{22}$ - 1 can be written as 71$^{22}$ - 1$^{22}$, which is of the form a$^n$ - b$^n$.
a$^n$ - b$^n$ is divisible by a+b when n is even.
So, 71$^{22}$ - 1$^{22}$ is divisible by 71+1, i.e 72.
Hence the remainder when 71$^{22}$ - 1 is divided by 72 is 0.
- What is the remainder when 146$^{49}$ + 254$^{49}$ is divided by 400?
- 0
- 1
- 100
- 399
Show the answer
Answer: 0
146$^{49}$ + 254$^{49}$ is in the form a$^n$ + b$^n$.
Now, a$^n$ + b$^n$ is divisible by a+b when n is odd.
So, 146$^{49}$ + 254$^{49}$ is divisible by 146+254, i.e 400.
Hence, the remainder when 146$^{49}$ + 254$^{49}$ is divided by 400 is 0.
- What is the remainder when 2$^{33}$ + 3$^{22}$ is divided by 17?
- 1
- 10
- 16
- 17
Show the answer
Answer: 17
2$^{33}$ + 3$^{22}$ = $(2^3)^{11} + (3^2)^{11}$
= $8^{11} + 9^{11}$
Now, a$^n$ + b$^n$ is divisible by a+b when n is odd.
Hence $8^{11} + 9^{11}$ leaves a remainder 0 when divided by 8+9, i.e 17
- What is the units digit in the product of 143 x 257 x 462 x 1569?
- 2
- 4
- 6
- 8
Show the answer
Answer: 8
Units digit can be obtained by multiplying the units digit in all the numbers.
i.e, ..3 x ..7 x ..2 x ..9
= ..1 x ..8 (product of 3 and 7 gives 1 as the units digit and, product of 2 and 9 gives 8 as the units digit)
= ..8
Hence the units digit in the product of 143 x 257 x 462 x 1569 is 8
- What is the units digit of 73$^{41}$?
- 1
- 3
- 7
- 9
Show the answer
Answer: 3
For a detailed explanation on how to find the units digit of powers, click here
In 73$^{41}$, base ends in 3 and the exponent when divided by 4 leaves remainder 1.
Hence the units digit of 73$^{41}$ will be the units digit of 3$^{1}$, which is 3.
