Problems on Numbers Questions

FACTS  AND  FORMULAE  FOR  PROBLEMS  ON  NUMBERS

I. Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number. A group of digits, denoting a number is called a numeral.

We represent a number, say 689745132 as shown below :

$\begin{array}{cccccc}Ten-Crores& Crores& Ten-Lacs& Lacs& Ten-Thousands& Thousands\\ 6& 8& 9& 7& 4& 5\end{array}\begin{array}{ccc}Hundreds& Tens& Units\\ 1& 3& 2\end{array}$

We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.

II. PlaceValue or Local Value of a Digit in a Numeral: In the above numeral : Place value of 2 is (2 x 1) = 2 Place value of 3 is (3 x 10) = 30 Place value of 1 is (1 x 100) = 100 and so on. Place value of 6 is $6×{10}^{8}$ = 600000000

III. Face Value : The face value of a digit in a numeral is the value of the digit itself at whatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3 is 3 and so on.

IV.TYPES OF NUMBERS

1. Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural numbers.

2. Whole Numbers : All counting numbers together with zero form the set of whole numbers. Thus,

(i) 0 is the only whole number which is not a natural number.

(ii) Every natural number is a whole number.

3. Integers : All natural numbers, 0 and negatives of counting numbers i.e., {…, -3,-2,-1, 0, 1, 2, 3,…..} together form the set of integers.

(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.

(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while{0, -1,-2,-3,…..} represents the set of non-positive integers.

4. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.

5. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.

6. Prime Numbers : - A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself. - Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. - Prime numbers Greater than 100 : Let p be a given number greater than 100. To find out whether it is prime or not, we use the following method : Find a whole number nearly greater than the square root of p. Let k > sqrt{p}. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime. e.g., We have to find whether 191 is a prime number or not. Now, 14 > sqrt{191}.Prime numbers less than 14 are 2, 3, 5, 7, 11, 13. 191 is not divisible by any of them. So, 191 is a prime number.

7. Composite Numbers : Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.

Note :

(i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

8. Co-primes :Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes

V. TESTS OF DIVISIBILITY

1. Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.

Ex. 84932 is divisible by 2, while 65935 is not.

2. Divisibility By 3 :A number is divisible by 3, if the sum of its digits is divisible by 3.

Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3. But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.

3. Divisibility By 4 : A number is divisible by 4, if the number formed by the last two digits is divisible by 4.

Ex. 892648 is divisible by 4, since the number formed by the last two digits is 48, which is divisible by 4. But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82, which is not divisible by 4.

4. Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5.

Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.

5. Divisibility By 6 : A number is divisible by 6, if it is divisible by both 2 and 3.

Ex. The number 35256 is clearly divisible by 2. Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.

6. Divisibility By 8 : A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8.

Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.

7. Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible by 9.

Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9. But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.

8. Divisibility By 10 : A number is divisible by 10, if it ends with 0. Ex. 96410, 10480 are divisible by 10, while 96375 is not.

9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Ex. The number 4832718 is divisible by 11, since : (sum of digits at odd places) - (sum of digits at even places) (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.

10. Divisibility By 12 : A number is divisible by 12, if it is divisible by both 4 and 3.

Ex. Consider the number 34632.

(i) The number formed by last two digits is 32, which is divisible by 4

(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3.

Hence, 34632 is divisible by 12.

11. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.

12. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.

13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.

Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.

14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.

15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.

16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.

Note : If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq. If p and q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q.

Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4 and 6 are not co-primes.

VI . MULTIPLICATION BY SHORT CUT METHODS

1. Multiplication By Distributive Law :

(i) a x (b + c) = a x b + a x c      (ii) a x (b-c) = a x b-a x c.

Ex.

(i) 567958 x 99999 = 567958 x (100000 - 1) = 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042.

(ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.

2. Multiplication of a Number By ${5}^{n}$ : Put n zeros to the right of the multiplicand and divide the number so formed by${2}^{n}$

Ex. 975436 x 625 = $975436×{5}^{4}$ = $\frac{9754360000}{{2}^{4}}$  = 609647600

VII. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM :  If we divide a given number by another number, then:Dividend = (Divisor x Quotient) + Remainder

VIII.

(i). $\left({x}^{n}-{a}^{n}\right)$ is divisible by (x - a) for all values of n

(ii). $\left({x}^{n}-{a}^{n}\right)$is divisible by (x + a) for all even values of n.

(iii). $\left({x}^{n}+{a}^{n}\right)$is divisible by (x + a) for all odd values of n.

IX. PROGRESSION A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.

1. Arithmetic Progression (A.P) : If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.

An A.P. with first term 'a' and common difference 'd' is given by a, (a + d), (a + 2d),(a + 3d),.....

1. nth term = a + (n - 1)d

2. Sum of n terms = $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$ = $\frac{n}{2}\left[a+l\right]$ , Where l is the last term.

SOME IMPORTANT RESULTS :

1. $\left(1+2+3+...+n\right)=\frac{1}{2}n\left(n+1\right)$

2. $\left({1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}\right)=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$

3.

2. Geometrical Progression (G.P) : A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression. The constant ratio is called the common ratio of the G.P.

A G.P. with first term 'a' and common ratio 'r' is : $a,ar,a{r}^{2},a{r}^{3},...$ In this G.P.

I. nth term = $a{r}^{n-1}$

II. Sum of n terms=

Q:

The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is :

 A) 100 B) 200 C) 300 D) 400

Explanation:

Let the numbers be x and y.

Then, xy = 9375 and  x/y = 15.

$xyxy=9375/15$

=>

=> y = 25

=> x = 15y = 15 x 25 = 375.

Sum of the numbers = 375 + 25 = 400.

49 68316
Q:

What is the place value of 6 in 64?

 A) 6 B) 60 C) 64 D) 10

Explanation:

We know that,

● Each digit has a fixed position called its place.

● Each digit has a value depending on its place called the place value of the digit.

● The face value of a digit for any place in the given number is the value of the digit itself

● Place value of a digit = (face value of the digit) × (value of the place).

Hence, the place value of 6 in 64 = 6 x 10 = 60.

263 61939
Q:

In a two-digit number, the digit in the unit's place is more than twice the digit in ten's place by 1. If the digits in the unit's place and the ten's place are interchanged,difference between the newly formed number and the oiginal number is less than the original number by 1. What is the original number ?

 A) 35 B) 36 C) 37 D) 39

Explanation:

Let the ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

60 47656
Q:

(3x + 2) (2x - 5) = ax² + kx + n .

What is the value of a - n + k ?

 A) 5 B) 8 C) 9 D) 10

Explanation:

This is a quadratic equation. To find n we multiply the last terms in each bracket together (2 times -5) = -10

To find ‘a’ we multiply the first terms in each bracket together (3x times 2x) = 6x2, So a = 6.

Thus a - n = 16

To find k we multiply the first term in the first bracket by the second term in the second bracket and add the result to the product of the second term in the first bracket and the first term in the second. (-15x + 4x) = -11x. So k = -11

16 - 11 = 5

6 40184
Q:

x and y are integers

x + y < 11 , and x > 6

What is the smallest possible value of x - y ?

 A) 1 B) 24 C) 4 D) -2

Explanation:

The smallest value will be obtained by giving x the smallest possible value while y gets the largest possible value. The smallest value for x = 7. This means that y will have to be less than or equal to 3 (so that x + y <11), but we want the largest value of y which is 3.

The difference, x - y will be 4.

4 38027
Q:

What approximate value will come in place of the question mark(?) in the below question?

(47% of 1442  - 36% of 1412) + 63 = ?

 A) 4 B) 5 C) 3 D) 6

Explanation:

23 34539
Q:

Three numbers are in the ratio 4 : 5 : 6 and their average is 25. The largest number is :

 A) 30 B) 40 C) 50 D) 60

Explanation:

Let the numbers be 4x, 5x and 6x, Then, (4x + 5x + 6x ) / 3 = 25

=> 5x = 25

=> x = 5.

Largest number  6x = 30.

41 29621
Q:

In a two-digit number, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is :

 A) 12 B) 24 C) 36 D) 48

Explanation:

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2

Sum of digits = x + (x + 2) = 2x + 2.

(11x + 2) (2x + 2) = 144

=> 22x^2+26x-140=0

=> (x - 2)(11x + 35) = 0

=> x = 2

Hence, Required Number = 11x + 2 = 24