1. NUMBERS
IMPORTANT FACTS AND
FORMULAE
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 :
Ten Crores (108)
|
Crores(107)
|
Ten Lacs (Millions) (106)
|
Lacs(105)
|
Ten Thousands (104)
|
Thousands (103)
|
Hundreds (102)
|
Tens(101)
|
Units(100)
|
6
|
8
|
9
|
7
|
4
|
5
|
1
|
3
|
2
|
We read it as : 'Sixty-eight crores,
ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.
II Place Value 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 x 108
= 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,
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 : Letp 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 > *jp. 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 > V191.
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 arid 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) ax(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 5n
: Put n zeros to the right of the
multiplicand and divide the number so formed by 2n
Ex. 975436 x
625 = 975436 x 54=
9754360000 = 609647600
16
VII. BASIC FORMULAE
1. (a + b)2
= a2 + b2 + 2ab 2. (a - b)2 =
a2 + b2 - 2ab
3. (a + b)2
- (a - b)2 = 4ab 4. (a + b)2 +
(a - b)2 = 2 (a2 + b2)
5. (a2 - b2) = (a + b) (a
- b)
6. (a + b + c)2 = a2 + b2
+ c2 + 2 (ab + bc + ca)
7. (a3 + b3) = (a +b) (a2
- ab + b2) 8. (a3
- b3) = (a - b) (a2 + ab + b2)
9. (a3
+ b3 + c3 -3abc) = (a + b + c) (a2 + b2
+ c2 - ab - bc - ca)
10. If a + b
+ c = 0, then a3 + b3 + c3 = 3abc.
VIII. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM
If we divide
a given number by another number, then :
Dividend = (Divisor x Quotient) + Remainder
IX. {i) (xn - an ) is
divisible by (x - a) for all values of n.
(ii) (xn - an)
is divisible by (x + a) for all even values of n.
(iii) (xn + an)
is divisible by (x + a) for all odd values of n.
X.
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),.....
The nth
term of this A.P. is given by Tn =a (n - 1) d.
The sum of
n terms of this A.P.
Sn
= n/2 [2a + (n - 1) d] = n/2 (first
term + last term).
SOME
IMPORTANT RESULTS :
(i) (1 + 2 + 3 +…. + n) =n(n+1)/2
(ii) (l2
+ 22 + 32 + ... + n2) = n (n+1)(2n+1)/6
(iii) (13 + 23 + 33 +
... + n3) =n2(n+1)2
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, ar2,
In this G.P.
Tn = arn-1
sum of the n
terms, Sn= a(1-rn)
(1-r)
SOLVED EXAMPLES
Ex. 1. Simplify
: (i) 8888 + 888 + 88 + 8
(ii) 11992 -
7823 - 456
Sol. i )
8888 ii)
11992 - 7823 - 456 = 11992 - (7823 + 456)
888
= 11992 - 8279 = 3713-
88 7823 11992
+ 8
+ 456 - 8279
9872
8279 3713
Ex. 2,
What value will replace the question mark in each of the following equations ?
(i) ? -
1936248 = 1635773 (ii) 8597 -
? = 7429 - 4358
Sol. (i) Let x
- 1936248=1635773.Then, x = 1635773 + 1936248=3572021. (ii) Let 8597 - x = 7429 -
4358.
Then, x =
(8597 + 4358) - 7429 = 12955 - 7429 = 5526.
Ex. 3. What could be the maximum value of Q in
the following equation? 5P9 +
3R7 + 2Q8 = 1114
Sol.
We may analyse the given equation as shown : 1 2
Clearly, 2 +
P + R + Q = ll.
5 P 9
So, the
maximum value of Q can be
3 R 7
(11 - 2)
i.e., 9 (when P = 0, R = 0); 2 Q 8
11 1 4
Ex. 4.
Simplify : (i) 5793405 x 9999 (ii)
839478 x 625
Sol.
i)5793405x9999=5793405(10000-1)=57934050000-5793405=57928256595.b
ii) 839478 x
625 = 839478 x 54 = 8394780000 = 524673750.
16
Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 - 983 x 107
Sol.
(i) 986 x 137
+ 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
(ii) 983 x
207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.
Ex. 6.
Simplify : (i) 1605 x 1605 ii) 1398 x
1398
Sol.
i) 1605 x
1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2
+ 2 x 1600 x 5
= 2560000 + 25 + 16000 = 2576025.
(ii) 1398 x
1398 - (1398)2 = (1400 - 2)2= (1400)2 + (2)2
- 2 x 1400 x 2
=1960000 + 4 - 5600 = 1954404.
Ex. 7.
Evaluate : (313 x 313 + 287 x 287).
Sol.
(a2 + b2) = 1/2 [(a + b)2
+ (a- b)2]
(313)2
+ (287)2 = 1/2 [(313 + 287)2 + (313 - 287)2] =
½[(600)2 + (26)2]
= 1/2 (360000
+ 676) = 180338.
Ex. 8.
Which of the following are prime numbers ?
(i)
241 (ii) 337 (Hi) 391 (iv) 571
Sol.
(i) Clearly,
16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241 is not divisible by any one of
them.
241 is a prime number.
(ii) Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5,
7, 11,13,17.
337 is not divisible by any one of
them.
337 is a prime number.
(iii) Clearly, 20 > Ö39l". Prime numbers less than 20 are 2,
3, 5, 7, 11, 13, 17, 19.
We find that 391 is
divisible by 17.
391 is not prime.
(iv) Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5,
7, 11, 13, 17, 19, 23.
571 is not divisible by
any one of them.
571 is a prime number.
Ex. 9. Find the unit's digit in the product
(2467)163 x (341)72.
Sol.
Clearly, unit's digit in the given product = unit's digit in 7153 x
172.
Now, 74 gives unit digit 1.
7152 gives unit digit 1,
\ 7153
gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence, unit's digit in the product = (7
x 1) = 7.
Ex. 10.
Find the unit's digit in (264)102 + (264)103
Sol.
Required unit's digit = unit's digit in (4)102 + (4)103.
Now, 42 gives unit digit 6.
\(4)102 gives unjt digit 6.
\(4)103 gives unit digit of the product (6 x 4) i.e., 4.
Hence, unit's digit in (264)m + (264)103
= unit's digit in (6 + 4) = 0.
Ex. 11. Find the total number of prime factors
in the expression (4)11 x (7)5 x (11)2.
Sol.
(4)11x (7)5 x (11)2 = (2 x 2)11 x
(7)5 x (11)2 = 211 x 211 x75x
112 = 222 x 75 x112
Total number of prime factors = (22 + 5
+ 2) = 29.
Ex.12.
Simplify : (i) 896 x 896 - 204 x 204
(ii) 387 x 387 +
114 x 114 + 2 x 387 x 114
(iii) 81 X 81 +
68 X 68-2 x 81 X 68.
Sol.
(i) Given exp
= (896)2 - (204)2
= (896 + 204) (896 - 204) = 1100 x 692 = 761200.
(ii) Given
exp = (387)2+ (114)2+
(2 x 387x 114)
= a2 + b2
+ 2ab, where a = 387,b=114
= (a+b)2 = (387 + 114 )2
= (501)2 = 251001.
(iii) Given
exp = (81)2 + (68)2 – 2x 81 x 68 = a2 + b2
– 2ab,Where a =81,b=68
= (a-b)2 = (81 –68)2 =
(13)2 = 169.
Ex.13. Which of the following numbers is divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.
(i) Sum of
digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.
Hence, 541326
is divisible by 3.
(ii) Sum of
digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.
Hence,
5967013 is not divisible by 3.
Ex.14.What
least value must be assigned to * so that the number 197*5462 is r 9 ?
Sol.
Let the
missing digit be x.
Sum of digits
= (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).
For (34 + x)
to be divisible by 9, x must be replaced by 2 .
Hence, the
digit in place of * must be 2.
Ex. 15.
Which of the following numbers is divisible by 4 ?
(i)
67920594 (ii) 618703572
Sol.
(i) The
number formed by the last two digits in the given number is 94, which is not
divisible by 4.
Hence,
67920594 is not divisible by 4.
(ii) The
number formed by the last two digits in the given number is 72, which is
divisible by 4.
Hence,
618703572 is divisible by 4.
Ex. 16.
Which digits should come in place of * and $ if the number 62684*$ is divisible
by both 8 and 5 ?
Sol.
Since the
given number is divisible by 5, so 0 or 5 must come in place of $. But, a
number ending with 5 is never divisible by 8. So, 0 will replace $.
Now, the
number formed by the last three digits is 4*0, which becomes divisible by 8, if
* is replaced by 4.
Hence, digits
in place of * and $ are 4 and 0 respectively.
Ex. 17.
Show that 4832718 is divisible by 11.
Sol. (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.
Hence, 4832718 is divisible by 11.
Ex. 18. Is
52563744 divisible by 24 ?
Sol. 24 = 3 x 8, where 3 and 8 are co-primes.
The sum of the digits in the given
number is 36, which is divisible by 3. So, the given number is
divisible by 3.
The number formed by the last 3 digits of
the given number is 744, which is
divisible by 8. So, the given number is divisible by 8.
Thus, the given number is divisible by
both 3 and 8, where 3 and 8 are co-primes.
So, it is
divisible by 3 x 8, i.e., 24.
Ex. 19.
What least number must be added to 3000 to obtain a number exactly divisible by
19 ?
Sol. On
dividing 3000 by 19, we get 17 as remainder.
\Number to be added = (19 - 17) = 2.
Ex. 20.
What least number must be subtracted from 2000 to get a number exactly
divisible by 17 ?
Sol. On
dividing 2000 by 17, we get 11 as remainder.
\Required number to be
subtracted = 11.
Ex. 21. Find the number which is nearest to 3105 and is
exactly divisible by 21.
Sol. On
dividing 3105 by 21, we get 18 as remainder.
\Number to be added to 3105
= (21 - 18) - 3.
Hence, required number = 3105 + 3 =
3108.
Ex. 22.
Find the smallest number of 6 digits which is exactly divisible by 111.
Sol.
Smallest number of 6 digits is 100000.
On
dividing 100000 by 111, we get 100 as remainder.
\Number to be added = (111 - 100) - 11.
Hence, required number = 100011.-
Ex. 23. On dividing 15968 by a certain number, the
quotient is 89 and the remainder is 37. Find the divisor.
Dividend - Remainder 15968-37
Sol. Divisor = -------------------------- =
------------- = 179.
.Quotient 89
Ex. 24. A
number when divided by 342 gives a remainder 47. When the same number ift
divided by 19, what would be the remainder ?
Sol. On dividing the given number by 342, let
k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k
+ 19 x 2 + 9) = 19 (18k + 2) + 9.
\The
given number when divided by 19, gives (18k + 2) as quotient and 9 as
remainder.
Ex. 25. A
number being successively divided by 3, 5 and 8 leaves remainders 1, 4
and 7
respectively. Find the respective remainders if the order of divisors be reversed,
Sol.
3
|
X
|
|
5
|
y
|
- 1
|
8
|
z
|
- 4
|
1
|
- 7
|
\z =
(8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1)
= 238.
Now,
8
|
238
|
|
5
|
29
|
- 6
|
3
|
5
|
- 4
|
1
|
- 9,
|
\Respective
remainders are 6, 4, 2.
Ex. 26. Find the remainder when 231 is divided
by 5.
Sol. 210 = 1024. Unit digit of 210
x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
\Unit digit of 231 is 8.
Now, 8 when divided by 5, gives 3
as remainder.
Hence, 231 when divided by 5, gives
3 as remainder.
Ex. 27.
How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are 14, 21, 28, 35, ....
77, 84.
This is an A.P. with a = 14 and d = (21
- 14) = 7.
Let it contain n terms.
Then, Tn = 84 =>
a + (n - 1) d = 84
=> 14 + (n - 1) x 7 = 84 or n = 11.
\Required number of terms =
11.
Ex. 28. Find the sum of all odd numbers upto 100.
Sol. The given numbers
are 1, 3, 5, 7, ..., 99.
This is an A.P. with a = 1 and d = 2.
Let it contain n terms. Then,
1 + (n - 1) x 2 = 99 or n = 50.
\Required sum = n
(first term + last term)
2
= 50 (1
+ 99) = 2500.
2
Ex. 29. Find the sum of all 2 digit numbers divisible by
3.
Sol. All 2 digit numbers divisible by 3
are :
12, 51, 18, 21, ..., 99.
This is an A.P. with a = 12 and d = 3.
Let it contain n terms. Then,
12 + (n - 1) x 3 = 99 or n = 30.
\Required sum = 30 x
(12+99) = 1665.
2
Ex.30.How
many terms are there in 2,4,8,16……1024?
Sol.Clearly
2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.
Let the number of terms be n . Then
2 x 2n-1 =1024 or 2n-1
=512 = 29.
\n-1=9 or n=10.
Ex. 31. 2
+ 22 + 23 + ... + 28 = ?
Sol.
Given series is a G.P. with a = 2, r = 2 and n = 8.
\sum = a(rn-1) = 2 x (28 –1)
= (2 x 255) =510
(r-1) (2-1)