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.
{…, -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 > *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
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)
