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,
(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 > *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)
