1. basic concepts (1)

Basic concepts, definitions and identities
Number System
Test of divisibility:
1. A number is divisible by ‘2’ if it ends in zero or in a digit which is a multiple of
‘2’i.e. 2,4, 6, 8.
2. A number is divisible by ‘3’, if the sum of the digits is divisible by ‘3’.
3. A number is divisible by ‘4’ if the number formed by the last two digits, i.e. tens
and units are divisible by 4.
4. A number is divisible by ‘5’ if it ends in zero or 5
5. A number is divisible by ‘6’ if it divisible by ‘2’ as well as by ‘3’.
6. A number is divisible by ‘8’ if the number formed by the last three digits, i.e,
hundreds tens and units is divisible by ‘8’.
7. A number is divisible by ‘9’ if the sum of its digit is divisible by ‘9’
8. A number is divisible by ‘10’ if it ends in zero.
9. A number is divisible by ‘11’ if the difference between the sums of the digits in
the even and odd places is zero or a multiple of ‘11’.
LCM:
LCM of a given set of numbers is the least number which is exactly divisible
by every number of the given set.
HCF:
HCF of a given set of numbers is the highest number which divides exactly
every number of the given set.
LCM, HCF:
1. Product of two numbers = HCF  LCM
2. HCF of fractions =
rsdenominatoofLCM
numeratorsofHCF
3. LCM of fractions =
rsdenominatoofHCF
numeratorsofLCM
4. One number =
number2nd
HCFLCM
5. LCM of two numbers=
HCF
numberstheofProduct
6.
LCM
numberstheofProduct
HCF 

Examples to Follow:
1. The square of an odd number is always odd.
2. A number is said to be a prime number if it is divisible only by itself and unity.
Ex. 1, 2, 3, 5,7,11,13 etc.
3. The sum of two odd number is always even.
4. The difference of two odd numbers is always even.
5. The sum or difference of two even numbers is always even.
6. The product of two odd numbers is always odd.
7. The product of two even numbers is always even.
Problems:
1. If a number when divided by 296 gives a remainder 75, find the remainder when
37 divides the same number.
Method:
Let the number be ‘x’, say
x = 296k + 75, where ‘k’ is quotient when ‘x’ is divided by ‘296’
= 37  8k + 37  2 + 1
= 37(8k + 2) + 1
Hence, the remainder is ‘1’ when the number ‘x’ is divided by 37.
2. If 232
+1 is divisible by 641, find another number which is also divisible by ‘641’.
Method:
Consider 296
+1 = (232
)3
+ 13
= (232
+1)(264
-232
+1)
From the above equation, we find that 296
+1 is also exactly divisible by 641,
since it is already given that 232
+1 is exactly divisible by ‘641’.
3. If m and n are two whole numbers and if mn
= 25. Find nm
, given that n  1
Method:
mn
= 25 = 52
m = 5, n = 2
nm
= 25
= 32

4. Find the number of prime factors of 610
 717
 5527
610
 717
 5527
= 210
310
717
527
1127
The number of prime factors = the sum of all the indices viz., 10 + 10 + 17 +
27 + 27 = 91
5. A number when successively divided by 9, 11 and 13 leaves remainders 8, 9 and
8 respectively.
Method:
The least number that satisfies the condition= 8 + (99) + (8911) = 8 + 81 +
792 = 881
6. A number when divided by 19, gives the quotient 19 and remainder 9. Find the
number.
Let the number be ‘x’ say.
x = 19  19 + 9
= 361 + 9 = 370
7. Four prime numbers are given in ascending order of their magnitudes, the
product of the first three is 385 and that of the last three is 1001. Find the
largest of the given prime numbers.
The product of the first three prime numbers = 385
The product of the last three prime numbers = 1001
In the above products, the second and the third prime numbers occur in
common.  The product of the second and third prime numbers = HCF of the
given products.
HCF of 385 and 1001 = 77
Largest of the given primes =
77
1001
= 13
Square root, Cube root, Surds and Indices
Characteristics of square numbers
1. A square cannot end with an odd number of zeros
2. A square cannot end with an odd number 2, 3, 7 or 8
3. The square of an odd number is odd

4. The square of an even number is even.
5. Every square number is a multiple of 3 or exceeds a multiple of 3 by unity.
Ex.
4  4 = 16 = 5  3 + 1
5  5 = 25 = 8  3 + 1
7  7 = 49 = 16  3 + 1
6. Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.
Ex.
5  5 = 25 = 6  4 + 1
7  7 = 49 = 12  4 + 1
7. If a square numbers ends in ‘9’, the preceding digit is even.
Ex.
7  7 = 49 ‘4’ is the preceding even numbers
27  27 = 729 ‘2’ is the preceding even numbers.
Characteristics of square roots of numbers
1. If a square number ends in ‘9’, its square root is a number ending in’3’ or ‘7’.
3. If a square number ends in ‘5’, its square root is a number ending in’5’
6. If a square number ends in ‘0’, its square root is a number ending in ‘0’.
Ex.
(i)
etc371369
331089
27729
23529




(ii)
19361
31961
981
11121




(iii)
onso&452025
351225
25625




(iv)
onso&28784
321024
864
22484




(v)
onso&676
576
256
196
26
24
16
14




(vi)
onso&10000
400
100
100
20
10




THEORY OF INDICES
1. am
 an
= am + n
2. (am
)n
= amn
3. nm
n
m
a
a
a 

4. (ab)m
= am
bm
5. a0
= 1
6. p/q
a = qth root of ap
=
q p
a
7. p1
a = pth root of a
8. m
mmm
c
ba
c
ab






9. a
= 
10. a-
= 0
1.Find the square root of 6561 (Factor method)
3
9
27
81
243
729
2187
6561
3
3
3
3
3
3
3
6561 = (33)(33)(33)(33)
=(99)(99)
= 8181
6561 = 81
2. Find the least number with which you multiply 882, so that the product may be
a perfect square.
First find the factors of 882.
882 = 2  3  3  7  7
Now, 882 has factors as shown above, ‘3’ repeated twice, ‘7’ repeated twice
and ‘2’ only once. So when one more factor ‘2’ is used, then it becomes a perfect
square.
882  2 = (2  2)  (3  3)  (7  7)
The least number required is ‘2’
7
49
147
441
882
7
3
3
2

3. Find the cube root of 2985984 (Factor method)
2985984 = 23
 23
 23
 23
 33
 33
3
2985984=
2  2  2  2  3  3 = 16  9 = 144
1
3
3
9
27
81
243
729
1458
2916
5832
11664
23328
46656
93312
186624
373248
746496
1492992
2985984
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
4. Find the value of
04.04
04.04


9.0
21
19
42
38
2.4
8.3
2.04
2.04
04.04
04.04






5. 30
3
90
3.0
9
09.0
81
09.0
81

6. Simplify
4
3
3
4

32
1
32
34
2
3
3
2
4
3
3
4



16
1
410
16
1
410 =
16
6561
=
16
6561
=
4
81
=
4
1
20

6.Find the least number with which you multiply 882. So that the product may be
a perfect square.
First find the factors of 882.
7
49
147
441
882
7
3
3
2
Now, 882 has factors as shown above.
‘3’ repeated twice, ‘7’ repeated twice and ‘2’ only
once. So, when one more factor ‘2’ is used, then
it becomes a perfect square.
8822=(22)(33)(77)
The least number required is ‘2’.
7. Find the cube root of 2985984 (Factor method)
3
9
27
81
243
729
1458
2916
5832
11664
23328
46656
93312
186624
373248
746496
1492992
2985984
3
3
3
3
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2985984 = 23
23
23
23
33
33
 3
2985984 = 332222  = 916 = 144
8.Find the value of
04.04
04.04


2.04
2.04
04.04
04.04





= 9.0
21
19
42
38
242.4
8.3
 (approx)
9.
09.0
81
=
09.0
81
=
3.0
9
=
3
90
= 30
10. Simplify
3
4
-
4
3

3
4
-
4
3
=
2
3
3
2
 =
32
34 
=
32
1
11. Find the least number with which 1728 may be added so that the
resulting number is a perfect square.
164
128
16
1728
42
82
4
Note:
Take the square root of 1728 by long division method. It comes
to 41.+ something. As shown, if 128 is made 164, we get the
square root as an integer. The difference between 164 and 128
i.e., 36 must be added to 1728, so that 1764 is a perfect
square 1764 =42
Theory of Indices Problems:
1. A certain number of persons agree to subscribe as many rupees each as there
are subscribers. The whole subscription is Rs.2582449. Find the number of
subscriber?
Let the number of subscribers be x, say since each subscriber agrees to
subscribe x rupees.
The total subscription = no. of persons  subscription per person
= x  x = x2
given x2
= 2582449
x = 1607
2. Simplify:
3 43
192 ba
Use the 2 formulas
 
  mnnm
mmmm
aa
cbaabc


 3
1
433 43
ba192ba192 

     
 
 
3 4
3
1
4
3
4
3
1
2
3
4
6
3
1
43
1
3
3
1
3b4a.
3b4a.
.ab.32
.ab32
ba.192





3. Simplify 3 129
yx
Sol.
 3
1
1293 129
yxyx 
   
43
3
1
123
1
9
yx
yx


4. Find the number whose square is equal to the difference between the squares of
75.12 and 60.12
Sol.
Let ‘x’ be the number required
x2
= (75.12)2
– (60.12)2
= (75.12 + 60.12) (75.12 – 60.12)
= 135.24 15 = 2028.60
  60.2028x 45.0399
5. Reduce to an equivalent fraction write rational denominator
Sol.
  
  3535
33353
35
353





1529
2
15418
35
31515315






10
9
55
7
2
44
21
14 
Sol.
26.5
10
9
55
117
44
637
10
9
55
7
2
44
21
14  approx.

7. An army general trying to draw his 16,160 men in rows so that there are as
many men as true are rows, found that he had 31 men over. Find the number of
men in the front row.
Let ‘a’ be the number of men in the front row.
a2
+ 31 = 161610
No. of men in the front row = 127
a2
= 161610 – 31 = 16129
a = 127
8. A man plants his orchid with 5625 trees and arranges them so that there are as
many rows as there are trees in a row. How many rows are there?
Sol.
Let ‘x’ be the number of rows and let the number of trees in a row be ‘x’ say
x2
= 5625
x = 75
There are 75 trees in a row and 75 rows are arranged.

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