Cheat Sheet: Number Properties | Quantitative for GMAT PDF Download

 Page 1


Cheat Sheet: Num b er Prop erties
The Num b er Systems c hapter in tro duces ho w all n um b ers—natural, whole, in tegers, r ationals,
and irrationals—fit together, teac hes y ou k ey prop erties (factors, m ultiples, divisibilit y , p arit y),
co v ers con v ersions among fractions, decimals, and p ercen ts, and giv es y ou quic k calc ulation
to ols (digit-sum tests, mo dular arithmetic, squares/cub es) to solv e problems faster and more
accurately .
Algebra F orm ulas
This section lists k ey algebraic iden tities useful for solving problems in the Num b er S ystem.
These form ulas simplify expressions and are often tested in comp etitiv e exams.
S.No. F orm ula Description Example
1 (a+b)(a-b) =a
2
-b
2
Difference of squares
iden tit y
(5+3)(5-3) = 5
2
-3
2
=
25-9 = 16
2 (a+b)
2
=a
2
+b
2
+2ab Square of a binomial sum (4+2)
2
= 4
2
+2
2
+2(4)(2) =
16+4+16 = 36
3 (a-b)
2
=a
2
+b
2
-2ab Square of a binomial
difference
(5-2)
2
= 5
2
+2
2
-2(5)(2) =
25+4-20 = 9
4 (a+b+c)
2
=
a
2
+b
2
+c
2
+2(ab+bc+ca)
Square of a trinomial (2+3+4)
2
=
2
2
+3
2
+4
2
+2(2·3+3·4+4·2) =
4+9+16+2(6+12+8) = 81
5 a
3
+b
3
= (a+b)(a
2
-ab+b
2
) Sum of cub es 2
3
+3
3
= (2+3)(2
2
-2·3+3
2
) =
5(4-6+9) = 5·7 = 35
6 a
3
-b
3
= (a-b)(a
2
+ab+b
2
) Difference of cub es 5
3
-3
3
= (5-3)(5
2
+5·3+3
2
) =
2(25+15+9) = 2·49 = 98
7 a
3
+b
3
+c
3
-3abc = (a+b+
c)(a
2
+b
2
+c
2
-ab-bc-ca)
Sum of cub es with three
terms
If a+b+c = 0 , then
a
3
+b
3
+c
3
= 3abc
8 (a+b)
n
=a
n
+
(
n
1
)
a
n-1
b+
(
n
2
)
a
n-2
b
2
+···+b
n
Binomial expansion (a+b)
2
=a
2
+2ab+b
2
Key P oin ts: These iden tities help simplify complex express ions and solv e problems e ?icien tly .
Practice applying them to n umerical examples to build confidence.
Num b er System F orm ulas
This section co v ers form ulas related to sums, factors, and prop erties of n um b ers, whic h are
essen tial for solving quan titativ e problems in the Num b er System.
S.No. F orm ula Description Example
1 1+2+···+n =
n(n+1)
2
Sum of first n natural
n um b ers
Sum of 1 to 35 =
35·36
2
= 630
1
Page 2


Cheat Sheet: Num b er Prop erties
The Num b er Systems c hapter in tro duces ho w all n um b ers—natural, whole, in tegers, r ationals,
and irrationals—fit together, teac hes y ou k ey prop erties (factors, m ultiples, divisibilit y , p arit y),
co v ers con v ersions among fractions, decimals, and p ercen ts, and giv es y ou quic k calc ulation
to ols (digit-sum tests, mo dular arithmetic, squares/cub es) to solv e problems faster and more
accurately .
Algebra F orm ulas
This section lists k ey algebraic iden tities useful for solving problems in the Num b er S ystem.
These form ulas simplify expressions and are often tested in comp etitiv e exams.
S.No. F orm ula Description Example
1 (a+b)(a-b) =a
2
-b
2
Difference of squares
iden tit y
(5+3)(5-3) = 5
2
-3
2
=
25-9 = 16
2 (a+b)
2
=a
2
+b
2
+2ab Square of a binomial sum (4+2)
2
= 4
2
+2
2
+2(4)(2) =
16+4+16 = 36
3 (a-b)
2
=a
2
+b
2
-2ab Square of a binomial
difference
(5-2)
2
= 5
2
+2
2
-2(5)(2) =
25+4-20 = 9
4 (a+b+c)
2
=
a
2
+b
2
+c
2
+2(ab+bc+ca)
Square of a trinomial (2+3+4)
2
=
2
2
+3
2
+4
2
+2(2·3+3·4+4·2) =
4+9+16+2(6+12+8) = 81
5 a
3
+b
3
= (a+b)(a
2
-ab+b
2
) Sum of cub es 2
3
+3
3
= (2+3)(2
2
-2·3+3
2
) =
5(4-6+9) = 5·7 = 35
6 a
3
-b
3
= (a-b)(a
2
+ab+b
2
) Difference of cub es 5
3
-3
3
= (5-3)(5
2
+5·3+3
2
) =
2(25+15+9) = 2·49 = 98
7 a
3
+b
3
+c
3
-3abc = (a+b+
c)(a
2
+b
2
+c
2
-ab-bc-ca)
Sum of cub es with three
terms
If a+b+c = 0 , then
a
3
+b
3
+c
3
= 3abc
8 (a+b)
n
=a
n
+
(
n
1
)
a
n-1
b+
(
n
2
)
a
n-2
b
2
+···+b
n
Binomial expansion (a+b)
2
=a
2
+2ab+b
2
Key P oin ts: These iden tities help simplify complex express ions and solv e problems e ?icien tly .
Practice applying them to n umerical examples to build confidence.
Num b er System F orm ulas
This section co v ers form ulas related to sums, factors, and prop erties of n um b ers, whic h are
essen tial for solving quan titativ e problems in the Num b er System.
S.No. F orm ula Description Example
1 1+2+···+n =
n(n+1)
2
Sum of first n natural
n um b ers
Sum of 1 to 35 =
35·36
2
= 630
1
2 1
2
+2
2
+···+n
2
=
n(n+1)(2n+1)
6
Sum of squares of first n
natural n um b ers
Sum of squares 1 to 40 =
40·41·81
6
=
20·41·27 = 22140
3 1
3
+2
3
+···+n
3
=
[
n(n+1)
2
]
2
Sum of cub es of first n
natural n um b ers
Sum of cub es 5 to 14: First 14 =
(
14·15
2
)
2
= 11025 ; First 4 =
(
4·5
2
)
2
=
100 ; 11025-100 = 10925
4 Sum of first n o dd n um b ers = n
2
Sum of first n o dd
n um b ers
Sum of o dd n um b ers 1 to 50 (n =
25 ) = 25
2
= 625 . Alternativ e:
S
25
=
25
2
·(1+49) = 625
5 Sum of first n ev en n um b ers =
n(n+1)
Sum of first n ev en
n um b ers
Sum of ev en n um b ers 2 to 50 (n =
25 ) = 25·26 = 650
6 Highest p o w er of n in m! =
?
m
n
?
+
?
m
n
2
?
+...
Highest p o w er of a prime
in factorial
Highest p o w er of 7 in 100! =
?
100
7
?
+
?
100
49
?
= 14+2 = 16
7 Num b er of zero es in n! =
Highest p o w er of 5 in n!
T railing zero es in factorial Zero es in 23! =
?
23
5
?
= 4
8 Sum of p erm utations of n
distinct digits = (n-1)!× (sum
of digits) ×(111...n times)
Sum of all n um b ers
formed b y n digits
Sum of n um b ers using 1,3,5,7 (n =
4 , sum=16): 16· 3!· 1111 = 16· 6·
1111 = 106656
9 Num b er of factors of N =
a
p
·b
q
·c
r
= (p+1)(q +1)(r+1)
Num b er of factors
F actors of 98 = 2
1
· 7
2
: (1+ 1)(2 +
1) = 2·3 = 6
10 Sum of factors of
N =
a
p+1
-1
a-1
×
b
q+1
-1
b-1
×...
Sum of all factors
Sum of factors of 98 = 2
1
·7
2
:
2
2
-1
2-1
×
7
3
-1
7-1
= 3·57 = 171
11 Pro duct of factors of N =N
a/2
Pro duct of factors
Pro duct of factors of 98 (6 factors):
98
6/2
= 98
3
= 941192
12 If N has n factors, n um b er of
factor pairs = n/2 (or (n+1)/2
if p erfect square)
Num b er of factor pairs
F or 36 (9 factors, p erfect square):
9+1
2
= 5 pairs
13 Ev en factors of
N = 2
p
·a
q
·b
r
=p(1+q)(1+r)
Num b er of ev en factors
F or 4500 = 2
2
·3
2
·5
3
: 2·(2+1)(3+
1) = 2·3·4 = 24
14 Odd factors of
N = (1+q)(1+r)
Num b er of o dd factors
F or4500 = 2
2
·3
2
·5
3
: (2+1)(3+1) =
3·4 = 12
2
Page 3


Cheat Sheet: Num b er Prop erties
The Num b er Systems c hapter in tro duces ho w all n um b ers—natural, whole, in tegers, r ationals,
and irrationals—fit together, teac hes y ou k ey prop erties (factors, m ultiples, divisibilit y , p arit y),
co v ers con v ersions among fractions, decimals, and p ercen ts, and giv es y ou quic k calc ulation
to ols (digit-sum tests, mo dular arithmetic, squares/cub es) to solv e problems faster and more
accurately .
Algebra F orm ulas
This section lists k ey algebraic iden tities useful for solving problems in the Num b er S ystem.
These form ulas simplify expressions and are often tested in comp etitiv e exams.
S.No. F orm ula Description Example
1 (a+b)(a-b) =a
2
-b
2
Difference of squares
iden tit y
(5+3)(5-3) = 5
2
-3
2
=
25-9 = 16
2 (a+b)
2
=a
2
+b
2
+2ab Square of a binomial sum (4+2)
2
= 4
2
+2
2
+2(4)(2) =
16+4+16 = 36
3 (a-b)
2
=a
2
+b
2
-2ab Square of a binomial
difference
(5-2)
2
= 5
2
+2
2
-2(5)(2) =
25+4-20 = 9
4 (a+b+c)
2
=
a
2
+b
2
+c
2
+2(ab+bc+ca)
Square of a trinomial (2+3+4)
2
=
2
2
+3
2
+4
2
+2(2·3+3·4+4·2) =
4+9+16+2(6+12+8) = 81
5 a
3
+b
3
= (a+b)(a
2
-ab+b
2
) Sum of cub es 2
3
+3
3
= (2+3)(2
2
-2·3+3
2
) =
5(4-6+9) = 5·7 = 35
6 a
3
-b
3
= (a-b)(a
2
+ab+b
2
) Difference of cub es 5
3
-3
3
= (5-3)(5
2
+5·3+3
2
) =
2(25+15+9) = 2·49 = 98
7 a
3
+b
3
+c
3
-3abc = (a+b+
c)(a
2
+b
2
+c
2
-ab-bc-ca)
Sum of cub es with three
terms
If a+b+c = 0 , then
a
3
+b
3
+c
3
= 3abc
8 (a+b)
n
=a
n
+
(
n
1
)
a
n-1
b+
(
n
2
)
a
n-2
b
2
+···+b
n
Binomial expansion (a+b)
2
=a
2
+2ab+b
2
Key P oin ts: These iden tities help simplify complex express ions and solv e problems e ?icien tly .
Practice applying them to n umerical examples to build confidence.
Num b er System F orm ulas
This section co v ers form ulas related to sums, factors, and prop erties of n um b ers, whic h are
essen tial for solving quan titativ e problems in the Num b er System.
S.No. F orm ula Description Example
1 1+2+···+n =
n(n+1)
2
Sum of first n natural
n um b ers
Sum of 1 to 35 =
35·36
2
= 630
1
2 1
2
+2
2
+···+n
2
=
n(n+1)(2n+1)
6
Sum of squares of first n
natural n um b ers
Sum of squares 1 to 40 =
40·41·81
6
=
20·41·27 = 22140
3 1
3
+2
3
+···+n
3
=
[
n(n+1)
2
]
2
Sum of cub es of first n
natural n um b ers
Sum of cub es 5 to 14: First 14 =
(
14·15
2
)
2
= 11025 ; First 4 =
(
4·5
2
)
2
=
100 ; 11025-100 = 10925
4 Sum of first n o dd n um b ers = n
2
Sum of first n o dd
n um b ers
Sum of o dd n um b ers 1 to 50 (n =
25 ) = 25
2
= 625 . Alternativ e:
S
25
=
25
2
·(1+49) = 625
5 Sum of first n ev en n um b ers =
n(n+1)
Sum of first n ev en
n um b ers
Sum of ev en n um b ers 2 to 50 (n =
25 ) = 25·26 = 650
6 Highest p o w er of n in m! =
?
m
n
?
+
?
m
n
2
?
+...
Highest p o w er of a prime
in factorial
Highest p o w er of 7 in 100! =
?
100
7
?
+
?
100
49
?
= 14+2 = 16
7 Num b er of zero es in n! =
Highest p o w er of 5 in n!
T railing zero es in factorial Zero es in 23! =
?
23
5
?
= 4
8 Sum of p erm utations of n
distinct digits = (n-1)!× (sum
of digits) ×(111...n times)
Sum of all n um b ers
formed b y n digits
Sum of n um b ers using 1,3,5,7 (n =
4 , sum=16): 16· 3!· 1111 = 16· 6·
1111 = 106656
9 Num b er of factors of N =
a
p
·b
q
·c
r
= (p+1)(q +1)(r+1)
Num b er of factors
F actors of 98 = 2
1
· 7
2
: (1+ 1)(2 +
1) = 2·3 = 6
10 Sum of factors of
N =
a
p+1
-1
a-1
×
b
q+1
-1
b-1
×...
Sum of all factors
Sum of factors of 98 = 2
1
·7
2
:
2
2
-1
2-1
×
7
3
-1
7-1
= 3·57 = 171
11 Pro duct of factors of N =N
a/2
Pro duct of factors
Pro duct of factors of 98 (6 factors):
98
6/2
= 98
3
= 941192
12 If N has n factors, n um b er of
factor pairs = n/2 (or (n+1)/2
if p erfect square)
Num b er of factor pairs
F or 36 (9 factors, p erfect square):
9+1
2
= 5 pairs
13 Ev en factors of
N = 2
p
·a
q
·b
r
=p(1+q)(1+r)
Num b er of ev en factors
F or 4500 = 2
2
·3
2
·5
3
: 2·(2+1)(3+
1) = 2·3·4 = 24
14 Odd factors of
N = (1+q)(1+r)
Num b er of o dd factors
F or4500 = 2
2
·3
2
·5
3
: (2+1)(3+1) =
3·4 = 12
2
15 P ositiv e in tegral solutions of
X
2
-Y
2
=N
V aries b y case: Case 1:
Odd, not p erfect square:
(factors of N )/2; Case 2:
Odd, p erfect square:
[(factors of N ) - 1]/2;
Case 3: Ev en, not p erfect
square: [factors of
(N /4)]/2; Case 4: Ev en,
p erfect square: {[factors of
(N /4)] - 1}/2
F orN = 135 : 8 factors, 8/2 = 4 sol.;
N = 121 : 3 factors, (3- 1)/2 = 1 ;
N = 160 : 8 factors of 40, 8/2 = 4 ;
N = 256 : 7 factors of 64, (7-1)/2 =
3
16 Num b er of digits in
a
b
=?blog
m
(a)?+1
Num b er of digits in a
n um b er
Num b er of digits in 2
10
=
?10log
10
(2)?+1˜ 4
17 Last t w o digits of
(50n±x)
2
= last t w o digits of x
2
Last t w o digits of a square
F or 268
2
, 268 = 50·5+18 , last t w o
digits = 18
2
= 324 , so 24
18 Last t w o digits of 2
10n
: o dd
n = 24 , ev en n = 76
Last t w o digits for p o w ers
of 2
F or 2
20
(n = 2 , ev en), last t w o digits
= 76
Key P oin ts: These form ulas are practical for summing sequences, finding factors, and solving
problems in v olving factorials and p erm utations. Examples mak e them easier to apply .
T yp es of Num b ers
This section defines differen t t yp es of n um b ers, whic h are fundamen tal to understanding the
Num b er System.
S.No. T yp e Definition Examples
1 Natural Num b ers P ositiv e in tegers from 1 to
infinit y
1, 2 , 3, 4, …
2 Whole Num b ers Natural n um b ers
including 0
0, 1 , 2, 3, …
3 In tegers Num b ers without decimals
(p ositiv e, negativ e, zero)
…, - 2, -1, 0, 1, 2, …
4 Real Num b ers All n um b ers on the
n um b er line
3.14,
v
2 , -5, 0
5 Rational Num b ers Num b ers of form a/b
(b?= 0 ), includes
terminating or rep eating
decimals
3/4 = 0.75 , -7/8 = -0.875 , 2/3 =
0.666...
6 Irrational Num b ers Non-rep eating,
non-terminating decimals
p = 3.14159... ,
v
2 = 1.41421...
3
Page 4


Cheat Sheet: Num b er Prop erties
The Num b er Systems c hapter in tro duces ho w all n um b ers—natural, whole, in tegers, r ationals,
and irrationals—fit together, teac hes y ou k ey prop erties (factors, m ultiples, divisibilit y , p arit y),
co v ers con v ersions among fractions, decimals, and p ercen ts, and giv es y ou quic k calc ulation
to ols (digit-sum tests, mo dular arithmetic, squares/cub es) to solv e problems faster and more
accurately .
Algebra F orm ulas
This section lists k ey algebraic iden tities useful for solving problems in the Num b er S ystem.
These form ulas simplify expressions and are often tested in comp etitiv e exams.
S.No. F orm ula Description Example
1 (a+b)(a-b) =a
2
-b
2
Difference of squares
iden tit y
(5+3)(5-3) = 5
2
-3
2
=
25-9 = 16
2 (a+b)
2
=a
2
+b
2
+2ab Square of a binomial sum (4+2)
2
= 4
2
+2
2
+2(4)(2) =
16+4+16 = 36
3 (a-b)
2
=a
2
+b
2
-2ab Square of a binomial
difference
(5-2)
2
= 5
2
+2
2
-2(5)(2) =
25+4-20 = 9
4 (a+b+c)
2
=
a
2
+b
2
+c
2
+2(ab+bc+ca)
Square of a trinomial (2+3+4)
2
=
2
2
+3
2
+4
2
+2(2·3+3·4+4·2) =
4+9+16+2(6+12+8) = 81
5 a
3
+b
3
= (a+b)(a
2
-ab+b
2
) Sum of cub es 2
3
+3
3
= (2+3)(2
2
-2·3+3
2
) =
5(4-6+9) = 5·7 = 35
6 a
3
-b
3
= (a-b)(a
2
+ab+b
2
) Difference of cub es 5
3
-3
3
= (5-3)(5
2
+5·3+3
2
) =
2(25+15+9) = 2·49 = 98
7 a
3
+b
3
+c
3
-3abc = (a+b+
c)(a
2
+b
2
+c
2
-ab-bc-ca)
Sum of cub es with three
terms
If a+b+c = 0 , then
a
3
+b
3
+c
3
= 3abc
8 (a+b)
n
=a
n
+
(
n
1
)
a
n-1
b+
(
n
2
)
a
n-2
b
2
+···+b
n
Binomial expansion (a+b)
2
=a
2
+2ab+b
2
Key P oin ts: These iden tities help simplify complex express ions and solv e problems e ?icien tly .
Practice applying them to n umerical examples to build confidence.
Num b er System F orm ulas
This section co v ers form ulas related to sums, factors, and prop erties of n um b ers, whic h are
essen tial for solving quan titativ e problems in the Num b er System.
S.No. F orm ula Description Example
1 1+2+···+n =
n(n+1)
2
Sum of first n natural
n um b ers
Sum of 1 to 35 =
35·36
2
= 630
1
2 1
2
+2
2
+···+n
2
=
n(n+1)(2n+1)
6
Sum of squares of first n
natural n um b ers
Sum of squares 1 to 40 =
40·41·81
6
=
20·41·27 = 22140
3 1
3
+2
3
+···+n
3
=
[
n(n+1)
2
]
2
Sum of cub es of first n
natural n um b ers
Sum of cub es 5 to 14: First 14 =
(
14·15
2
)
2
= 11025 ; First 4 =
(
4·5
2
)
2
=
100 ; 11025-100 = 10925
4 Sum of first n o dd n um b ers = n
2
Sum of first n o dd
n um b ers
Sum of o dd n um b ers 1 to 50 (n =
25 ) = 25
2
= 625 . Alternativ e:
S
25
=
25
2
·(1+49) = 625
5 Sum of first n ev en n um b ers =
n(n+1)
Sum of first n ev en
n um b ers
Sum of ev en n um b ers 2 to 50 (n =
25 ) = 25·26 = 650
6 Highest p o w er of n in m! =
?
m
n
?
+
?
m
n
2
?
+...
Highest p o w er of a prime
in factorial
Highest p o w er of 7 in 100! =
?
100
7
?
+
?
100
49
?
= 14+2 = 16
7 Num b er of zero es in n! =
Highest p o w er of 5 in n!
T railing zero es in factorial Zero es in 23! =
?
23
5
?
= 4
8 Sum of p erm utations of n
distinct digits = (n-1)!× (sum
of digits) ×(111...n times)
Sum of all n um b ers
formed b y n digits
Sum of n um b ers using 1,3,5,7 (n =
4 , sum=16): 16· 3!· 1111 = 16· 6·
1111 = 106656
9 Num b er of factors of N =
a
p
·b
q
·c
r
= (p+1)(q +1)(r+1)
Num b er of factors
F actors of 98 = 2
1
· 7
2
: (1+ 1)(2 +
1) = 2·3 = 6
10 Sum of factors of
N =
a
p+1
-1
a-1
×
b
q+1
-1
b-1
×...
Sum of all factors
Sum of factors of 98 = 2
1
·7
2
:
2
2
-1
2-1
×
7
3
-1
7-1
= 3·57 = 171
11 Pro duct of factors of N =N
a/2
Pro duct of factors
Pro duct of factors of 98 (6 factors):
98
6/2
= 98
3
= 941192
12 If N has n factors, n um b er of
factor pairs = n/2 (or (n+1)/2
if p erfect square)
Num b er of factor pairs
F or 36 (9 factors, p erfect square):
9+1
2
= 5 pairs
13 Ev en factors of
N = 2
p
·a
q
·b
r
=p(1+q)(1+r)
Num b er of ev en factors
F or 4500 = 2
2
·3
2
·5
3
: 2·(2+1)(3+
1) = 2·3·4 = 24
14 Odd factors of
N = (1+q)(1+r)
Num b er of o dd factors
F or4500 = 2
2
·3
2
·5
3
: (2+1)(3+1) =
3·4 = 12
2
15 P ositiv e in tegral solutions of
X
2
-Y
2
=N
V aries b y case: Case 1:
Odd, not p erfect square:
(factors of N )/2; Case 2:
Odd, p erfect square:
[(factors of N ) - 1]/2;
Case 3: Ev en, not p erfect
square: [factors of
(N /4)]/2; Case 4: Ev en,
p erfect square: {[factors of
(N /4)] - 1}/2
F orN = 135 : 8 factors, 8/2 = 4 sol.;
N = 121 : 3 factors, (3- 1)/2 = 1 ;
N = 160 : 8 factors of 40, 8/2 = 4 ;
N = 256 : 7 factors of 64, (7-1)/2 =
3
16 Num b er of digits in
a
b
=?blog
m
(a)?+1
Num b er of digits in a
n um b er
Num b er of digits in 2
10
=
?10log
10
(2)?+1˜ 4
17 Last t w o digits of
(50n±x)
2
= last t w o digits of x
2
Last t w o digits of a square
F or 268
2
, 268 = 50·5+18 , last t w o
digits = 18
2
= 324 , so 24
18 Last t w o digits of 2
10n
: o dd
n = 24 , ev en n = 76
Last t w o digits for p o w ers
of 2
F or 2
20
(n = 2 , ev en), last t w o digits
= 76
Key P oin ts: These form ulas are practical for summing sequences, finding factors, and solving
problems in v olving factorials and p erm utations. Examples mak e them easier to apply .
T yp es of Num b ers
This section defines differen t t yp es of n um b ers, whic h are fundamen tal to understanding the
Num b er System.
S.No. T yp e Definition Examples
1 Natural Num b ers P ositiv e in tegers from 1 to
infinit y
1, 2 , 3, 4, …
2 Whole Num b ers Natural n um b ers
including 0
0, 1 , 2, 3, …
3 In tegers Num b ers without decimals
(p ositiv e, negativ e, zero)
…, - 2, -1, 0, 1, 2, …
4 Real Num b ers All n um b ers on the
n um b er line
3.14,
v
2 , -5, 0
5 Rational Num b ers Num b ers of form a/b
(b?= 0 ), includes
terminating or rep eating
decimals
3/4 = 0.75 , -7/8 = -0.875 , 2/3 =
0.666...
6 Irrational Num b ers Non-rep eating,
non-terminating decimals
p = 3.14159... ,
v
2 = 1.41421...
3
7 Complex Num b ers Num b ers of form a+bi (i
= imaginary unit)
3+4i , -2-i
8 Imaginary Num b ers Square ro ots of negativ e
n um b ers
v
-1 =i
9 Ev en Num b ers Num b ers divisible b y 2 2, 6 , 8, 14
10 Odd Num b ers Num b ers not divisible b y 2 3, 7 , 9, 15
11 Prime Num b ers Num b ers > 1 with exactly
t w o factors (1 and itself )
2, 3 , 5, 7, 11
12 Comp osite Num b ers Num b ers > 1 that are not
prime
4, 6 , 8, 10
Key P oin ts: Kno wing these t yp es helps classify n um b ers in problems and apply relev an t prop-
erties, lik e divisibilit y or prime factorization.
Divisibilit y R ules
This section pro vides rules to quic kly c hec k if a n um b er is divisible b y certain v alues, sa ving
time in calculations.
Divisor R ule Example
2 Last digit is ev en (0, 2, 4, 6, 8) 7248 (last digit 8) is divisible b y 2
3 Sum of digits divisible b y 3 123 (1+2+3=6) is divisible b y 3
4 Last t w o digits divisible b y 4 o r 00 7248 (48 ÷ 4 = 12) is divisible b y 4
5 Last digit is 0 or 5 235 (last digit 5) is divisible b y 5
6 Divisible b y b oth 2 and 3 36 (ev en and 3+6=9) is divisibl e b y 6
7 T wice last digit subtracted fr om rest
is divisible b y 7
343 (34 - 2· 3 = 28, divis ible b y 7)
8 Last three digits divisible b y 8 or 000 7248 (248 ÷ 8 = 31) is divisible b y 8
9 Sum of digits divisible b y 9 998 (9+9+8=26, not divisible b y 9 )
10 Last digit is 0 1230 (last digit 0) is divisibl e b y 10
11 Difference of sum of o dd and ev en
p osition digits is 0 or divisible b y 11
1782 (o dd: 1+8=9, ev en: 7+2=9,
9-9=0) is divisible b y 11
12 Divisible b y b oth 3 and 4 72 (72 ÷ 3 = 24, 72 ÷ 4 = 18) is
divisible b y 12
13 F our times last digit added to rest,
rep eat un til t w o-digit n um b er divisible
b y 13
104 (10 + 4· 4 = 26, divisible b y 13)
4
Page 5


Cheat Sheet: Num b er Prop erties
The Num b er Systems c hapter in tro duces ho w all n um b ers—natural, whole, in tegers, r ationals,
and irrationals—fit together, teac hes y ou k ey prop erties (factors, m ultiples, divisibilit y , p arit y),
co v ers con v ersions among fractions, decimals, and p ercen ts, and giv es y ou quic k calc ulation
to ols (digit-sum tests, mo dular arithmetic, squares/cub es) to solv e problems faster and more
accurately .
Algebra F orm ulas
This section lists k ey algebraic iden tities useful for solving problems in the Num b er S ystem.
These form ulas simplify expressions and are often tested in comp etitiv e exams.
S.No. F orm ula Description Example
1 (a+b)(a-b) =a
2
-b
2
Difference of squares
iden tit y
(5+3)(5-3) = 5
2
-3
2
=
25-9 = 16
2 (a+b)
2
=a
2
+b
2
+2ab Square of a binomial sum (4+2)
2
= 4
2
+2
2
+2(4)(2) =
16+4+16 = 36
3 (a-b)
2
=a
2
+b
2
-2ab Square of a binomial
difference
(5-2)
2
= 5
2
+2
2
-2(5)(2) =
25+4-20 = 9
4 (a+b+c)
2
=
a
2
+b
2
+c
2
+2(ab+bc+ca)
Square of a trinomial (2+3+4)
2
=
2
2
+3
2
+4
2
+2(2·3+3·4+4·2) =
4+9+16+2(6+12+8) = 81
5 a
3
+b
3
= (a+b)(a
2
-ab+b
2
) Sum of cub es 2
3
+3
3
= (2+3)(2
2
-2·3+3
2
) =
5(4-6+9) = 5·7 = 35
6 a
3
-b
3
= (a-b)(a
2
+ab+b
2
) Difference of cub es 5
3
-3
3
= (5-3)(5
2
+5·3+3
2
) =
2(25+15+9) = 2·49 = 98
7 a
3
+b
3
+c
3
-3abc = (a+b+
c)(a
2
+b
2
+c
2
-ab-bc-ca)
Sum of cub es with three
terms
If a+b+c = 0 , then
a
3
+b
3
+c
3
= 3abc
8 (a+b)
n
=a
n
+
(
n
1
)
a
n-1
b+
(
n
2
)
a
n-2
b
2
+···+b
n
Binomial expansion (a+b)
2
=a
2
+2ab+b
2
Key P oin ts: These iden tities help simplify complex express ions and solv e problems e ?icien tly .
Practice applying them to n umerical examples to build confidence.
Num b er System F orm ulas
This section co v ers form ulas related to sums, factors, and prop erties of n um b ers, whic h are
essen tial for solving quan titativ e problems in the Num b er System.
S.No. F orm ula Description Example
1 1+2+···+n =
n(n+1)
2
Sum of first n natural
n um b ers
Sum of 1 to 35 =
35·36
2
= 630
1
2 1
2
+2
2
+···+n
2
=
n(n+1)(2n+1)
6
Sum of squares of first n
natural n um b ers
Sum of squares 1 to 40 =
40·41·81
6
=
20·41·27 = 22140
3 1
3
+2
3
+···+n
3
=
[
n(n+1)
2
]
2
Sum of cub es of first n
natural n um b ers
Sum of cub es 5 to 14: First 14 =
(
14·15
2
)
2
= 11025 ; First 4 =
(
4·5
2
)
2
=
100 ; 11025-100 = 10925
4 Sum of first n o dd n um b ers = n
2
Sum of first n o dd
n um b ers
Sum of o dd n um b ers 1 to 50 (n =
25 ) = 25
2
= 625 . Alternativ e:
S
25
=
25
2
·(1+49) = 625
5 Sum of first n ev en n um b ers =
n(n+1)
Sum of first n ev en
n um b ers
Sum of ev en n um b ers 2 to 50 (n =
25 ) = 25·26 = 650
6 Highest p o w er of n in m! =
?
m
n
?
+
?
m
n
2
?
+...
Highest p o w er of a prime
in factorial
Highest p o w er of 7 in 100! =
?
100
7
?
+
?
100
49
?
= 14+2 = 16
7 Num b er of zero es in n! =
Highest p o w er of 5 in n!
T railing zero es in factorial Zero es in 23! =
?
23
5
?
= 4
8 Sum of p erm utations of n
distinct digits = (n-1)!× (sum
of digits) ×(111...n times)
Sum of all n um b ers
formed b y n digits
Sum of n um b ers using 1,3,5,7 (n =
4 , sum=16): 16· 3!· 1111 = 16· 6·
1111 = 106656
9 Num b er of factors of N =
a
p
·b
q
·c
r
= (p+1)(q +1)(r+1)
Num b er of factors
F actors of 98 = 2
1
· 7
2
: (1+ 1)(2 +
1) = 2·3 = 6
10 Sum of factors of
N =
a
p+1
-1
a-1
×
b
q+1
-1
b-1
×...
Sum of all factors
Sum of factors of 98 = 2
1
·7
2
:
2
2
-1
2-1
×
7
3
-1
7-1
= 3·57 = 171
11 Pro duct of factors of N =N
a/2
Pro duct of factors
Pro duct of factors of 98 (6 factors):
98
6/2
= 98
3
= 941192
12 If N has n factors, n um b er of
factor pairs = n/2 (or (n+1)/2
if p erfect square)
Num b er of factor pairs
F or 36 (9 factors, p erfect square):
9+1
2
= 5 pairs
13 Ev en factors of
N = 2
p
·a
q
·b
r
=p(1+q)(1+r)
Num b er of ev en factors
F or 4500 = 2
2
·3
2
·5
3
: 2·(2+1)(3+
1) = 2·3·4 = 24
14 Odd factors of
N = (1+q)(1+r)
Num b er of o dd factors
F or4500 = 2
2
·3
2
·5
3
: (2+1)(3+1) =
3·4 = 12
2
15 P ositiv e in tegral solutions of
X
2
-Y
2
=N
V aries b y case: Case 1:
Odd, not p erfect square:
(factors of N )/2; Case 2:
Odd, p erfect square:
[(factors of N ) - 1]/2;
Case 3: Ev en, not p erfect
square: [factors of
(N /4)]/2; Case 4: Ev en,
p erfect square: {[factors of
(N /4)] - 1}/2
F orN = 135 : 8 factors, 8/2 = 4 sol.;
N = 121 : 3 factors, (3- 1)/2 = 1 ;
N = 160 : 8 factors of 40, 8/2 = 4 ;
N = 256 : 7 factors of 64, (7-1)/2 =
3
16 Num b er of digits in
a
b
=?blog
m
(a)?+1
Num b er of digits in a
n um b er
Num b er of digits in 2
10
=
?10log
10
(2)?+1˜ 4
17 Last t w o digits of
(50n±x)
2
= last t w o digits of x
2
Last t w o digits of a square
F or 268
2
, 268 = 50·5+18 , last t w o
digits = 18
2
= 324 , so 24
18 Last t w o digits of 2
10n
: o dd
n = 24 , ev en n = 76
Last t w o digits for p o w ers
of 2
F or 2
20
(n = 2 , ev en), last t w o digits
= 76
Key P oin ts: These form ulas are practical for summing sequences, finding factors, and solving
problems in v olving factorials and p erm utations. Examples mak e them easier to apply .
T yp es of Num b ers
This section defines differen t t yp es of n um b ers, whic h are fundamen tal to understanding the
Num b er System.
S.No. T yp e Definition Examples
1 Natural Num b ers P ositiv e in tegers from 1 to
infinit y
1, 2 , 3, 4, …
2 Whole Num b ers Natural n um b ers
including 0
0, 1 , 2, 3, …
3 In tegers Num b ers without decimals
(p ositiv e, negativ e, zero)
…, - 2, -1, 0, 1, 2, …
4 Real Num b ers All n um b ers on the
n um b er line
3.14,
v
2 , -5, 0
5 Rational Num b ers Num b ers of form a/b
(b?= 0 ), includes
terminating or rep eating
decimals
3/4 = 0.75 , -7/8 = -0.875 , 2/3 =
0.666...
6 Irrational Num b ers Non-rep eating,
non-terminating decimals
p = 3.14159... ,
v
2 = 1.41421...
3
7 Complex Num b ers Num b ers of form a+bi (i
= imaginary unit)
3+4i , -2-i
8 Imaginary Num b ers Square ro ots of negativ e
n um b ers
v
-1 =i
9 Ev en Num b ers Num b ers divisible b y 2 2, 6 , 8, 14
10 Odd Num b ers Num b ers not divisible b y 2 3, 7 , 9, 15
11 Prime Num b ers Num b ers > 1 with exactly
t w o factors (1 and itself )
2, 3 , 5, 7, 11
12 Comp osite Num b ers Num b ers > 1 that are not
prime
4, 6 , 8, 10
Key P oin ts: Kno wing these t yp es helps classify n um b ers in problems and apply relev an t prop-
erties, lik e divisibilit y or prime factorization.
Divisibilit y R ules
This section pro vides rules to quic kly c hec k if a n um b er is divisible b y certain v alues, sa ving
time in calculations.
Divisor R ule Example
2 Last digit is ev en (0, 2, 4, 6, 8) 7248 (last digit 8) is divisible b y 2
3 Sum of digits divisible b y 3 123 (1+2+3=6) is divisible b y 3
4 Last t w o digits divisible b y 4 o r 00 7248 (48 ÷ 4 = 12) is divisible b y 4
5 Last digit is 0 or 5 235 (last digit 5) is divisible b y 5
6 Divisible b y b oth 2 and 3 36 (ev en and 3+6=9) is divisibl e b y 6
7 T wice last digit subtracted fr om rest
is divisible b y 7
343 (34 - 2· 3 = 28, divis ible b y 7)
8 Last three digits divisible b y 8 or 000 7248 (248 ÷ 8 = 31) is divisible b y 8
9 Sum of digits divisible b y 9 998 (9+9+8=26, not divisible b y 9 )
10 Last digit is 0 1230 (last digit 0) is divisibl e b y 10
11 Difference of sum of o dd and ev en
p osition digits is 0 or divisible b y 11
1782 (o dd: 1+8=9, ev en: 7+2=9,
9-9=0) is divisible b y 11
12 Divisible b y b oth 3 and 4 72 (72 ÷ 3 = 24, 72 ÷ 4 = 18) is
divisible b y 12
13 F our times last digit added to rest,
rep eat un til t w o-digit n um b er divisible
b y 13
104 (10 + 4· 4 = 26, divisible b y 13)
4
14 Divisible b y b oth 2 and 7 28 (ev en and 28 ÷ 7 = 4) is divisible
b y 14
16 Last four digits divisible b y 16 123456 (3456 ÷ 16 = 2 16) is divisible
b y 16
27 Sum of blo c ks of 3 digits (righ t to left)
divisible b y 27
123456 (123 + 456 = 579, not
divisible b y 27)
Key P oin ts: These rules help quic kly determine divisibilit y without p erforming full divis ion,
whic h is useful for factoring and simplifying calculations.
HCF and LCM
This section explains ho w to find the Highest Common F actor (HCF) and Least Com mon
Multiple (LCM) of n um b ers, along with related prop erties. The explanations for finding HCF
and LCM b y prime factorization and LCM b y long division ha v e b een simplified for clarit y .
S.No. Concept F orm ula/R ule Example
1 HCF × LCM = Pro duct
of t w o n um b ers
F or t w o n um b ers a and b ,
their HCF m ultiplied b y
their LCM equals their
pro duct
F or 42 and 70: HCF = 14, LCM
= 210, 42·70 = 14·210 = 2940
2 If N is divisible b y X and
Y with HCF(X , Y ) = 1
If a n um b er N is divisible
b y t w o n um b ers X and Y ,
and X and Y share no
common factors other
than 1, then N is divisible
b y their pro duct X·Y
If N is divisible b y 3 and 5
(HCF=1), N is d ivisible b y 15
3 HCF b y prime
factorization
Break do wn eac h n um b er
in to its prime factors (lik e
2, 3, 5). T ak e the common
prime factors with their
lo w est p o w ers and
m ultiply them to get the
HCF
F or 96 = 2
5
·3
1
, 36 = 2
2
·3
2
,
18 = 2
1
·3
2
: Common factors are
2
1
and 3
1
, so HCF = 2
1
·3
1
= 6
4 LCM b y prime
factorization
Break do wn eac h n um b er
in to its prime factors.
T ak e eac h prime factor
with its highest p o w er
across all n um b ers and
m ultiply them to get the
LCM
F or 96 = 2
5
·3
1
, 36 = 2
2
·3
2
,
18 = 2
1
·3
2
: Highest p o w ers are
2
5
and 3
2
, so LCM =
2
5
·3
2
= 32·9 = 288
5