NUMBER SYSTEM1
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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.
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) NonPositive and
NonNegative Integers: 0
is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of
nonnegative integers, while {0, 1,2,3,…..} represents the set of
nonpositive 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 up to 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.
Coprimes: Two
numbers a and b are said to be coprimes, if their H.C.F. is 1. e.g., (2, 3),
(4, 5), (7, 9), (8, 11), etc. are coprimes
Basic
Formulas:
1.
(a + b)^{2} = a^{2}^{ }+ b^{2} + 2ab
2.
(a – b)^{2}^{ }= a^{2}^{ }+ b^{2} – 2ab
3.
(a + b)^{2} – (a
– b)^{2} =
4ab
4.
(a + b)^{2}^{ }+ (a
– b)^{2} = 2
(a^{2} + b^{2})
5.
(a^{2}^{ } b^{2})
= (a + b) (a – b)
6.
(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca)
7.
(a^{3} + b^{3})
= (a +b) (a^{2} – ab
+ b^{2})
8.
(a^{3} – b^{3})
= (a – b) (a^{2} + ab
+ b^{2})
9.
(a^{3} + b^{3} + c^{3} 3abc) = (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)
10.
If a + b + c = 0, then a^{3} + b^{3}^{ }+ c^{3} = 3abc.
Natural numbers (Positive integers) : 1, 2, 3, 4,....
Whole numbers
(Nonnegative integers) :
0, 1, 2, 3,....
Negative
integers: − 1, −
2, − 3,....
Integers: ...., −2, − 1, 0,1, 2,.....
Even numbers: ...., − 2, 0, 2, 4,.... (2n)
Odd numbers: ....., −3, −1,1, 3,... (2n +
1)
Prime
numbers (exactly 2 factors) : 2, 3, 5,
7, 11,....
Composite
numbers (more than 2 factors) : 4, 6, 8, 9,10,....
Perfect
numbers (Sum of all the factors is twice the number) : 6, 28, 496, ..
Coprimes
(relative primes) (Two numbers whose HCF is 1) : 2 & 3, 8 & 9,..
Twin primes (Two prime
numbers whose difference is 2) : 3 & 5, 5 & 7,.
Rational
numbers (qp form, p & q are integers, q ≠ 0 ) : 32, 23 , 2, 0.5,..
Irrational
numbers (which cannot be represented in the form of a fraction)
:2, 35, e, π , 0.231764735...)
:2, 35, e, π , 0.231764735...)
Decimal Numbers: 0.2, 1.25, 0.3333….
Terminating
Decimal Numbers (which terminates): 0.23, 2.374, ….
Non
Terminating Decimal Numbers (Which doesn’t terminate): 0.33…., 0.121212…, 0.2317…
Pure
Recurring Decimals (All the figures after decimal point repeats) : 0.33…., 0.121212…
Mixed
Recurring Decimals (Some figures after decimal repeats): 0.245555…, 0. 2343434…
Pure
recurring decimal to fraction conversion
Ex. 0.ababab ….. = 99ab
Mixed
recurring decimal to fraction conversion
Ex. 0.abcbcbc … = 990aabc−
→ 1 is
the neither prime, nor composite.
→ 2 is the only even prime.
→ If x & y are two integers, then (x + y) ! is divisible by x !. y!
→ The product of ‘n’ consecutive numbers is divisible by n!.
→ (xn + yn) is divisible by (x + y), when n is an odd number.
→ (xn – yn) is divisible by (x + y)(x – y), when n is an even number.
→ (xn – yn) is divisible by (x – y), when n is an odd number.
→ 2 is the only even prime.
→ If x & y are two integers, then (x + y) ! is divisible by x !. y!
→ The product of ‘n’ consecutive numbers is divisible by n!.
→ (xn + yn) is divisible by (x + y), when n is an odd number.
→ (xn – yn) is divisible by (x + y)(x – y), when n is an even number.
→ (xn – yn) is divisible by (x – y), when n is an odd number.
9.
Some Important points:
→ Every
number ‘N’ can be written as N = ap × bq × cr …. . (a, b, c,…. are prime
numbers.)
→ If p,
q, r ……. are even, ‘N’ is a perfect square.
→ If p,
q, r are multiples of 3, ‘N’ is a perfect cube.
→
Number of factors of N = (p+ 1) (q + 1) (r + 1) …..
→ Sum
of the factors of N = ((a^(p+1)1)/(a1))((b^(q+1)1)/(b1)).........
→
Number of co – primes of ‘N’ , which are less than N = N (1 – 1/a) (1 – 1/b)….
→ Sum
of these coprimes = N/2 × N (1 – 1/a) (1 – 1/b)….
→
Numbers of ways of writing ‘N’ as a product of 2 coprimes = 2 n – 1 , n is the
number of different prime numbers in ‘N’
→ If n
is a prime number, (n –1)! +1 is divisible by n.
→ If n
is a natural number and p is a prime number, then (np –n) is divisible by p
→ The
last digit of the powers of 2,3,7,8 repeats after every 4th power.
→ The
last digit of any power of 0,1, 5,6 is always 0,1,5, 6 respectively.
→ The
last digit of the powers of 4 and 9 repeats after every 2nd power.
→ The
last two digits of any number is the remainder obtained by dividing that number
by 100.
 Decimal Fractions:
Fractions in which denominators are powers of 10 are known as decimal
fractions.
Thus,

1

= 1 tenth = .1;

1

= 1 hundredth =
.01;

10

100

99

= 99 hundredths =
.99;

7

= 7 thousandths =
.007, etc.;

100

1000

 Conversion of a Decimal into Vulgar Fraction:
Put 1 in the denominator under the decimal point and annex
with it as many zeros as is the number of digits after the decimal point. Now,
remove the decimal point and reduce the fraction to its lowest terms.
Thus, 0.25 =

25

=

1

;
2.008 =

2008

=

251

.

100

4

1000

125

 Annexing Zeros and Removing Decimal Signs:
Annexing zeros to the extreme right of a decimal fraction
does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.
If numerator and denominator of a fraction contain the
same number of decimal places, then we remove the decimal sign.
Thus,

1.84

=

184

=

8

.

2.99

299

13

 Operations on Decimal Fractions:
 Addition
and Subtraction of Decimal Fractions: The given numbers are so placed under each
other that the decimal points lie in one column. The numbers so arranged
can now be added or subtracted in the usual way.
 Multiplication
of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as
many places as is the power of 10.
Thus, 5.9632 x 100 = 596.32; 0.073 x 10000 = 730.
 Multiplication
of Decimal Fractions: Multiply the given numbers considering them without decimal
point. Now, in the product, the decimal point is marked off to obtain as
many places of decimal as is the sum of the number of decimal places in
the given numbers.
Suppose we have to find the product (.2 x 0.02 x .002).
Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) =
6.
.2 x .02 x .002 = .000008
 Dividing
a Decimal Fraction By a Counting Number: Divide the given number without considering
the decimal point, by the given counting number. Now, in the quotient,
put the decimal point to give as many places of decimal as there are in
the dividend.
Suppose we have to find the quotient (0.0204 Ã· 17). Now,
204 Ã· 17 = 12.
Dividend contains 4 places of decimal. So, 0.0204 Ã· 17 =
0.0012
 Dividing
a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor
by a suitable power of 10 to make divisor a whole number.
Now, proceed as above.
Thus,

0.00066

=

0.00066 x 100

=

0.066

= .006

0.11

0.11 x 100

11

 Comparison of Fractions:
Suppose some fractions are to be arranged in ascending or
descending order of magnitude, then convert each one of the given fractions in
the decimal form, and arrange them accordingly.
Let us to arrange
the fractions

3

,

6

and

7

in descending
order.

5

7

9

Now,

3

= 0.6,

6

= 0.857,

7

= 0.777...

5

7

9

Since, 0.857 >
0.777... > 0.6. So,

6

>

7

>

3

.

7

9

5

 Recurring Decimal:
If in a decimal fraction, a figure or a set of figures is
repeated continuously, then such a number is called a recurring decimal.
n a recurring decimal, if a single figure is repeated, then it
is expressed by putting a dot on it. If a set of figures is repeated, it is
expressed by putting a bar on the set.
Thus,

1

= 0.333... = 0.3;

22

=
3.142857142857.... = 3.142857.

3

7

Pure Recurring Decimal: A decimal fraction, in which all the figures after the
decimal point are repeated, is called a pure recurring decimal.
Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once
in the numerator and take as many nines in the denominator as is the number of
repeating figures.
Thus, 0.5
=

5

; 0.53 =

53

; 0.067 =

67

, etc.

9

99

999

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat
and some of them are repeated, is called a mixed recurring decimal.
Eg. 0.1733333.. = 0.173.
Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the
difference between the number formed by all the digits after decimal point
(taking repeated digits only once) and that formed by the digits which are not
repeated. In the denominator, take the number formed by as many nines as there
are repeating digits followed by as many zeros as is the number of
nonrepeating digits.
Thus, 0.16
=

16  1

=

15

=

1

; 0.2273
=

2273  22

=

2251

.

90

90

6

9900

9900

 Some Basic Formulae:
 (a + b)(a  b) =
(a^{2} + b^{2})
 (a + b)^{2} = (a^{2}
+ b^{2} + 2ab)
 (a  b)^{2} = (a^{2}
+ b^{2}  2ab)
 (a + b + c)^{2}
= a^{2} + b^{2} + c^{2} + 2(ab
+ bc + ca)
 (a^{3} + b^{3})
= (a + b)(a^{2}  ab + b^{2})
 (a^{3}  b^{3})
= (a  b)(a^{2} + ab + b^{2})
 (a^{3} + b^{3}
+ c^{3}  3abc) = (a + b + c)(a^{2}
+ b^{2} + c^{2}  ab  bc  ac)
 When a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc.
 Laws of Indices:
 a^{m} x a^{n} = a^{m}^{
+ n}
a^{m}

= a^{m}^{
 n}

a^{n}

 (a^{m})^{n} = a^{mn}
 (ab)^{n} = a^{n}b^{n}

a


n

=

a^{n}

b

b^{n}

 a^{0} = 1
Surds:
Let a be rational number and n be a positive
integer such that a^{(1/n)} = a
Then, a is called a surd of order n.
 Laws of Surds:
 a = a^{(1/n)}
 ab = a x b
 (a)^{m} = a^{m}
Square of numbers in 100
> Choose a number over 100 (START
WITH SMALLER NUMBER).
> The last two places will be the square of
the last two digits (keep if any carry) _ _ _ X X.
> The first three places will be the number plus
the last two digits plus any carry: X X X _ _.
Here is an Example:
> let the number be 108:
2. Square the last two digits (no carry): 8 × 8 = 64: _ _ _ 64
3. Add the last two digits to the number: 108 + 08= 116:
so 1 1 6 _ _
4. So 108 × 108 = 11664.
Steps to find square of numbers
in 200's
> Choose a number in the 200s (start with numbers under 210, then try for larger ones).
>The first digit of the square is 4: 4 _ _ _ _
> The next two digits will be 4 times the last 2 digits: _ X X _ _
> The last two places will be the square of the last digit: _ _ _ X X
here we take an Example:
> let the number be 207:
> Choose a number in the 200s (start with numbers under 210, then try for larger ones).
>The first digit of the square is 4: 4 _ _ _ _
> The next two digits will be 4 times the last 2 digits: _ X X _ _
> The last two places will be the square of the last digit: _ _ _ X X
here we take an Example:
> let the number be 207:
> The first digit is 4
so 4 _ _ _ _
> The next two digits are 4 times the last digit:
4 × 7 = 28
so _ 2 8 _ _
Square the last digit: 7× 7 = 49
so _ _ _ 49
So finally we get 206 × 206 = 42849.
and For larger numbers work right to left:
> Square the last two digits (keep the carry): _ _ _ X X
> 4 times the last two digits + carry: _ X X _ _
> Square the first digit + carry: X _ _ _ _
example
>If the number to be squared is 225:
> Square last two digits (keep carry):
25x25 = 625 (keep 6): _ _ _ 2 5
> 4 times the last two digits + carry:
4x25 = 100; 100+6 = 106 (keep 1): _ 0 6 _ _
> Square the first digit + carry:
2x2 = 4; 4+1 = 5: 5 _ _ _ _
> So 225 × 225 = 50625.
for
this example 37:
> Look for the nearest 10 boundary
> In this case up 3 from 37 to 40.
> Since you went UP 3 to 40 go DOWN 3 from 37 to 34.
> Now mentally multiply 34x40
> The way I do it is 34x10=340;
> Double it mentally to 680
> Double it again mentally to 1360
> This 1360 is the FIRST interim answer.
> 37 is "3" away from the 10 boundary 40.
> Square this "3" distance from 10 boundary.
> 3x3=9 which is the SECOND interim answer.
> Add the two interim answers to get the final answer.
> Answer: 1360 + 9 = 1369
> Look for the nearest 10 boundary
> In this case up 3 from 37 to 40.
> Since you went UP 3 to 40 go DOWN 3 from 37 to 34.
> Now mentally multiply 34x40
> The way I do it is 34x10=340;
> Double it mentally to 680
> Double it again mentally to 1360
> This 1360 is the FIRST interim answer.
> 37 is "3" away from the 10 boundary 40.
> Square this "3" distance from 10 boundary.
> 3x3=9 which is the SECOND interim answer.
> Add the two interim answers to get the final answer.
> Answer: 1360 + 9 = 1369
Square
of 3 digit number
LET THE NUMBER BE ABC
SQ (ABC) is calculated like this
STEP 1. Last digit = last digit of SQ(C)
SQ (ABC) is calculated like this
STEP 1. Last digit = last digit of SQ(C)
STEP 2. Second Last Digit = 2*B*C + any carryover from STEP 1.
STEP 3. Third Last Digit 2*A*C+ Sq(B) + any carryover from STEP2.
STEP 4. Fourth last digit is 2*A*B + any carryover from STEP 3.
STEP 5 . In the beginning of result will be Sq(A) + any carryover
from Step 4.
EXAMPLE:
SQ (431)
STEP 1. Last digit = last digit of SQ(1) =1
STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP1.= 6
STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP
2.= 2*4*1 +9= 17. so 7 and 1 carryover
STEP 4. Fourth last digit is 2*4*3 + any carryover (which is 1) . =
24+1=25. So 5 and carry over 2.
STEP 5 . In the beginning of result will be Sq(4) + any carryover
from Step 4. So 16+2 =18.
So the result will be 185761.
Squaring the numbers:
1.
Squaring the numbers ending with 5.
35^{2}
=
Ø
Separate the 5 from the digits in front. in
this case there is only a 3 in front of the 5. Add 1 to 3 get 4 (3+1= 4)
Ø
Multiply these numbers together: 3 x 4 = 12
Ø
Write the square of 5 (25) after 12. We will
get 1225.
135 ^{2} = ??
Ø Take 13, add
1 to it we will get 14.
Ø Then 13 x 14
= 182
Ø Add the
square of 5 next to it. We will get 18225.
SQUARING THE NUMBERS
NEAR TO 50:
1.
46^{2 }=
Ø
Forty six squared means 46 x 46. Rounding
upwards, 50 x 50 = 2500.
Ø
Take 50 and 2500 as our reference points.
50 46^{2}
^{ } 4
46 = 504,
so 4 is a minus number.
Ø
So we take 4 from the 25 hundreds.
Ø
(254) x
100= 2100
Ø
To get the rest of the answer, we square the
number in the minus. ( 4 ^{2}= 16)
Ø
Add 2100 and 16 we will get 2116 is the answer.
56^{2} =
Ø
Fifty six squared means 56 x 56. Rounding
upwards, 50 x 50 = 2500.
Ø
Take 50 and 2500 as our reference points.
50 56^{2}
^{ }+6
56 = 50+4,
so 6 is a positive number.
Ø
So we add 6 to 25 hundreds.
Ø
(25+6) x
100= 3100
Ø
To get the rest of the answer, we square the
number in the minus. ( 6^{2}= 36)
Ø
Add 3100 and 36 we will get 3136 is the answer.
SQUARING NUMBERS NEAR
TO 500:
Ø This is
similar to our strategy for squaring numbers near 50.
Ø Five hundred
times 500 is 250000, we take 500 and 250000 as our reference number.
1.
506^{2} =
500 506^{2}
^{ }+6
500^{2}
= 250000
Ø
Five
hundred and six is greater than 500,
Ø
Square of 500 is 250000
Ø
The number 6 is added to the thousands
Ø
(250+6) x1000 = 256000
Ø
Square 6 is 36.
Ø
256000+36 =256036 is the answer.
Square the number ends
with 1:
1.
31^{2} =
Ø
First, subtract 1 from the number. The number
now ends in zero and should be easy to square. (30^{2 } = 3 x 3 x 10 x 10) = 900
Ø
Add 30 and its next number 31 (30+31). We will
get 61)
Ø
Add (900 + 61) = 961.
2.
351^{2}=?
Ø
350^{2} = 122500
Ø
350 +351 = 701
Ø
122500 +701 = 123201
We can also use the method for squaring numbers
ending in 1 for those ending in 6.
3.
86^{2}
=
Ø 85^{2} =7225
Ø 85+86 = 171
Ø 7225+171 = 7396
Squaring
numbers ending with 9
1.
29^{2}
=
Ø Add 1 to the number. The number now ends in
zero and is easy to square.
Ø 30^{2} = 900
Ø Now add 30 with given number29 (30+29 =59)
Ø Then sub (900  59 = 841)
2.
349^{2}
=
Ø 350^{2} = 122500
Ø 350+349 = 699
Ø Sub (122500  699 = 121801)
Square a 2 Digit Number Ending in 5
For this example we
will use 25
• Take the
"tens" part of the number (the 2 and add 1)=3
• Multiply the
original "tens" part of the number by the new number (2x3)
• Take the result
(2x3=6) and put 25 behind it. Result the answer 625.
Try a few more 75
squared ... = 7x8=56 ... put 25 behind it is 5625.
55 squared = 5x6=30
... put 25 behind it ... is 3025. Another easy one! Practice it on paper first!
SINGLE STEP:
35 square
(3x[3+1]) / (5x5) = 12 / 25 = 1225
(3x[3+1]) / (5x5) = 12 / 25 = 1225
Square 2 Digit Number: UPDOWN
Method
Square a 2 Digit
Number, for this example 37:
• Look for the
nearest 10 boundary
• In this case up 3
from 37 to 40.
• Since you went UP
3 to 40 go DOWN 3 from 37 to 34.
• Now mentally
multiply 34x40
• The way I do it
is 34x10=340;
• Double it
mentally to 680
• Double it again
mentally to 1360
• This 1360 is the
FIRST interim answer.
• 37 is
"3" away from the 10 boundary 40.
• Square this
"3" distance from 10 boundary.
• 3x3=9 which is
the SECOND interim answer.
• Add the two
interim answers to get the final answer.
• Answer: 1360 + 9
= 1369
ADDITION
The basic
rule for mental addition:
To add 9,
add 10 and subtract 1: to add 8, add 10 and subtract 2; to add 7 add 10 and
subtract 3, and so on.
Ø If you wanted to add 47, you would add 50 and
subtract 3,
Ø To add 196, add 200 and subtract 4.
Ø To add 38 to a number, add 40 and subtract 2,
TWO DIGIT
MENTAL ADDITIONS:
If the
unit’s digit is high, round off to the next ten and then subtract the
difference. If the units’ digit is low, add the tens then the units.
Ø With two digit mental addition you add the tens
digit of each number first, then the units. If the unit’s digit is high, round
off the number upwards and then subtract the difference. If you are adding47,
add 50, and then subtract 3.
Ø To add 35, 67, and 43 together you would begin
with 34, add 70 to get 105, subtract 3 to get 102, add 10 to get 142 then the 3
to get your answer of 145.
ADDING THREE
DIGIT NUMBERS:
355+752+694 =?
Ø 355+700 = 1055
Ø 1055+50+2 = 1107
Ø 1107+7006 = 18076 = 1801
OR
Ø You may prefer to add from left to right; adding
the hundreds first, then the tens and then the units.
ADDITING
LARGER NUMBERS:
8461
+5678
Ø We begin with the thousands column.8+5 = 13,
since we are dealing with thousands, our answer is 13 thousand.
Ø Observe that the numbers in the hundreds column
conveniently add to 10, so that gives us another thousand. Then answer is
14000.
Ø Then add 61 to 14000, we getting 14061.
Ø Add 80 to and subtract 2. To add 80 add 100 and
subtract 20, (14061+100202) = 14161202=141412=14139 is the answer.
Ø An easy rule is: when adding a column of
numbers add pairs of digits to make tens first, then add the other digits.
SUBTRACTION:
To subtract mentally, try and round off the number you are
subtracting and then correct the answer.
To subtract 9, take 10 and add 1: to subtract 8, take 10 and
add 2; to subtract 7, take 10 and add 3,
1. Eg: 569 =
1
(To take 9 from 56 in your head, the easiest and fastest method
is to subtract 10, (46) and add1 we get 47.)
2.
5438 = 16
+2
Ø 4440, plus 2
makes 16
3.
43687 =
+13
Ø Take 100 to get
336. Add 13 and we will get 349 easy.
Ø SUBTRACTING ONE NUMBER BELOW A HUNDREDS VALUE FROM
ANOTHER WHICH IS JUST ABOVE THE SAME HUNDREDS NUMBERS.
Three digit SUBTRACTIONS:
1.
461 275 =
25
161+25 = 160+20+5+1 = 186
2.
834 – 286 =
14
534+14 = 530+10+4+4 = 540+8 = 548
Subtraction method
one:
1.
72543897 =
6
1 4
7
2 5 4
3 3
9 7
3
3 5 7
Ø Subtract 7 from 4. We can’t, so we borrow 1
from the tens column.
Ø Cross out the 5 and write 4.
Ø Don’t say 7 from 14, we have to say 7 from 10
and add 4 we getting 3+4 = 7 ( the first digit of the answer)
Ø Nine from 4 won’t go, so borrow again. Nine
from 10 is 1, plus 4, the next digit answer.
Ø Eight from 1 won’t go, so borrow again. Eight
from 10 is 2, plus 1 is 3, three from 6 is 3, the final digit of the answer.
Subtraction method two:
7
2 5 4
3
8 9 7
3
3 5 7
Ø Subtract 7 from 4. We can’t, so we borrow 1
from the tens column. Put a 1 in front of the 4 to make 14 and write a small 1
alongside the 9 in the tens column.
Don’t say 7 from 14, but 7 from 10, add 4 on top gives 7, the first
digit of the number.
Ø Ten ( 9+1) from 5 won’t go so borrow again in a
similar fashion. Ten from 15 is 5 or 10 is zero, plus 5 is 5.
Ø Nine from 2 won’t go, so borrow again. Nine
from 10 is 1 plus 2 is 3.
Ø Four from 7 is 3. You have your answer.
Subtraction from a power of 10:
The rule is
: SUBTRACT THE UNITS DIGIT FROM 10, THEN EACH SUCCESSIVE DIGIT FROM 9, THEN
SUBTRACT 1 FROM THE DIGIT ON THE LEFT.
1.
1000
574
Ø 104=6,
Ø 9  7 = 2,
Ø 9  5 = 4,
Ø 1  1 = 0
The answer is 0426
Subtracting smaller numbers:
If
the number we are subtracting has fewer digits than the one you are subtracting
from, then add zeros before the number (at least, mentally) to make the
calculation:
For
instance:
23
000 – 46 =
23 000
0 046
22 954
Use
the same principle as subtraction method 2.
Shortcut for subtraction
Ø What is the easiest way to take 90 from a
number?
Take 100 and give back 10
Ø What is the easiest way to take 80 from a
number?
Take 100 and give back 20
Ø What is the easiest way to take 70 from a
number?
Take 100 and give back 30
100 98x135 = 1330070 = 13230
2 35 30
Ø How do we take 70 from 13,330?
Ø Take away 100 and give back 30
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