Beauties of Numbers
Beauties of Numbers
The objective of this section is to get rid of the unwanted fear of numbers and to create
an enjoyment with numbers and thereby to enhance quantitative aptitude.
Be familiar with numbers and enjoy numbers
1 ⨯ 9 = 9
2 ⨯ 9 = 18
3 ⨯ 9 = 27
4 ⨯ 9 = 36
5 ⨯ 9 = 45
6 ⨯ 9 = 54
7 ⨯ 9 = 63
8 ⨯ 9 = 72
9 ⨯ 9 = 81
10 ⨯ 9 = 90
What can we observe in the multiplication table of 9 ?
6 ⨯ 9 = 5 4
Observation : 5 = 6 - 1, 4 = 9 - 5 or 4 = 10 - 6.
Why ?
6 ⨯ 9 = 6 (10 -1) = (5 + 1)(10 - 1) = 5 x 10 - 5 + (10 - 1) = 5 ⨯ 10 + (9 - 5) = 5 ⨯ 10 + (10 - 6)
6 ⨯ 9 = 5 ⨯ 10 + (9 - 5) = 5 4
Thinking Beyond :
9 = 10 -1
99 = 100 - 1 = 102 - 1
999 = 1000 - 1 = 103 - 1
…………………..
Can we extend ?
What is 53 ⨯ 99 = _ _ _ _
Yes
53 ⨯ 99 = 52 47
52 = 53 -1 and 47 = 99 - 52 or 4 = 9 - 5, 2 = 9 - 7
53 ⨯ 99 = 52⨯100 + (100 - 53) = 52⨯100 + (99 - 52)
7 ⨯ 99 = 06 93 = 693
279 ⨯ 999 = 278 721
79 ⨯ 999 = 078 921
Now we observe multiplication by 11.
8 ⨯ 11 = 088 0 8 8 0+8 = 8
10 ⨯ 11 = 110 1 1 0 1+0 = 1
11 ⨯ 11 = 121 1 2 1 1+1 = 2
. . . . . . . . . ..
18 ⨯ 11 = 198 1 9 8 1+8 = 9
19 ⨯ 11 = 209 1 10 9 1+9 = 10 1+1 = 2
38 ⨯ 11 = 418 3 11 8 3+8 = 11 3+1 = 4
99 ⨯ 11 = 1089 9 18 9 9+9 = 18 9+1 = 10
(To get 18 ⨯ 11 write 1 in 100th place and 8 in the unit place then write112 = 121
1112 = 12321
11112 = 1234321
. . . . . .
11…12 = 123….(n-1)n(n-1)....21
(n 1s)
332 = 9 ⨯ 121 = 1089, since 33 = 3 ⨯ 11 and (ab)2 = a2b2
3332 = 9 ⨯ 12321 = 110889
33332 = 9 ⨯ 1234321 = 11108889
333332 = 9 ⨯ 123454321 = 1111088889
. . . . . . . .
333…32 = 9 ⨯ 123….(n-1)n(n-1)....21 = 11….10888….89
(n 3s) (n-1 1s n-1 8s)
112 = 121
1012 = 10201
10012 = 1002001
100012 = 100020001
. . . . . . . . .
772 = 72 ⨯ 112 = 49 ⨯ 121 = 50 ⨯ 121 - 121 = 6050-121 = 5929
What about division by 9 ?
1/9 = 0.111111………
2/9 = 0.222222………
3/9 = 0.333333………
4/9 = 0.444444………
5/9 = 0.555555………
6/9 = 0.666666………
7/9 = 0.777777………
8/9 = 0.888888………
Why ?
Let 0 < a < 10 and x = 0.aaaaa….. . Then 10x = a.aaaaaa…. = a + x.
i.e. 9x = a and x = a/9.
Hence, 1 = 9/9 = 0.999999……… (Note that the decimal representation of 1 is not unique)
0.0999999……… = 0.1
0.4999999……… = 0.5
Extension :
23/99 = 0.23232323…………
87/99 = 0.87878787………….
7/99 = 0.07070707………….
895/999 = 0.895895895…….
84/999 = 0.084084084………
8/999 = 0.008008008………
Think
136/333 = ? 28/33 = ? 7/33 = ?
103/111 = ? 9/11 = ?
136/333 = (136⨯333)/(3⨯3) = 408/999 = 0.408408408…………….
We know that (x+b)(x+c) = xx + xc + bx + bc = x(x+b+c)+bc.
This can be used for some multiplications.
87 ⨯ 85 = (80+7)(80+5) = 80(80+7+5)+7⨯5 = 80 ⨯ 92 + 35 = 7360 + 35 = 7395
63 ⨯ 63 = (60+3)(60+3) = 60 ⨯ 66 + 3 ⨯ 3 = 3960 + 9 = 3969
67 ⨯ 67 = (60+7)(60+7) = 60 ⨯ 74 + 49 = 4440 + 49 = 4489
68 ⨯ 64 = 60 ⨯ 72 + 32 = 4320 + 32 = 4352
312 ⨯ 286 = (300+12)(300 - 14) = (300 + 12 - 14) + 12 ⨯ (-14) = 300 ⨯ 298 + 12 ⨯ (-14)
= 89400 - (10 ⨯ 16 + 8) = 89232
96 ⨯ 107 = (100-4)(100+7) = 100 ⨯ 103 + (-4) ⨯ 7 = 10300 - 28 = 10272
96 ⨯ 98 = (100-4)(100-2) = 94 ⨯ 100 + (-4)(-2) = 9408
96 ⨯ 98 = 90 ⨯ 104 + 48 = 9360 + 48 = 9408
If b+c = 10 then (x+b)(x+c) = x(x+10)+bc
If x = 80 then x+b+c = 90 and (x+b)(x+c) = 80 ⨯ 90 + bc = 8 ⨯ 9 ⨯ 100 + bc
Hence if b+c = 10 then ab ⨯ ac = a(a+1) ⨯ 100 + b ⨯ c , first write the value of a(a+1)
then next to it write the value of bc
87 ⨯ 83 = 7200 + 21 = 7221 (first 8 ⨯ 9 then 7 ⨯ 3)
68 ⨯ 62 = 4216 (first 6 ⨯ 7 then 8 ⨯ 2)
65 ⨯ 65 = 4225 (first 6 ⨯ 7 then 5 ⨯ 5)
85 ⨯ 85 = 7225 (first 8 ⨯ 9 then 5 ⨯ 5)
Also, (a+b)(a+c) = a2 + bc+a(b+c) and (a+b)2 = a2+b2+2ab.
Hence 68 ⨯ 64 = 4352
3632 a2 + bc
+ 720 a(b+c)
63 ⨯ 63 = 3969
3609 a2+b2
+ 360 2ab
63 ⨯ 63 = (60+3)(60+3) = 60 ⨯ 60 + 2 ⨯ 6 ⨯ 3 ⨯ 10 + 9
156 ⨯ 156 = 150 ⨯ 150 + 2 ⨯ 150 ⨯ 6 + 6 ⨯ 6 = 22500 + 1800 + 36 = 24336
And (a+b)(a-b) = a2 - b2 , so 68 ⨯ 72 = (70-2)(70+2) = 702 - 22 = 4900 - 4 = 4896
1+2+3+. . . +100 = (1+100)+(2+99)+ . . . . +(50+51) = 101 + 101 + . . . +101, 50 times
= 50 ⨯ 101 = 5050
In general, 1+2+3+. . . +n = n(n+1)/2
1+3+5+ . . . +(2n-1) = n2
1+3+5+ . . . +n = ((n+1)/2)2 , when n is odd
Since 1+3+5+ . . . +(2n-1) = n2,
12 = 1,
22 = 1 + 3 = 4
32 = 4 + 5 = 9
42 = 9 + 7 = 16
52 = 16 + 9 = 25
62 = 25 + 11 = 36
72 = 36 + 13 = 49
and so on and in general, n2 = (n - 1)2 + (2n - 1) .
2+4+6+ . . . +2n = n(n+1)
n2 + (n+1)2 + (n(n+1))2 = (n(n+1)+1)2
(n(n-1)+1) + (n(n-1)+3) + (n(n-1)+5) + . . . + (n(n-1)+(2n-1)) = n2(n-1)+n2 = n3
142857 ⨯ 1 = 142857
142857 ⨯ 2 = 285714
142857 ⨯ 3 = 428571
142857 ⨯ 4 = 571428
142857 ⨯ 5 = 714285
142857 ⨯ 6 = 857142
Note that the digits in the product are the same as the digits of 142857,
each occurring in the product exactly once and in some cyclic order.
But 142857 ⨯ 7 = 999999
If ab is a two digit number then ab+ba = 10a+b+10b+a = 11(a+b), a multiple of 11.
26+62 = 88 = 11(2+6)
A perfect square ends with 0, 1, 4, 5, 6 and 9.
But not every number ends with 0, 1, 4, 5, 6 and 9 is a perfect square.
In the case of 0, a number ending with an even number of 0s is a perfect square.
The square or cube of an odd number is odd and the even number is even.
Square root of an n digit number is a n/2 or (n+1)/2 digit number when n is even or odd
respectively.
1729 = 123 + 13 = 103 + 93
1729 = 7 ⨯ 13 ⨯ 19 = 91 ⨯19 , 1 + 7 + 2 + 9 = 19 and in the reverse order of digits 19 is 91.
One day when Hardy met Ramanujan and told that he came by a taxi which is not comfortable
and its registration number is 1729. Immediately Ramanujan reacted that 1729 is a significant
number because 1729 is the smallest number which can be expressed as a sum of cubes of two
numbers in two different ways. Hence, this number is called Ramanujan number.
To be familiar with numbers, express numbers in different ways.
For example, 29 = 30 - 1 = 7 ⨯ 4 + 1 = 52 + 22 = 33 + 2 = 5 ⨯ 4 + 32 = . . . .
Familiarity with numbers in these ways can help to get answers quickly for the number series
problems in competitive examinations.
The following equivalent of some of the multiples of 5 can be used in multiplication.
5 = 10/2, 25 = 100/4, 50 = 100/2, 125 = 1000/8, 250 = 1000/4, 625 = 10000/16.
In general, 5n = (10/2)n = 10n/2n.
For example, 529 × 125 = 529 × 1000/8 = 529000/8 = 66125.
GCD ⨯ LCM = product of the numbers
GCD of two consecutive numbers = 1
GCD of two consecutive odd numbers = 1
GCD of two consecutive even numbers = 2
The product of three consecutive numbers is divisible by 6.
(one is a multiple of 3 and one is even)
The sum of two consecutive even numbers is not a multiple of 4, since 2n + (2n + 2) = 4n + 2.
But the sum of two consecutive odd numbers is a multiple of 4, since (2n - 1) + (2n + 1) = 4n.
If a number has no prime factor less than or equal to the square root of it then it is a prime.
1 million = 106 = 10 lakhs and 1 billion = 109 = 100 crores = 1000 million
Divisibility Rule
1, 2, 3, . . . . . are the natural numbers or the counting numbers and N = { 1, 2, 3, . . . . }.
If n is a natural number then n+1 is the next natural number and there is no natural number
between n and n+1.
The set of all integers Z = { . . . . , -3, -2, -1, 0, 1, 2, 3, . . . . } (German Zahl, plural Zahlen).
We can speak of consecutive natural numbers or integers and distance between them is 1.
If z is an integer then z+1 is the next integer and there is no integer between z and z+1.
The set of all rational numbers Q = {m/n / m and n are integers with n ≠ 0} and
a/b = c/d if and only if ad = bc.
If p and q are two rational numbers then (p+q)/2 is also a rational number and
it lies between p and q.
So, there are no consecutive rational numbers by value and rational numbers are dense.
Negative integers are the solutions of the equations x + n = 0, n ∈ N.
Rational numbers are the solutions of the equations nx + m = 0, where m and n are
integers with n ≠ 0.
The equation x2 = 2 have no solution in Q.
i.e. 
is not rational.
In general, for any positive integer n ≠ m2, for any integer m,
the equation x2 = n has no solution in Q.
i.e. in general, 
where n is not a perfect square, is not rational.
Roman number system
Basic symbols used in the Roman number system are I, V, X, L, C, D and M and
they represent 1, 5, 10, 50, 100, 500 and 1000 respectively.
Repetition of a symbol means addition of the value of the symbol and
a symbol can be repeated at most three times.
For example II is 2, CCC is 300.
But CCCC is not a valid representation for 400. It is represented by CD.
In general, if a symbol with lower value is written to the left of a higher value symbol then
the smaller value has to be subtracted from the higher value.
And if a symbol with lower value is written to the right of a higher value symbol
then the smaller value has to be added with the higher value.
For example, IV = 5 - 1 = 4, CM = 1000 - 100 = 900, XI = 10 + 1 = 11,
LXXX = 50 + 10 + 10 + 10 = 80, XC = 100 - 10 = 90.
A bar above any symbol multiplies its value by 1000.
For example,
Love numbers
Explore the Beauties of Numbers
Enjoy
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