Observe this
table and there after we will find some interesting facts about numbers:
Basing on the above table the following facts can be conceptualized:
unit digits of 1st 10
natural numbers

1

2

3

4

5

6

7

8

9

0

unit digits of squares of
1st 10 natural numbers

1

4

9

6

5

6

9

4

1

0

unit digits of cubes of 1st
10 natural numbers

1

8

7

4

5

6

3

2

9

0

unit digits of 4th powers of
1st 10 natural numbers

1

6

1

6

5

6

1

6

1

0

unit digits of 5th powers of
1st 10 natural numbers

1

2

3

4

5

6

7

8

9

0

unit digits of 6th powers of
1st 10 natural numbers

1

4

9

6

5

6

9

4

1

0

ConceptI
The horizontal pattern
repeats for every 4 powers (vertically
downwards) and for every 10 consecutive numbers (Horizontally rightwards) such that we can find
the pattern for any power.
For example, the unitdigit
series of 112th^{} powers is same as that of 4^{th} powers,
because 112 (= 4*28) is a multiple of 4 as the series is repeated for
every fourth power.
We see one more
example to emphasize this concept. We will find the unit digit of 117^{123}:
unit digit of 117^{123
}=> unit digit of 7^{123} => unit digit of 7^{(4*30+3)}
=> unit digit of 7^{3}
=> Answer is
“3”
Concept II
All the numbers
from 0 to 9 exist in the sequence of unit digits of cubes (and likewise for 7^{th}
powers, 11^{th} powers etc..) of natural numbers. Each number from 0 to
9 has an occurrence in the series of unit digits of cube roots of first 10
natural numbers. And of course this pattern repeats for the next 10 natural
numbers (ie., from 11 to 20) and so on..
Here my point is:
n being a natural number, for a set of ‘10n’ consecutive natural numbers, if we
consider unit digits of cubes of all these numbers, the probability of ‘0’ being
the unit digit = probability of ‘1’ being the unit digit = …… = probability of ‘9’ being the unit digit = 1/10
Concept III
Unit digits of
fourth powers (likewise 8^{th} powers, 12^{th} powers etc.,) of
natural numbers must be from the set {0,1,5,6}. And in any set of unitdigits of
4^{th} powers of 10 consecutive natural numbers, there exist one 0, one
5, four 1’s and four 6’s.
Now I quickly relate to a problem which can be built from this basic concept.
Find the number of ways of selecting x and y from the set of
natural numbers {1,2,3,4,5,…. up to 10n} such that x^{4}y^{4}
is divisible by 5?
For any number to
be divisible by 5, the unit digit must be 0 or 5.
Here as x^{4}
and y^{4} are fourth powers of natural numbers, their unit digit must
be one of {0,1,5,6}
So for x^{4}y^{4}
to be divisible by 5, the possible pairs of unit digits of x^{4} and y^{4}
are:
(0,0),(1,1),(5,5),(6,6),(0,5),(1,6)
as in all these cases, the unit digit of difference is either 0 or 5.
As per the
concept, in each consecutive 10 numbers, there exists one ‘0’
=> For the
given 10n numbers, there exist ‘n’ number of ‘0’s as the unit digits of fourth
powers
As per the
concept, in each consecutive 10 numbers, there exists one ‘5’
=> For the
given 10n numbers, there exist ‘n’ number of ‘5’s as the unit digits of fourth
powers
In each
consecutive 10 numbers, there exist four ‘1’s
=> For the
given 10n numbers, there exist ‘4n’ number of ‘1’s as the unit digits of fourth
powers
In each
consecutive 10 numbers, there exist four ‘6’s
=> For the
given 10n numbers, there exist ‘4n’ number of ‘6’s as the unit digits of fourth
powers
Now we will find
the number of ways:
(0,0):
The number of ways
of having 2 out of n zeros = nC2 = n(n1)/2
(1,1):
The number of
ways of having 2 out of 4n ones = 4nC2 = 4n(4n1)/2
(5,5):
The number of
ways of having 2 out of n fives = nC2 = n(n1)/2
(6,6):
The number of
ways of having 2 out of 4n sixes = 4nC2 = 4n(4n1)/2
(0,5):
The number of
ways of having 1 out of n zeroes and 1 out of n fives = nC1 * nC1 = n^{2}
(1,6):
The number of
ways of having 1 out of 4n ones and 1 out of 4n sixes = 4nC1 * 4nC1 = 16n^{2}
Summing up all
the above, the total number of ways = 34n^{2}5n
Link for the first post in the Magic Series is here:
http://mathbyvemuri.blogspot.in/2011/10/magicofnumbers1unitdigitpatterns.html
Link for the first post in the Magic Series is here:
http://mathbyvemuri.blogspot.in/2011/10/magicofnumbers1unitdigitpatterns.html
In the above concept the fact of interchangeability of x and y is not used.
ReplyDeletee.g. put n=2 we have numbers from 1 to 20. Now according to above concept(III) no. of ways of choosing 2 out of 2 zeros would be one, but actually it is 2 i.e. (10,20),(20,10) and this is the case where x and y are distinct, if repetition is allowed(x and y are same) it also include (10,10),(20,20)
and then it would be simply n^2 and others would be (4n)^2,n^2,(4n)^2,2(n^2),32(n^2) summing to 68(n^2) which is exactly same according to probability concept
consider 2 events
1st probability of choosing 0 or 5 as unit digit
2nd probability of choosing 1 or 6 as unit digit
p(1st event)=0.2, p(2nd event)=0.8
and required probability is
p(1st event)*p(1st event)+p(2nd event)*p(2nd event)= 0.2*0.2+0.8*0.8 =0.68