## Sunday, 16 December 2012

### Magic of Numbers-II

Observe this table and there after we will find some interesting facts about numbers:

 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

Basing on the above table the following facts can be conceptualized:

Concept-I
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 unit-digit series of 112th powers is same as that of 4th 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 117123:
unit digit of 117123 => unit digit of 7123 => unit digit of 7(4*30+3) => unit digit of 73
=> Answer is “3”

Concept -II
All the numbers from 0 to 9 exist in the sequence of unit digits of cubes (and likewise for 7th powers, 11th 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 8th powers, 12th powers etc.,) of natural numbers must be from the set {0,1,5,6}. And in any set of unit-digits of 4th 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 x4-y4 is divisible by 5?
For any number to be divisible by 5, the unit digit must be 0 or 5.
Here as x4 and y4 are fourth powers of natural numbers, their unit digit must be one of {0,1,5,6}
So for x4-y4 to be divisible by 5, the possible pairs of unit digits of x4 and y4 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(n-1)/2
(1,1):
The number of ways of having 2 out of 4n ones = 4nC2 = 4n(4n-1)/2
(5,5):
The number of ways of having 2 out of n fives = nC2 = n(n-1)/2
(6,6):
The number of ways of having 2 out of 4n sixes = 4nC2 = 4n(4n-1)/2
(0,5):
The number of ways of having 1 out of n zeroes and 1 out of n fives = nC1 * nC1 = n2
(1,6):
The number of ways of having 1 out of 4n ones and 1 out of 4n sixes = 4nC1 * 4nC1 = 16n2
Summing up all the above, the total number of ways = 34n2-5n

Link for the first post in the Magic Series is here:
http://mathbyvemuri.blogspot.in/2011/10/magic-of-numbers-1-unit-digit-patterns.html

#### 1 comment:

1. In the above concept the fact of interchangeability of x and y is not used.
e.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