World Applied Sciences Journal 18 (6): 786-787, 2012
ISSN 1818-4952
© IDOSI Publications, 2012
DOI: 10.5829/idosi.wasj.2012.18.06.306
A New Algorithm for Divisibility of Numbers
H. Khosravi, P. Jafari and V.T. Seifi
Department of Mathematics, Mashhad Branch,
Islamic Azad University, Mashhad, 91735-413, Iran
Abstract: We give a new algorithm for divisibility of numbers. In the other words, we study divisibility of
numbers with a new algorithm about decreasing the num-bers of digits. In this paper, this algorithm called
reduce digits algorithm.
Key words: Divisibility
Partition of integer
Primes
INTRODUCTION
Prime factors
positive regardless of their signs (as same as 62 in
follow Example). If we don't know whether a new
number is divisible by prime divisor,we should apply
the new algorithm again. In this case ,the new numbers
are dividend.
There are several different methods with many
variants and some of them can be found in [1-6].
For example in [5, 7] presented that numbers which
are dividable to 2 should have the last digit is even
[7] or in presented that numbers which are dividable
to 5 should have the last digit is 0 or 5. Similarly some
studies are presented for special numbers such
as 7, 11, 13, etc ([2, 3, 5]). In this paper, we suppose that
z=an an–1...a1 and w=bm bm–1...b1 are dividend and divisor
respectively.
Example 2.1: Is 899 divisible by 31?
Solution: [(3*9)-89] = -62
But 62 is divisible by 31. So 899 is divisible by 31.
Remark 2.2: The above theorem is true for odd numbers
with the above condition.
Theorem 1.1: (Fundamental Theorem of Arithmetic) Every
natural number is a product of primes [8]
Example 2.2: Is 175321152973124 divisible by 19258231?
A New Algorithm for Divisibility and Results
Theorem 2.1: If z=anan–1...a1 is dividend and w=bmbm–1...b1
is prime divisor such that b1=1, then w|z if
w|(bmbm–1...b2)a1 – anan–1...a2.
Solution:
[(4*1925823)
-17532115297312]
=
17532107594020.
But the divisibility 17532107594020 by 19258231 is not
clear, using above theorem
for 17532107594020 to 19258231,we have [(0*1925823) 1753211529731] = 1753211529731.
But the divisibility 1753211529731 by 19258231 is not
clear, using above theorem
for 1753211529731 to 19258231,we have [(1*1925823) 175321152973] = 175319227150.
But the divisibility 175319227150 by 19258231 is not clear,
using above theorem
for 175319227150 to 19258231,we have [(0*1925823) 17532115297] = 17532115297.
But the divisibility 17532115297 by 19258231 is not clear,
using above theorem
Proof: If w|(bmbm–1...b2)a1 – anan–1...a2. then there is a
integer k such that kw=(bmbm–1...b2)a1 – anan–1...a2.
Therefore, 10kw=(bmbm–1...b2)10a1 – (a na n–1...a 2)10, hence
we have
10kw-a1=(bmbm–1...b2)101-z , thus
a1w-10kw=w(a1-10k), so
w|z.
Remark 2.1: In this paper,with using theorems for
dividend and prime divisor ,we can introduce the new
numbers.We always consider these numbers to be
Corresponding Author: H. Khosravi, Department of Mathematics, Mashhad Branch, Islamic Azad University,
Mashhad, 91735-413, Iran.
786
World Appl. Sci. J., 18 (6): 786-787, 2012
Theorem 2.4: If z=an an–1...a2 is dividend and w=b m
bm–1...b2 is prime divisor such that b1=9 , then w|z if w|w((9w-1)/10)a1 + (an an–1...a2)
for 17532115297 to 19258231,we have [(7*1925823) 1753211529] = 1739730768.
But the divisibility 1739730768 by 19258231 is not clear,
using above theorem
for 1739730768 to 19258231,we have [(8*1925823) 175321152] = 159914568.
We have 159914568 is not divisible by 19258231, so
175321152973124 is not divisible by
19258231.
Proof: If w| w-((9w-1)/10) a1 + (an an–1...a2) then there is
a integer k such that
kw= w-((9w-1)/10)a1 + (an an–1...a2). Therefore,
10kw=10w-(9w-1)a1 + (an an–1...a2) 10 ,then we have
10kw-10w+9w=z ,so
w|z.
Theorem 2.2: If z=an an–1...a1 is dividend and w=bm
bm–1...b1 is prime divisor such that b 1=3, then w|z if w|(w
– (7w – 1)/10) a1 + (an an–1...a2)
Example 2.5: Is 1292 divisible by 19?
Solution: [19-34+129]=114.
We have 38 is divisible by 19.So 1292 is divisible by 19.
Proof: If w|(w – (7w – 1)/10) a1 + an an–1...a2) then there is
a integer k such that
kw=(w – (7w – 1)/10) a1 + (an an–1...a2). Therefore,
10kw=10w-(7w-1)a1 + (an an–1...a2), then we have
10kw=10w-7w+z, so
w|z.
Remark 2.5: The above theorem is true for odd numbers
with the above condition.
Theorem 2.5: If z=an an–1...a1 is dividend and w = 5 is
prime divisor,divisibility is clear [6].
Example 2.3: Is 975 divisible by 13?
Theorem 2.6: If z=an an–1...a1 is dividend and w=b m
bm–1...b1 is composite divi-sor (coprime divisor), with using
of fundamental theorem of arithmetic proof is obvious [5].
Solution: [(13-9) *5+97] = 117, using above theorem for
117 to 13.We have 39 is divisible by 13.So 975 is divisible
by 13.
ACKNOWLEDGEMENTS
Remark 2.3: The above theorem is true for odd numbers
with the above condition.
The authors thank the research council of Mashhad
Branch, (Islamic Azad University). Also,We would like to
thank the referee for his/her many helpful suggestions.
Theorem 2.3: If z=an an–1...a2 is dividend and w=bm
bm–1...b1 is prime divisor such that b1=7 , then w|z if w| w((3w – 1)/10)a1 + (an an–1...a2).
REFERENCES
1.
Proof: If w| w-((3w – 1)/10)a1 + (an an–1...a2), then there is
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10kw=10w-3w+z , so
w|z.
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Example 2.4: Is 1513 divisible by 17?
5.
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Remark 2.4: The above theorem is true for odd numbers
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