Download Extension of Reduce Digits Algorithm for Divisibility of Even Numbers

World Applied Sciences Journal 18 (12): 1760-1763, 2012
ISSN 1818-4952
© IDOSI Publications, 2012
DOI: 10.5829/idosi.wasj.2012.18.12.1496
Extension of Reduce Digits Algorithm for Divisibility of Even Numbers
H. Khosravi , P. Jafari and E. Faryad
Department of Mathematics, Faculty of Science, Mashhad Branch,
Islamic Azad University, Mashhad, P.O. Box: 91735-413, Iran
Abstract: Let z = an an-1 …a1 is dividend and w = bmb m-1 …b1 is even divisor. In this paper, we study
divisibility of even numbers with extension of reduce digits algorithm. In the other words, we study
divisibility of even numbers with a new algorithm entitled reduce digits algorithm such that this algorithm
reduced the numbers of digits for z.
AMS subject classification: 13AXX 13F15
•
Key words: Divisibility Partition of Numbers Reduce Digits Algorithm
•
•
INTRODUCTION
In number theory, divisibility methods of whole numbers are very useful because they help us to quickly
determine if a number can be divided by n. There are several different methods for divisibility of
numbers with many variants and some of them can be found in [1,7]. For example in [5] presented that
numbers which are dividable to 3 should have the sum of the digits is divisible by 3 [5] or in [2] presented that
numbers which are dividable to 5 should have the last digits is 0 or 5 or in [6] presented that numbers which
are dividable to 11 should have the (sum of the odd numbered digits)-(sum of the even numbered digits) is divisible
by 11. Similarly some studies are presented for special numbers such as 15, 17, 19, [5, 7, 8]. In this paper, we
suppose that z = an an-1 …a1 and w = bn b n-1 …b1 are dividend and even divisor respectively. Also, we show that in [1]
(with using reduce digits algorithm)if z = an an-1 …a1 and w = bmb m-1 …b1 are dividend and prime divisor respectively,
then
1)
2)
3)
4)
5)
6)
If w|( b mb m-1 …b2 ) a1 -( an an-1 … a2 ) and b 1 = 1 then w|z.
If w|w-(7w-1)/10)a 1 + ( an an-1 … a2 ) and b 1 = 3 then w|z.
If w|w-(3w-1)/10)a 1 +( an an-1 … a2 ) and b 1 = 7 then w|z.
If w|w-(9w-1)/10)a 1 +( an an-1 … a2 ) and b 1 = 9 then w|z.
If w = 5 is prime divisor then the proof of w|z is clear.
If w = bmb m-1 …b1 is composite divisor, then with using of fundamental theorem of arithmetic, the proof of w|z
is obvious.
Now, in this paper we study divisibility of even numbers with extension of reduce digits algorithm.
Theorem 1.1: (Euclid’s Lemma) If a|c, b|c,(a,b) = 1 then ab|c [4].
Theorem 1.2: Let m,n are two integer. m divisible by 2n if the last n digits of m are divisible by 2n [6].
MAIN TEXT AND RESULTS
Definition 2.1: Let w be a even number. We define w1 = w/2 such that wj = (wj-1 )/2: j = {2,3,4,… }.
Corresponding Author: H. Khosravi, Department of Mathematics, Faculty of Science, Mashhad Branch, Islamic Azad
University, Mashhad, P.O. Box: 91735-413, Iran
1760
World Appl. Sci. J., 18 (12): 1760-1763, 2012
Definition 2.2: Let w be a even number then we define w|n = wn such that w|n is a first odd number that product
from above definition. For examp le, if w = 120 then w1 = w/2 =120/2 = 60, w2 = w/4=30, w3 = w/3= 15. Therefore,
w|3 = 15.
Lemma 2.3: (2n , w|n) = 1.
Theorem 2.4: Let z = an an-1 …a1 is dividend and w|n = ct ct-1 …c1 is odd divisor such that c1 = 1, then w|n |z if
w|n |(c t ct-1 …c2 )a1 -(an an-1 …a2 )| [1].
Theorem 2.5: Let z = an an-1 …a1 is dividend and w = bmb m-1 …b1 is even divisor such that w|n = ct ct-1 …c1 and c1 = 1,
then w|z if w|n |(c t ct-1 …c2 )a 1 -(an an-1 …a2 )| and 2 n |z.
Proof: If w|n |(c t ct-1 …c2 )a1 -(an an-1 …a2 ), then there exists an integer k such that kw|n = (ct ct-1 …c2 )a 1 -(an an-1 …a2 ).
Therefore, 10kw|n = (w|n -1)a1 -(an an-1 …a2 )*10, hence we have (a 1 -10k)w|n =|-z| = z. Therefore, w|n |z. (I)
But 2n |z. (II)
With using (I),(II) we have w|z .
Example 2.6: Is 11286 divisible by 22?
Solution: With using above theorem, we have 22|11286 if w|1 =11|11286 and 21 |11286 . But, it is true to say that
21 |11286 because 21 |6. Also, 11|11286 because with using of theorem (2.4.) 11(1*6)-(1128) = |-1122| = 1122.
Therefore, 11286 is divisible by 22.
Example 2.10: Is 887783426 divisible by 90112?
Solution: With using above theorem, we have 90112|887783426 if w|13 =11|887783426 and 213 |887783426 .But
887783426 is not divisible by 11.We proof this case by contradiction. In the other words, we assume that
11|887783426. Now, with using above theorem 11(1*6)-(88778342) = -| 88778336| = 88778336. But the divisibility
88778336 by 11 is not clear, hence with using above theorem 11(1*6)-(8877833) = |-8877827| = 8877827. But the
divisibility 8877827 by 11 is not clear, hence with using above theorem 11(1*7)-(887782) = |-887775| = 887775. But
the divisibility 887775 by 11 is not clear, hence with using above theorem 11(1*5)-(88777) = |-88772| = 88772. But
the divisibility 88772 by 11 is not clear, hence with using above theorem 11(1*2)-(8877) = |-8875| = 8875. But the
divisibility 8875 by 11 is not clear, hence with using above theorem 11(1*5)-(887) = -| 882| = 882. But the
divisibility 882 by 11 is not clear, hence with using above theorem 11(1*2)-(88) = -| 86| = 86.This is contradict.
Therefore, 887783426 is not divisible by 90112.
Theorem 2.7: Let z = an an-1 …a1 is dividend and w = bmb m-1 …b1 is even divisor such that w|n = ct ct-1 …c1 and c1 = 3,
then w|z if w|n | | (7w|n-1)/10)a 1 -( an an-1 … a2 ) | and 2n |z.
Proof: If w|n | | (7w|n-1)/10)a 1 -( an an-1 … a2 ) |,then there exists an integer k such that kw|n = (7w|n-1)/10)a 1 -( an an-1 … a2 ).
Therefore, 10kw|n = (7w|n -1))a 1 -( an an-1 … a2 )*10, hence we have (10k-7a 1 )w|n = |-z| = z. Therefore, w|n |z. (I)
But 2n |z. (II)
With using (I),(II) we have w|z .
Example 2.8: Is 226428288 divisible by 384?
Solution: With using above theorem, we have 384|226428288 if w|7 =3|226428288 and 27 | 226428288. But, it is true
to say that 27 |226428288 because 27 |26428288. Also, 3|226428288 because with using above theorem 226428288
divisible by 3. Therefore, 226428288 is divisible by 384.
Example 2.10: Is 5052065470 divisible by 7232?
1761
World Appl. Sci. J., 18 (12): 1760-1763, 2012
Solution: With using above theorem, we have 7232|5052065470 if w|6 =11|5052065470 and 26 |5052065470.But
5052065470 is not divisible by 11.We proof this case by contradiction. In the other words, we assume that
113|5052065470. Now, with using above theorem 113|(79*0)-(505206547) = |-505206547| = 505206547. But the
divisibility 505206547 by 113 is not clear, hence with using above theorem 113|(79*7)-(50520654) = |-50520101| =
50520101. But the divisibility 50520101 by 113 is not clear, hence with using above theorem 113|(79*1)-(5052010)
= |-5051931| = 5051931. But the divisibility 5051931 by 113 is not clear, hence with using above theorem
113|(79*1)-(505193) = -| 505114| = 505114. But the divisibility 505114 by 113 is not clear, hence with using above
theorem 113|(79*4)-(50511) = -50195|
|
= 50195. But the divisibility 50195 by 113 is not clear, hence with using
above theorem 113|(79*5)-(5019) = -| 4624| = 4624. But the divisibility 4624 by 113 is not clear, hence with using
above theorem 113|(79*4)-(462) = |-146| = 146.This is contradict. Therefore, 5052065470 is not divisible by 7232.
Theorem 2.9: Let z = an an-1 …a1 is dividend and w = bmb m-1 …b1 is even divisor such that w|n = ct ct-1 …c1 and c1 = 7,
then w|z if w|n | | (3w|n-1)/10)a 1 -( an an-1 … a2 ) | and 2n |z.
Proof: If w|n | | (3w|n-1)/10)a 1 -( an an-1 … a2 ) | ,then there exists an integer k such that kw|n = | (3w|n -1)/10)a 1 -( an an-1 …
a2 ) Therefore, 10kw|n = | (3w|n -1))a 1 -( an an-1 … a2 )*10, hence we have (10k-3a 1 )w|n = |-z| = z.
Therefore, w|n |z. (I)
But 2n |z. (II)
With using (I),(II) we have w|z .
Example 2.10: Is 3766656 divisible by 544?
Solution: With using above theorem, we have 544|3766656 if w|5 =17|3766656 and 25 |3766656.But,it is true to say
that 25 |3766656 because 25 |66656. Also, 17|3766656 because with using above theorem 17|(5*6)-(376665) = -|
376635| = 376635. But the divisibility 376635 by 17 is not clear, hence with using above theorem 17|(5*5)-(37663)
= |-37638| = 37638. But the divisibility 37638 by 17 is not clear, hence with using above theorem 17|(5*8)-(3763) =
|-3723| = 3723. But the divisibility 3723 by 17 is not clear, hence with using above theorem 17|(5*3)-(372) = |-357|
= 357. But the divisibility 357 by 17 is not clear, hence with using above theorem 17|(5*7)-(35) = 0. Therefore,
3766656 is divisible by 544.
Example 2.10: Is 3464727552 divisible by 734208?
Solution: With using above theorem, we have 734208|3464727552 if w|10 =717|3464727552 and 210 |3464727552.
But 3464727552 is not divisible by 717.We proof this case by contradiction. In the other words, we assume that
717|3464727552. Now, with using above theorem 717|(215*2)-(346472755) = |346742325| = 346742325. But the
divisibility 346742325 by 717 is not clear, hence with using above theorem 717|(215*2)-(34674232) = |34673157| =
34673157. But the divisibility 34673157 by 717 is not clear, hence with using above theorem 717|(215*7)(3467315) =-3465810 = 3465810. But the divisibility 3465810 by 717 is not clear, hence with using above theorem
717|(215*0)-(346581) = -| 346581| = 346581. But the divisibility 346581 by 717 is not clear, hence with using above
theorem 717|(215*1)-(34658) = -| 34443| = 34443. But the divisibility 34443 by 717 is not clear, hence with using
above theorem 717|(215*3)-(3444) = |-2799| = 2799. But the divisibility 2799 by 717 is not clear, hence with using
above theorem 717|(219*9)-(279) = |-1656| = 1656.This is contradict. Therefore, 3464727552is not divisible by
734208.
Theorem 2.11: Let z = an an-1 …a1 is dividend and w = bmb m-1 …b1 is even divisor such that w|n = ct ct-1 …c1 and c1 = 9,
then w|z if w|n | | (9w|n-1)/10)a 1 -( an an-1 … a2 ) | and 2n |z.
Proof: If w|n | | (3w|n-1)/10)a 1 -( an an-1 … a2 ) | ,then there exists an integer k such that kw|n = (9w|n -1)/10)a 1 -( an an-1 …
a2 ). Therefore, 10kw|n = (9w|n-1))a 1 -( an an-1 … a2 )*10, hence we have (10k-9a 1 )w|n = |-z| = z, Therefore, w|n |z. (I)
But 2n |z. (II)
With using (I),(II) we have w|z .
1762
World Appl. Sci. J., 18 (12): 1760-1763, 2012
Example 2.12: Is 2055483936 divisible by 2064?
Solution: With using above theorem, we have 2064|2055483936 if w|4 =1297|2055483936 and 24 |2055483936 . But,
it is true to say that 24 |2055483936 becaus e 24 |3936. Also, 129|2055483936 because with using above theorem
129|(116*6)-(205548393) = |-205547697| = 205547697. But the divisibility 205547697 by 129 is not clear, hence
with using above theorem 129|(116*7)-(20554769) = -| 20553957| = 20553957. But the divisibility 20553957 by 129
is not clear, hence with using above theorem 129|(116*7)-(2055395) = -| 2054583| = 2054583. But the divisibility
2054583 by 129 is not clear, hence with using above theorem 129|(116*3)-(205458) = |-205110| = 205110. But the
divisibility 205110 by 129 is not clear, hence with using above theorem 129|(116*0)-(20511) = -| 20511| = 20511.
But the divisibility 20511 by 129 is not clear, hence with using above theorem 129|(116*1)-(2051) = |-1935| = 1935.
But the divisibility 1935 by 129 is not clear, hence with using above theorem 129|(116*5)-(193) = -| 387| = 387.
Therefore, 2055483936 is divisible by 2064.
ACKNOWLEDGEMENTS
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.
REFERENCES
1.
2.
3.
4.
5.
6.
7.
8.
H. Khosravi, , P. Jafari and V.T. Seifi, 2012. A New Algorithm For Divisibility of Numbers. World Applied
Sciences Journal, Vol: 18.
Edwin Clark, W., 2002. Elementary Number Theory. University of South Florida.
Everest, G. and T. Ward, 2005. An Introduction to Number Theory. Graduate Text 232, Springer.
Stein, W., 2009. Elementary Number Theory, Springer.
Peterson, I., 2002. Testing for divisibility, MAA Online 2002.
Sandor, J. and B. Crstici, 2006. Handbook of Number Theory, Springer.
Hardy, G.H., 1979. An Introduction to the Theory of Numbers, 5th Edn. Oxford, England: Clarendon Press.
Guy, R.K., 1981. A12 in, Unsolved Problems in Number Theory. New York, Springer-Verlag.
1763