نتایج جستجو برای: محصول mod 10

تعداد نتایج: 1044044  

2000
Nico F. Benschop

The finite ring Zk = Z(+, .) mod p k of residue arithmetic with odd prime power modulus is analysed. The cyclic group of units Gk in Zk(.) has order (p − 1).p , implying product structure Gk ≡ Ak.Bk with |Ak| = p− 1 and |Bk| = p , the ”core” and ”extension subgroup” of Gk respectively. It is shown that each subgroup S ⊃ 1 of core Ak has zero sum, and p+1 generates subgroup Bk of all n ≡ 1 mod p...

1999
Ouri Wolfson A. Prasad Sistla Bo Xu Jutai Zhou Sam Chamberlain Yelena Yesha Naphtali Rishe

1 Background Consider a database that represents information about moving objects and their location. For example, for a database representing the location of taxi-cabs a typical query may be: retrieve the free cabs that are currently within 1 mile of 33 N. Michigan Ave., Chicago (to pickup a customer); or for a trucking company database a typical query may be: retrieve the trucks that are curr...

Journal: :Cell 1995
Tatlana I. Gerasimova David A. Gdula Denis V. Gerasimov Olga Simonova Victor G. Corces

The suppressor of Hairy wing (su(Hw)) protein inhibits the function of transcriptional enhancers located distally from the promoter with respect to the location of su(Hw)-binding sites. This polarity is due to the ability of the su(Hw)-binding region to form a chromatin insulator. Mutations in modifier of mdg4 (mod(mdg4)) enhance the effect of su(Hw) by inhibiting the function of enhancers loca...

1998
Thomas Arts Jürgen Giesl

The framework of dependency pairs allows automated termination and innermost termination proofs for many TRSs where such proofs were not possible before. In this paper we present a reenement of this framework in order to prove termination in a modular way. Our mod-ularity results signiicantly increase the class of term rewriting systems where termination resp. innermost termination can be prove...

2010
Zhi-Wei Sun ZHI-WEI SUN

Let p > 3 be a prime. We show that p−1 k=0 −1/(p + 1) k p+1 ≡ 0 (mod p 5) and p−1 k=0 1/(p − 1) k p−1 ≡ 0 (mod p 4). For any positive integer m ≡ 0 (mod p), we prove that p−1 k=0 (−1) km p/m − 1 k m ≡ 0 (mod p 4), and p−1 k=1 (−1) km k 2 p/m − 1 k m ≡ 1 p p−1 k=1 1 k (mod p 3) if p > 5. The paper also contains some open conjectures.

1999
John Kelsey Bruce Schneier David A. Wagner

We introduce “mod n cryptanalysis,” a form of partitioning attack that is effective against ciphers which rely on modular addition and bit rotations for their security. We demonstrate this attack with a mod 3 attack against RC5P, an RC5 variant that uses addition instead of xor. We also show mod 5 and mod 257 attacks against some versions of a family of ciphers used in the FireWire standard. We...

Journal: :Communications in Algebra 2021

Gabriel showed that for a unital ring R, there exists bijective correspondence between the set of filters R and Giraud subcategories Mod(R). In this paper we prove result analogous to one given by Gabriel: small preadditive category C, left C Mod(C).

Journal: :Journal of biomedical materials research. Part B, Applied biomaterials 2006
Sema Belli Funda Kont Cobankara Oğuz Eraslan Gürcan Eskitascioglu Vistasp Karbhari

In this study, the effect of flowable composite reinforced with a leno wave ultra high modulus (LWUHM) polyethylene fiber (Ribbond) on fracture resistance of endodontically treated molars with MOD cavity and lingual cuspal fracture was evaluated. Sixty sound extracted human mandibular molars were randomly assigned to six groups (n = 10). Group 1 served as control. Teeth in groups 2-6 received r...

2010
Nik Lygeros Olivier Rozier Jean-Pierre Serre

The known solutions to the equation τ(p) ≡ 0 (mod p) were p = 2, 3, 5, 7, and 2411. Here we present our method to compute the next solution, which is p = 7758337633. There are no other solutions up to 10.

2007

Therefore 0 + l)* s P + 1 mod p\ Now put t=<r + vp, (0 < c r < p 1). Then ^ = cr̂ , 0 + l)*' s (er + 1)*" mod p. Therefore (er + l)* = a + 1 mod p, (0 < a < p 1). This is relation (7) of my previous note; from this follows (1) as in the earlier treatment. Hence (1) is satisfied by all primes of the form Qn + 1. Therefore the test can be useful only when the exponent p is 3 or is of the form 6n — 1.

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