This book deals with the topics taught in a first course in Number Theory.
A short tutorial on number theory. Highly readable piece on a very difficult subject!
Elementary Number Theory - SolutionsDescription complète
Elementary Number Theory - SolutionsDescripción completa
After an eclipse of some 50 years, Number Theory, that is to say the study of the properties of the integers, has regained in France a vitality worthy of its distinguished past. More 'and more rese...
Elementary Number Theory - SolutionsFull description
Algebraic Number Theory J. S. Milne
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Divisibility, Prime numbers, GCD, LCM, Euclidean Algorithm, Congruencies, Chinese Remainder theorem, Wilsons theorem, Fermat's theorem, Arithmetic functions, quadratic residues, jacobi symbol conti...
jee advanced
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Number TheoryFull description
number theory
Problems to solve in number theory
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MULTIPLE CHOICE QUESTIONS INTRODUCTION TO NUMBER THEORY Total marks 40. Each right answer will be awarded 4 and wrong answer will attract a penalty of -1 marks. Time 15 min. ……………………………………………………………………………………..
1. Small trees are to be planted planted in rows rows such that that each row contains contains equal equal but odd odd Number of tree. If there are total of 3500 3 500 plants then the no of possible way is 1. 24 2. 8 3. 16 4. 112 Solution: 3500 = 2^2 * 5^3 * 7^1 As each row must contain odd no of tree so ans is equal to total of odd divisor = (3+1)*(1+1) = 8. he hence nce 2 is ans.
2. A number number of statem statements ents are are given below below with respect respect to to two numbers numbers 2835 and and 2800. Chose the correct staement among the following: a) 2835 2835 and 2800 2800 have have equal equal number number of of diviso divisor r b) 2835 2835 has a total total of of 18 disti distinct nct fact factor or c) 2835 2835 and 2800 2800 have have equal equal number number of of odd divis divisor or d) 2800 2800 has a total total of 23 23 even even fact factors ors e) 2835*28 2835*2800 00 have have 41 distin distinct ct facors facors 1. b and d 2. a ,c and e 3. b, c and e 4. a and b Solution: ans is 1. As all other statement are incorrect 2835 = 3^4 * 5 * 7 and 2800 = 2^4 *5^2 *7 Hence 2835 has 18 distinct factors (all odd) And 2800 has 28 distinct factors, 5 odd and 23 even factors. 3. 32400 is is decomposed decomposed into N pair of of factor factor such such that that each pair when Multiplied together yields 32400 itself. Then the maximum value of N is: 1. 37 2. 38 3. 39 4. 41 Solution: 32400 = 2^4 * 3^4 * 5^2. Hence possible no of such pair is Given by 0.5*[(4+1)*(4+1)*(2+1) +1] = 38. Hence 2. 4. The no 1111 111111 1111 1111 1111 11 is perfectly perfectly divis divisible ible by: 1. 11 2. 37 3. 407 4. all all of of the the abov abovee Solution: If no of one is even => divisible by 11 If no of one is multiple of three=> divisible by 37. Hence also divisible by 37*11=407 .Ans is 4.
5. What is is the remainder remainder when when 234*235*236 234*235*236*237*238 *237*238*239 *239 is is divided divided by 120? 1. 119 2. 1 3. 0 4. 2 Solution: any six consecutive no is divisible by 6! So it is also divisible By 5! or120. Hence the ans is 3. 6. Let a three three digit digit number number be in form of of xyz such such that 2x is arithm arithmetic etic mean mean of y and z , also y is arithmetic mean of x and z ,then the no can be 1. 131 2. 357 3. 471 4. 253 Solution: Ans is 2 as it fulfills above condition. 7. If the number number 78 in decimal decimal base base system system is represent represented ed as 243 in some some other base base Notation then 442 in the other base notation is eqivalent to: 1. 122 122 in in deci decima mall syst system em 2. 172 172 in in oct octal al sys system tem 3. 7A in hexad hexadec ecim imal al syst system em 4. all all of of the the abov abovee 5. none of th these Solution: Ans is 4. 78 in decimal system is 243 in base 5 system. 442 in base 5 system is 122 in decimal, 172 in octal and 7A in hexadecimal. 8. How many many zero zeross are ther ther at the the end end of 126! 126! 1. 25 2. 30 3. 31 4. 29 Solution: Zero is obtained by multiplication of 2 and 5. Total no. of 2 in 126! Is greater than total no of 5 in 126!. Hence No. of zeros is equal to no. of times 5 comes in 126! The value is equal to (126/5) + (126/25) + (126/125) = 25 + 5 +1 = 31. Ans is 3. 9. Assume Assume n to be a positive positive integer integer and p and q are are prime number number.. It is also also given That n is greater than 2 and p, q is greater than 6 then which of the following statement is true: a) Seventh Seventh power of of n substract substracted ed by n ( n^7-n n^7-n ) is necessar necessarily ily divisible by 42. b) Square Square of p substra substracted cted by square square of q ( p^2 - q^2 q^2 ) is necessarily divisible by 48 c) (p-1 (p-1)! )!+1 +1 is is divi divisi sible ble by by p
d) (pq-1) (pq-1)!+1 !+1 is divisi divisible ble by pq pq 1. a , b and d 2. a and c 3. c and d 4. a, b and c Solution: n^7-n is divisible by 6 and 7 so is also divisible by 42 Given n is greater than 2. p^2-q^2 is necessarily divisible by 24 but not by 48. c is true but d is false. Hence 2 is the answer. 10. There are 16 ways in which students students of a class can be grouped such that each Group contains equal number of student. Then the minimum no of student in the Class is: 1. 120 2. 180 3. 150 4. 80 Solution: Answer is 2. As total no of distinct factor of 180 is 16 excluding 180 and 1.