SOLUTIONS FOR PROBLEM SET 4
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1 SOLUTIONS FOR PROBLEM SET 4 A. A certain integer a gives a remainder of 1 when divided by 2. What can you say about the remainder that a gives when divided by 8? SOLUTION. Let r be the remainder that a gives when divided by 8. Then a r (mod 8) and 0 r 7. Since 2 8, we also have a r (mod 2). We are given that a 1 (mod 2). Hence r 1 (mod 2). Since 0 r 7, we can conclude that r {1,3,5,7}. B. A certain integer b satisfies the congruence b 3 (mod 19). What can you say about the remainder that b gives when divided by 95? SOLUTION. Let r be the remainder that b gives when divided by 95. Then 0 r 94 and b r (mod 95). Note that It follows that b r (mod 19). We are given that b 3 (mod 19). Therefore, r 3 16 (mod 19). Since 0 r 94, we find the following possibilities for r: r = 16, r = 35, r = 54, r = 73, r = 92. C. A certain integer c gives a remainder of 5 when divided by 15. What can you say about the remainder that c gives when divided by 91? SOLUTION. Suppose that r is any integer satisfying the inequality 0 r 90. We are given that c 5 (mod 15). Consider the pair of congruences c 5 (mod 15), c r (mod 91). Since (15,91) = 1, the Chinese Remainder Theorem implies that the above pair of congruences is equvalent to a single congruence c a (mod 15 91) for some a Z. Infinitely many integers c satisfying this congruence exist. For any such integer c, the remainder that c gives when divided by 15 is 5 and the remainder that c gives when divided by 91 is r. Therefore, based on the information about c given in the problem, every integer r in the interval 0 r 90 is a possible remainder when c is divided by 91. 1
2 D. Suppose that a, b Z and that a 2 18b 2 = 34. Prove that 17 a. SOLUTION. Suppose to the contrary that 17 a. Thus, a 0 (mod 17). We will obtain a contradiction. Note that we have 18b 2 = a 2 34 and hence 18b 2 a (mod 17). Therefore, 17 18b 2. That is, b b. The extended version of Euclid s Lemma implies that or 17 b or 17 b. Since 17 18, it follows that 17 b. Thus, we have 17 a and 17 b. That is, a = 17c and b = 17d, wnere c,d Z. It follows that 34 = a 2 18b 2 = (17c) 2 18(17d) 2 = 17 2 (c 2 18d 2 ) = 17 2 q where q = c 2 18d 2. Obviously, q Z. Therefore, it follows that 17 2 = 289 divides 34. This is not true. Therefore, we have obtained a contradiction. It follows that 17 a, as we wanted to prove. E. Do problem 1(b) on page 53. SOLUTION. One finds easily that (33,105) = 3. Note that Therefore, the stated congruence has no solutions. F. Do problem 3 on page 53. SOLUTION. It is always good to look for shortcuts. Notice that 1 1 (mod 2) and 2 1 (mod 3). Hence, the first two congruences are equivalent to x 1 (mod 2) and x 1 (mod 3). UsingCongruenceProperties9and10,since(2,3) = 1,itfollowsthatthispairofcongruences is equivalent to x 1 (mod 6). Notice also that 3 2 (mod 5) and 5 2 (mod 7). Since (5,7) = 1, it again follows from Congruence Properties 9 and 10 that the third and fourth congruences in the problem are equivalent to the single congruence x 2 (mod 35). And so the problem is equivalent to finding the solutions to the following pair of congruences x 1 (mod 6) and x 2 (mod 35). 2
3 Since (6,35) = 1 and 6 35 = 210, the Chinese Remainder Theorem implies that the above pair of congruences is equivalent to a single congruence of the form x a (mod 210) for some a Z. We must just find an a satisfying a 1 (mod 6) and a 2 (mod 35). Equivalently, we want a = 35k 2 where 35k 2 1 (mod 6). We want to have 35k 1 (mod 6). Since 35 1 (mod 6), it suffices to have k 1 (mod 6). We can take k = 1. Thus, we can take a = 37. by Therefore, the solutions to the system of congruences given in the problem are described x 37 (mod 210). G. Let n = Find the remainder that n gives when divided by 99. SOLUTION. Note that (mod 9) and so (mod 9). Furthermore, 5 3 = (mod 9) and hence we have Also, 10 1 (mod 9) and hence 5 21 = (5 3 ) 7 ( 1) 7 1 (mod 9) (mod 9). Finally, notice that 2 6 = 64 1 (mod 9). Therefore, 2 42 = (2 6 ) (mod 9). Therefore, (mod 9). It follows that and hence that n 2 (mod 9). n 1 ( 1) 1 2 (mod 9) Note that (mod 11) and (mod 11). Hence (3 5 ) (mod 11). Furthermore, (mod 11) by Fermat s Little Theorem and hence we have 5 21 = (mod 11). 3
4 Also, 10 1 (mod 11) and hence ( 1) 19 1 (mod 11). Finally, notice that (mod 11) by Fermat s Little Theorem. Therefore, It follows that and hence that n 15 4 (mod 11) (2 10 ) (mod 11). n 1 5 ( 1) ( 3) (mod 11) In summary, we have found that n 2 (mod 9) and n 4 (mod 11). Thus, we have n = k, where k Z. We also have n 2 (mod 9) and hence 4+11k 2 (mod 9). Equivalently, we want to find the integers k satisfying 11k 6 (mod 9). As discussed in class, since (11,9) = 1, the above congruence has exactly one incongruent solution modulo 9. One easily finds that k = 6 satisfies the congruence and therefore the general solution is given by k 6 (mod 9). Thus, k = 6+9q, where q Z, and hence n = 4+11k = 4+11(6+9q) = 70+99q. It follows that n gives a remainder of 70 when divided by 99. H. Do problems 5(a) and 5(c) on page 61. SOLUTION. For problem 5(a), let P(x) = 3x 2 5x+7. If x Z, then x a (mod 13) for some integer a in the interval 6 a 6. We have P(0) = 7, P(1) = 5, P( 1) = 15, P(2) = 9, P( 2) = 29, P(3) = 19, P( 3) = 49, P(4) = 35, P( 4) = 75, P(5) = 57, P( 5) = 107, P(6) = 85, P( 6) = 145. None of these values are divisible by 13. Hence, the congruence P(x) 0 (mod 13) has no solutions. 4
5 Forproblem5(c), weconsiderthepolynomialp(x) = x 2 +7x+10. Wecouldapproachthis problem just as in part (a). It would suffice to compute the values P(a), where 5 a 5. However, notice that P(x) = (x+5)(x+2). Hence, P(x) 0 (mod 11) means that (x+5)(x+2) 0 (mod 11). Since 11 is a prime, we know that (x+5)(x+2) 0 (mod 11) x+5 0 (mod 11) or x+2 0 (mod 11) We are using Congruence Property 13 here. We can now conclude that the solutions to the congruence P(x) 0 (mod 11) are given by x 2 (mod 11) or x 5 (mod 11). I. Are there positive integers n with the following two properties: The last three digits of n (in base 10) are 111 and n gives a remainder of 32 when divided by 49. What can you say about the number of such integers n in the interval 0 < n < 200,000? SOLUTION. The two properties are n 111 (mod 1000) and n 32 (mod 49). Note that (10,7) = 1. By Divisibility Proposition 14, it follows that (10 3,7) = 1. The same proposition implies that (10 3,7 2 ) = 1. Thus, (1000,49) = 1. According to the Chinese Remainder Theorem, the above two properties can be expressed in the following equivalent form: n a (mod 49,000) where a is some integer. We will not need to find a specific choice for a. However, we have a r (mod 49,000), where 0 r < 49,000, and hence the above congruence can be expressed equivalently as n r (mod 49,000) for a certain value of r in the above interval. This means that n = r + 49,000q, where q Z. We can now conclude that such integers n do indeed exist. In fact, there are infinitely many such integers n, parametrized by the integer q. Note that r 0 because r 111 (mod 1000), but (mod 1000). Hence 0 < r < 49, 000. For the final part of the question, we must determine the number of integers n = r + 49,000q, where q Z, such that 0 < n < 200,000. That is, we want (1) 0 < r +49,000q < 200,000. 5
6 Obviously, this inequality implies that q 0. Also, if q 0, then r +49,000q r > 0. If q 3, then r +49,000q < 49,000+49,000 3 = 196,000 < 200,000. Hence the inequality (1) is satisfied when 0 q 3. If q 5, then r +49,000q > 0+49,000 5 = 245,000 > 200,000 and hence the inequality (1) is not satisfied. It remains to consider q = 4. Then r+49,000q = r+196,000. The inequality (1) will be satisfied if and only if r < However, we must have r 111 (mod 1000). If r < 4000, then the possibilities for r are r = 111, r = 1111, r = 2111, or r = Using the fact that (mod 49), one finds that (mod 49), (mod 49), (mod 49), (mod 49) and hence the above possible values of r don t satisfy r 32 (mod 49). Therefore, we must have r > 4000 and inequality (1) is not satisfied when q = 4. We have shown that the inequality (1) holds only when q {0,1,2,3} and hence the numbers of integers n with the stated properties and in the interval 0 < n < 200,000 is exactly four. J. Prove that there are infinitely many positive integers n with the following two properties: All of the digits of n in base 10 are 1 s and n is divisible by 49. This problem will be part of problem set 5. 6
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