CS280, Spring 2004: Final
1. [4 points] Which of the following relations on {0,1,2,3}is an equivalence relation. (If it is,
explain why. If it isn’t, explain why not.) Just saying “Yes” or “No” with no explanation
gets 0 points. The explanations can be very short though.)
(a) {(0,0,),(1,1),(2,2),(3,3)}
(b) {(0,0),(0,1),(1,0),(1,1)}.
Solution: (a) is an equivalence relation, since it is reflexive, symmetric, and transitive.
(b) is not an equivalence relation, since it is not reflexive; neither (2,2) nor (3,3) are in
the relation. (It is symmetric and transitive; you got one point if you said that, even if
you didn’t get the right answer.)
Grading: Basically, 1 point for the correct answer and 1 point for the reason. For (b), 1
point was given for the wrong answer if you said it was symmetric and transitive.
2. [5 points] Recall that a composite nis a Carmichael number if for any brelatively prime to
n,bn−1≡1 (mod n). Show that 1105 is a Carmichael number. [Hint: 1105 = 5 ·13 ·17.]
Solution: We want to show that if b is relatively prime to 1105, then b1104 ≡1
(mod 1105). By the Chinese Remainder Theorem, it suffices to show that b1104 ≡1
(mod 5), b1104 ≡1 (mod 13), and b1104 ≡1 (mod 17). By Fermat’s Little Theorem,
ap−1≡1 (mod p) if ais relatively prime to p. By assumption, bis relatively prime
to 1105, so bis also relatively prime to 5, 13, and 17. Thus b4≡1 (mod 5), b12 ≡1
(mod 13), and b16 ≡1 (mod 17). Thus, b4n≡1 (mod 5) for all n; similarly b12n≡1
(mod 13) and b16n≡1 (mod 17) for all n. Since 4, 12, and 16 all divide 1104 evenly, it
follows that b1104 ≡1 (mod 5), b1104 ≡1 (mod 13), and b1104 ≡1 (mod 17).
Grading: You lost one point if you didn’t explicitly state that since bis relatively prime
to 1105, it is relatively prime to each of 5, 13, and 17. You also lost one point if you
didn’t justify the fact that bp−1≡1 (mod p) for each of p= 5,13,17 (i.e., you had to
mention Fermat’s Little Theorem). You also lost one point for not mentioning the Chinese
Remainder Theorem.
3. [4 points] Consider any six natural numbers n1, . . . , n6. Show that the sum of some
subsequence of consecutive numbers is divisible by 6 (e.g., perhaps n3+n4+n5is divisible
by 6, or n4itself is divisible by 6, or n2+n3is divisible by 6). [Hint: Look at the sums
0, n1,n1+n2,n1+n2+n3, . . . , n1+n2+·· · +n6, and think in terms of mod 6.]
Solution: Following the hint, consider each of the sums 0, n1,n1+n2,n1+n2+n3,
. . . , n1+n2+· ·· +n6mod 6. If any of them is 0 mod 6, we’re done (since that sum is
divisible by 6). By the pigeonhole principle, two of the sums must be the same mod 6.
(The “pigeons” are the sums, the “holes” are 1, . . . , 5, since all the sums must be in the
range 1, . . . , 5 mod 6.) If two sums are the same mod 6, their difference must be 0 mod
6, hence must be divisible by 6. But the difference between two such sequences is a sum
of consecutive integers. (The difference between n1+·· · +nkand n1+· ·· +nm(where
k > m), is nm+1 +· · · +nk.) Thus, this sum of consecutive integers is divisible by 6.
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