Discrete Math, Spring 2012 - Problems VI
Please turn in by April 10.
Write clearly and show your work.
1. Are there positive integers a, b, c such that 2a · 3b · 5c = 730120? Explain your answer.
No. Clearly 3 divides 2a · 3b · 5c . But 3 does not divide 730120, so 2a · 3b · 5c = 730120 is
impossible.
2. Consider the set S of all integers of the form 3n + 1 as n varies through the natural
numbers. Thus S = {4, 7, 10, 13, 16, . . .}. An element of S is said to be an S-prime if it
cannot be written as product of two smaller integers in S. Hence 10 is an S-prime but
16 = 4 · 4 is not.
a. Prove that any element of S is a product of S-primes.
b. Give an example to show that an integer in S can sometimes be factored into S-primes
in more than one way (i.e., uniqueness of factorization does not hold in S).
a. We use strong induction to show that the n-th entry on the list is a product of Sprimes for any positive integer n. For the base case, 4 is an S-prime and so is surely
a product of S-primes. Let n > 1 and assume that any entry on the list before the
n-th is a product of S-primes. We need to show that the n-th entry is a product of
S-primes. If it’s an S-prime, we’re done. If not, then it’s a product ab where a and b
occur earlier on the list. By hypothesis, a and b are products of S-primes and so ab
is a product of S-primes. Hence, by (strong) induction, every number on the list is a
product of S-primes.
b. 4, 10, 25 are S-primes and 4 × 25 = 10 × 10.
3. Use Euclid’s algorithm to find the greatest common divisor d of 277 and 355 and to find
integers x and y such that d = 277x + 355y.
355 = 1 × 277 + 78
277 = 3 × 78 + 43
78 = 1 × 43 + 35
43 = 1 × 35 + 8
35 = 4 × 8 + 3
8=2×3+2
3=1×2+1
Hence the greatest common divisor of 277 and 355 is 1. From the last equation, we have
1=3−1×2
= 3 − 1 × (8 − 2 × 3)
(from the second to last equation)
=3×3−1×8
= 3 × (35 − 4 × 8) − 1 × 8
= 3 × 35 − 13 × 8.
(from the third to last equation)
2
Continuing in this way, we eventually obtain
1 = 103 × 355 − 132 × 277.
4. Let a, b, c be integers such that a and b and also a and c are relatively prime.
a. Prove that a and bc are relatively prime.
b. Give an example to show that a and b + c need not be relatively prime.
a. If a prime p divides bc then we know from class that p must divide b or p must divide
c. Thus any common prime divisor of a and bc is a common prime divisor of a and b
or a common prime divisor of a and c. But (a, b) = 1 and (a, c) = 1. So a and b and
also a and c have no common prime divisor. Hence a and bc have no common prime
divisor and so (a, bc) = 1.
b. Take a = 2, b = 3, c = 5. Clearly (2, 3) = 1 and (2, 5) = 1 but (2, 3 + 5) = 2.
5. Let a and b be integers and let n be a positive integer.
a. Show that (a, b) = 1 implies (an , bn ) = 1.
b. Show that if an | bn then a | b.
a. Suppose (an , bn ) 6= 1. Then there is a prime number p such that p | an and p | bn .
Since p is prime, it follows that p | a and p | b and thus (a, b) 6= 1. This shows (a, b) 6= 1
whenever (an , bn ) 6= 1. Equivalently, (a, b) = 1 implies (an , bn ) = 1.
b. Here’s an alternative proof to the one we gave in class. Observe first that if A, B, D
are integers such that D | A and D | B then (A/D, B/D) = (A, B)/D. Indeed,
from class (A/D, B/D) is the least positive integer than can be written in the form
(A/D)x + (B/D)y, for some integers x and y. But this is 1/D times the least positive
integer of the form Ax + By (for some x and y). Hence (A/D, B/D) = (A, B)/D.
There is no harm in assuming a and b are positive. Write (a, b) = d. By what we’ve
just proved,
(a/d, b/d) = (a, b)/d = 1.
Part a then implies (an /dn , bn /dn ) = 1. Hence (an , bn )/dn = 1 or (an , bn ) = dn . But
an | bn and so (an , bn ) = an . Thus an = dn and a = d or equivalently a | b.
6. If today is Thursday, what day will it be a googol (= 10100 ) days from now?
We need to find the remainder when 10100 is divided by 7. Now 10 ≡ 3 (mod 7) and
so 10100 ≡ 3100 (mod 7). By Fermat’s little theorem, 36 ≡ 1 (mod 7). Thus, using
100 = 6 × 16 + 4, we have
3100 ≡ 36×16+4
(mod 7)
6 16 4
(mod 7)
≡ (3 ) 3
4
≡3
(mod 7).
Finally, 32 ≡ 2 (mod 7), and so
34 ≡ 32 × 32
≡2×2
≡4
(mod 7)
(mod 7)
(mod 7).
3
Therefore 10100 ≡ 4 (mod 7). So if today is Thursday then it will be Monday (= 4 days
later) in 10100 days.
7. Find the units digit of 31,000,000 .
We need to determine 31,000,000 (mod 10). Note 32 ≡ −1 (mod 10) and thus
31,000,000 ≡ (32 )500,000
≡ (−1)
≡1
(mod 10)
500,000
(mod 10)
(mod 10).
The units digit of 31,000,000 is 1.
8. A positive integer is said to be a base 10 palindrome if its decimal representation reads
the same forwards and backwards (e.g., 1221 and 13431 are palindromes in this sense).
Show that any base 10 palindrome that has an even number of digits is divisible by 11.
[Hint: 10 ≡ −1 (mod 11).]
To see the main idea, consider the palindrome abba. Note 10 ≡ −1 (mod 11) and thus
10m ≡ (−1)m (mod 11), for any positive integer m. Hence
abba = a + b × 10 + b × 102 + a × 103
≡ a − b + b − a (mod 11)
≡0
(mod 11).
That is, 11 | abba. The general argument is the same but requires more notation. Thus
consider the palindrome a0 a1 . . . an an . . . a1 a0 . This equals
a0 + a1 × 10 + · · · + an × 10n + an × 10n+1 + · · · + a1 × 102n + a0 × 102n+1
and hence is congruent mod 11 to
a0 − a1 + · · · + (−1)n an + (−1)n+1 an + · · · + a1 − a0 = 0.
In other words, 11 | a0 a1 . . . an an . . . a1 a0 .
9. a. Show that if a is an odd integer then a2 ≡ 1 (mod 8).
b. More generally, show that, for any odd integer a and any integer n ≥ 1,
n
a2 ≡ 1
(mod 2n+2 ).
a. Since a is odd, a = 2b + 1 for some integer b. Thus
a2 − 1 = (2b + 1)2 − 1
= 4b2 + 4b
= 4b(b + 1).
Note b or b + 1 is even, i.e., is divisible by 2. Hence 4 × 2 = 8 divides 4b(b + 1). This
says exactly that a2 ≡ 1 (mod 8).
4
b. We use induction. The base case is given by Part a. For the inductive step, suppose
k
k
a2 ≡ 1 (mod 2k+2 ) for some positive integer k, equivalently 2k+2 | a2 − 1. Consider
k+1
a2
k
− 1 = a2
×2
−1
2k 2
= (a ) − 1
k
k
= (a2 − 1)(a2 + 1).
k
k
k
By hypothesis, 2k+2 | a2 − 1. Further, a2 is odd (since a is odd) and so a2 + 1 is
k
k
k+1
even. Thus 2k+3 = 2k+2 ×2 divides (a2 −1)(a2 +1) and hence a2
≡ 1 (mod 2k+3 ).
n
By induction, a2 ≡ 1 (mod 2n+2 ) for all integers n ≥ 1.
10. Use Fermat’s little theorem to show that 13 divides 270 + 370 .
By Fermat’s little theorem, 212 ≡ 1 (mod 13). Now 70 = 12 × 5 + 10 and thus
270 ≡ (212 )5 210
10
≡2
(mod 13)
(mod 13)
Note 24 ≡ 3 (mod 13). Therefore
210 ≡ (24 )2 4
2
≡3 4
≡ 10
70
In all, 2
(mod 13)
(mod 13)
(mod 13).
≡ 10 (mod 13). In the same way, we show 370 ≡ 3 (mod 13). Hence
270 + 370 ≡ 10 + 3
≡0
that is, 13 | 270 + 370 .
(mod 13)
(mod 13),