Download Problem Set 2 Solutions: Number Theory

Problem Set 2 Solutions: Number Theory
1. There are 4 one-digit reversibly prime numbers, namely all the primes less than 10. The other reversibly prime numbers
are 11, 13, 17, 31, 37, 71, 73, 79, and 97. Therefore, there are 13 reversibly prime numbers.
2. Since τ (n) is odd, n must be a perfect square. Furthermore, it must be the fourth power of a prime number; if it were
the square of a composite number, it would have at least two prime factors p1 and p2 , but then it would have at least
9 factors. Thus, n2010 is the 8040th power of a prime number, and therefore has 8041 divisors.
3. The discriminant of this quadratic equation is b2 +144; if this is a square, then the given quadratic has integer solutions.
Thus we must solve b2 + 144 = a2 , or a2 − b2 = 144. This has 9 solutions, since there are nine ways to distribute the
factors of 144 such that a + b = x and a − b = y have x and y being the same parity.
4. By Euler’s Theorem, since φ(1000) = 400, we have that 72010 ≡ 710 (mod 1000). We can exponentiate by squares to
get an answer of 249 .
5. τ (N ) = 22010 , since we can choose whether each prime is a potential factor or not of a given divisor. φ(22010 ) = 22009 ,
since exactly the odd numbers are relatively prime to a power of two. τ (22009 ) = 2010 , since 2x divides 22009 for
0 ≤ x ≤ 2009.
6. Note that if 2010! ends in k zeroes in base b, then bk must divide 2010!. We first calculate the power of 2 that exactly
divides 2010!. The desired exponent is 1005 + 502 + 251 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 2002. We now calculate the
power of 3 that exactly divides 2010!. The desired exponent is 670 + 223 + 74 + 24 + 8 + 2 = 1001. If 12n divides 2010!,
we must have 2n ≤ 2002 and n ≤ 1001, so clearly, the maximal valid n is 1001 .
7. We analyze this equation on four cases.
Case 1: n is odd. If n is odd, we need to find a prime p such that p − 1 is a factor of 12. The primes that satisfy this
are 3, 5, 7, and 13. Note that if 3 is a factor, then 7 must also be a factor, so our first solution is 3 · 7 = 21. Note
furthermore that 13 is prime, so φ(13) = 12, so 13 is also a solution.
Case 2: n is exactly divisible by two. We just multiply the solutions we get from case 1 each by 2, giving us 26 and 42.
Case 3: n is exactly divisible by four. We now need a factor of 3 in φ(n), so either 32 divides n or we have a prime that
is one greater than a multiple of 3. Testing the former case, we note that n = 36 is satisfactory. Checking the second
case, n = 28 is valid.
Case 4: n is exactly divisible by eight. Note, however, that φ(n) = 3 has no solutions because φ(n) is even for all n ≥ 2.
Thus, we have 6 solutions.
8. Note that z must be even, otherwise the LHS is odd and there are no solutions. Furthermore, z can range from 0 to
670. Thus, 2x + 4y = k, where k is a multiple of 6 between 0 and 2010. Since k is even, we can divide both sides of
this equation by 2, giving
us x + 2y = k,
where k is a multiple of 3 between 0 and 1005. From here, we note that y can
k
k
range from 0 to
, giving us 1 +
possibilites for x + 2y = k. To simplify the analysis, we consider k odd and k
2
2
even separately.
k+1
Case 1: k is odd. Therefore, there are
possibilities for x + 2y = k. Thus, we are calculating the sum of the
2
505 · 168
arithmetic sequence 2 + 5 + 8 + · · · + 503. This sum is
= 42420.
2
k
Case 2: k is even. Thus, there are + 1 possibilities for x + 2y = k. Thus, we are calculating the sum of the arithmetic
2
sequence 1 + 4 + 7 + · · · + 502. This is just 42420 − 168 = 42252.
Thus the answer is 42420 + 42256 = 84672 .
9. We can factor the expression into (x − 1)(y − 2)(z − 3) = 6. In effect, we are now calculating the number of ordered
triples (a, b, c) such that abc = 6. We first compute unordered triplets and then compute the number of ways these can
be ordered and the number of ways signs can be inverted. In fact, for any arbitrary ordered triplet of positive integers,
there are 3 more ordered triplets that are valid by inverting the signs of 2 of the numbers present.
The unordered triplets are (1, 1, 6) and (1, 2, 3). The first one has 12 ordered triplets correspond to it; the second one
24. Thus there are 36 valid ordered triplets.
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10. Every positive integer is a 2010-binary multiplier. Consider an arbitrary integer k. It is obvious that there exist two
integers a and b such that 2010a ≡ 2010b (mod k); apply the pigeonhole principle over the first k + 1 positive integers,
for example. Since we have a repeated value, just take 2010a , 2010b , and k − 2 more iterations. You have a number
that is obviously a multiple of k, since it is written as the sum of k integers that have the same residue modulo k, and
clearly, when written in base 2010, it has ones if and only if the corresponding power was used in the sum, and zeroes
otherwise. Thus, 2010 of the first 2010 positive integers are 2010-binary multipliers.
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