Number Theory: Divisibility and Prime Numbers: Problems with Solutions

Solution:
For [tex]n\ge 5[/tex] the product [tex]n!=1\times 2\times 3\times 4\times 5\times \ldots\times n[/tex] contains the factors [tex]2[/tex] and [tex]5[/tex], so it is divisible by [tex]10[/tex] and ends in [tex]0[/tex]. Hence the terms from [tex]5![/tex] to [tex]2018![/tex] do not affect the last digit of [tex]P[/tex], and it equals the last digit of the sum
[tex]1!+2!+3!+4!=1+2+6+24=33[/tex]
Therefore the last digit of [tex]P[/tex] is [tex]3[/tex].

Solution:
The last digit of a product depends only on the last digits of the factors, so [tex]237^{127}[/tex] ends in the same digit as [tex]7^{127}[/tex]. The first powers of [tex]7[/tex] are
[tex]7^1=7,\quad 7^2=49,\quad 7^3=343,\quad 7^4=2401[/tex]
Their last digits are [tex]7[/tex], [tex]9[/tex], [tex]3[/tex], [tex]1[/tex], and after that they repeat with period [tex]4[/tex], because multiplying a number ending in [tex]1[/tex] by [tex]7[/tex] again gives a number ending in [tex]7[/tex]. Since [tex]127=4\times 31+3[/tex], we have
[tex]7^{127}=\left(7^4\right)^{31}\times 7^3[/tex]
The factor [tex]\left(7^4\right)^{31}[/tex] ends in [tex]1[/tex] and [tex]7^3=343[/tex] ends in [tex]3[/tex], so [tex]7^{127}[/tex] ends in [tex]3[/tex]. Hence the last digit of [tex]237^{127}[/tex] is [tex]3[/tex].

Solution:
Let [tex]k[/tex] be the required number of years. Today the age [tex]1[/tex] is not prime, so [tex]k\ge 1[/tex]. All present ages are odd. If [tex]k[/tex] is odd, all six new ages are even, and [tex]3+k\ge 4[/tex] is an even number greater than [tex]2[/tex], so it is not prime. Hence [tex]k[/tex] is even.
For [tex]k=2[/tex] the ages are [tex]3, 5, 9, 11, 15, 17[/tex], and [tex]9=3\times 3[/tex] is not prime.
For [tex]k=4[/tex] the ages are [tex]5, 7, 11, 13, 17, 19[/tex], and all of them are prime.
Therefore the smallest number of years is [tex]4[/tex].

Solution:
Factor out the smallest power of [tex]3[/tex]:
[tex]3^{2016}+3^{2017}+3^{2019}=3^{2016}\left(1+3+3^{3}\right)=31\times 3^{2016}[/tex]
Since [tex]31[/tex] is prime, every divisor of the number has the form [tex]3^{k}[/tex] or [tex]31\times 3^{k}[/tex], where [tex]0\le k\le 2016[/tex]. The two-digit divisors of the form [tex]3^{k}[/tex] are [tex]27[/tex] and [tex]81[/tex]. The two-digit divisors of the form [tex]31\times 3^{k}[/tex] are [tex]31[/tex] and [tex]93[/tex], because already [tex]31\times 9=279[/tex] has three digits.
The largest of them is [tex]93[/tex].

Solution:
The pages are numbered from [tex]1[/tex], so every leaf has an odd page on its front side and the next even page on its back side. The torn block starts with page [tex]387[/tex] (odd, the front of a leaf) and ends with the back side of a leaf, so the last torn page has an even number greater than [tex]387[/tex].
The other numbers written with the digits [tex]3[/tex], [tex]8[/tex] and [tex]7[/tex] are [tex]378, 738, 783, 837, 873[/tex]. The number [tex]378[/tex] is less than [tex]387[/tex], and [tex]783[/tex], [tex]837[/tex], [tex]873[/tex] are odd, so the last torn page is [tex]738[/tex]. Peter tore out all pages from [tex]387[/tex] to [tex]738[/tex] inclusive, and their number is
[tex]738-387+1=352[/tex]

Solution:
Write the number as [tex]a=2007q+99[/tex], where [tex]q\ge 0[/tex] is an integer. Both [tex]a[/tex] and [tex]99[/tex] are odd, so [tex]2007q=a-99[/tex] is even; since [tex]2007[/tex] is odd, [tex]q[/tex] is even, [tex]q=2s[/tex]. Then
[tex]a=4014s+99=18\times 223s+18\times 5+9=18\left(223s+5\right)+9[/tex]
Since [tex]0\le 9\lt 18[/tex], the remainder of [tex]a[/tex] on division by [tex]18[/tex] is [tex]9[/tex].

Solution:
A number with [tex]k[/tex] digits has digit sum at most [tex]9k[/tex], and [tex]9\times 223=2007\lt 2015[/tex], so the number has at least [tex]224[/tex] digits. Numbers with [tex]224[/tex] digits and digit sum [tex]2015[/tex] exist (one is given below), and every number with more digits is larger, so the smallest such number has exactly [tex]224[/tex] digits.
Of two numbers with the same number of digits, the one with the smaller first digit is smaller. The other [tex]223[/tex] digits have sum at most [tex]2007[/tex], so the first digit is at least [tex]2015-2007=8[/tex]. The first digit equals [tex]8[/tex] only if all the other [tex]223[/tex] digits are [tex]9[/tex], which gives exactly one number:
[tex]8\underbrace{99\ldots 9}_{223}[/tex]
Its units digit is [tex]9[/tex].

Solution:
Assume the contrary: each of the ten digits [tex]0, 1, \ldots, 9[/tex] occurs at most [tex]10[/tex] times. Then [tex]A[/tex] has at most [tex]10\times 10=100[/tex] digits. Since [tex]A[/tex] has exactly [tex]100[/tex] digits, equality holds, so every digit occurs exactly [tex]10[/tex] times. Then the digit sum of [tex]A[/tex] is
[tex]10\times \left(0+1+2+\ldots+9\right)=10\times 45=450[/tex]
This sum is divisible by [tex]3[/tex], so by the divisibility rule for [tex]3[/tex] the number [tex]A[/tex] is divisible by [tex]3[/tex]. But [tex]A=2^{332}[/tex] has no prime divisor other than [tex]2[/tex], a contradiction. Hence at least one digit occurs at least [tex]11[/tex] times.

Solution:
Let [tex]p[/tex] be such a prime. Since [tex]p[/tex] is a sum of two primes, [tex]p\ge 2+2=4[/tex], so [tex]p[/tex] is odd and [tex]p\ge 5[/tex]. A sum of two primes is odd only if one of them is the even prime [tex]2[/tex]; hence [tex]p=q+2[/tex] with [tex]q[/tex] prime, i.e. [tex]p-2[/tex] is prime.
From [tex]p=s-t[/tex] with primes [tex]s[/tex] and [tex]t[/tex] we get [tex]s=p+t\gt p\ge 5[/tex], so [tex]s[/tex] is odd; then [tex]t=s-p[/tex] is even, i.e. [tex]t=2[/tex], and [tex]p+2=s[/tex] is prime.
Thus [tex]p-2[/tex], [tex]p[/tex] and [tex]p+2[/tex] are all prime. Since [tex]p-2\equiv p+1\pmod 3[/tex], these three numbers give the same remainders on division by [tex]3[/tex] as the consecutive numbers [tex]p[/tex], [tex]p+1[/tex], [tex]p+2[/tex], i.e. the remainders [tex]0[/tex], [tex]1[/tex] and [tex]2[/tex]. So one of them is divisible by [tex]3[/tex] and, being prime, equals [tex]3[/tex]. As [tex]p\ge 5[/tex], the only possibility is [tex]p-2=3[/tex], i.e. [tex]p=5[/tex]. Check:
[tex]5=2+3,\quad 5=7-2[/tex]
So [tex]5[/tex] is the only such prime.

Solution:
Every fraction has the form [tex]\frac{k}{n+k+2}[/tex] with [tex]k=19, 20, \ldots, 91[/tex]. A common divisor of [tex]k[/tex] and [tex]n+k+2[/tex] also divides their difference [tex]n+2[/tex], and conversely, so
[tex]\gcd(k, n+k+2)=\gcd(k, n+2)[/tex]
Hence all fractions are irreducible exactly when [tex]n+2[/tex] is coprime with each of the numbers [tex]19, 20, \ldots, 91[/tex].
Let [tex]p[/tex] be a prime divisor of [tex]n+2[/tex] (it exists, because [tex]n+2\ge 3[/tex]). If [tex]19\le p\le 91[/tex], then [tex]p[/tex] itself is one of the numerators. If [tex]p\lt 19[/tex], then among the [tex]73[/tex] consecutive numbers [tex]19, 20, \ldots, 91[/tex] there is a multiple of [tex]p[/tex]. In both cases some fraction can be reduced by [tex]p[/tex].
Therefore every prime divisor of [tex]n+2[/tex] is greater than [tex]91[/tex], and the smallest prime greater than [tex]91[/tex] is [tex]97[/tex]. Hence
[tex]n+2\ge 97\Rightarrow n\ge 95[/tex]
For [tex]n=95[/tex] we have [tex]n+2=97[/tex], a prime that divides none of the numbers [tex]19, 20, \ldots, 91[/tex], so all fractions are irreducible. The answer is [tex]n=95[/tex].