Problems Involving Divisibility Rules: Very Difficult Problems with Solutions

Solution:
[tex]28=4\times 7[/tex], so the number must be divisible by [tex]4[/tex] and by [tex]7[/tex].
Divisibility by [tex]4[/tex]: [tex]\overline{4b}[/tex] is divisible by [tex]4[/tex], so [tex]b\in\{0, 4, 8\}[/tex].
[tex]\overline{ab}+1=10a+b+1=9a+(a+b+1)[/tex], so [tex]a+b+1[/tex] is divisible by [tex]9[/tex]: [tex]a+b=8[/tex] or [tex]a+b=17[/tex].
[tex]b=0 \Rightarrow a=8[/tex]: [tex]63840[/tex]
[tex]b=4 \Rightarrow a=4[/tex]: [tex]63444[/tex]
[tex]b=8 \Rightarrow a=0[/tex] or [tex]a=9[/tex]: [tex]63048[/tex], [tex]63948[/tex]
Divisibility by [tex]7[/tex]:
[tex]63840=7\times 9120[/tex]
[tex]63444=7\times 9063+3[/tex]
[tex]63048=7\times 9006+6[/tex]
[tex]63948=7\times 9135+3[/tex]
Answer: [tex]63840[/tex]

Solution:
The number [tex]10^{2007}+5=100\ldots 005[/tex] has digit sum [tex]1+5=6[/tex], so it is divisible by [tex]3[/tex].
The second factor has [tex]2008[/tex] terms (an even number), so we group them in pairs:
[tex](2+2^2)+(2^3+2^4)+\ldots+(2^{2007}+2^{2008})=2(1+2)+2^3(1+2)+\ldots+2^{2007}(1+2)=3(2+2^3+\ldots+2^{2007})[/tex]
So the second factor is also divisible by [tex]3[/tex]. The product of two numbers, each divisible by [tex]3[/tex], is divisible by [tex]3\times 3=9[/tex].

Solution:
A number and the sum of its digits leave the same remainder when divided by [tex]9[/tex]. Since [tex]N[/tex] is divisible by [tex]9[/tex], so are [tex]a[/tex], [tex]b[/tex] and [tex]c[/tex] (and they are positive).
Each of the [tex]1998[/tex] digits is at most [tex]9[/tex], so [tex]a\le 9\times 1998=17982[/tex].
So [tex]a[/tex] has at most five digits, and if it has five, its first digit is [tex]1[/tex]. Hence [tex]b\le 1+9+9+9+9=37[/tex].
[tex]b[/tex] is a multiple of [tex]9[/tex] not greater than [tex]37[/tex], so [tex]b\in\{9, 18, 27, 36\}[/tex], and in every case [tex]c=9[/tex].

Solution:
For every natural number [tex]x[/tex], the difference [tex]x-S(x)[/tex] is divisible by [tex]9[/tex]. Indeed, if [tex]x=\overline{d_k\ldots d_1d_0}[/tex], then
[tex]x-S(x)=d_k(10^k-1)+\ldots+d_2(100-1)+d_1(10-1)[/tex]
and every number [tex]10^m-1=99\ldots 9[/tex] is divisible by [tex]9[/tex].
Then both [tex]a-S(a)[/tex] and [tex]2a-S(2a)[/tex] are divisible by [tex]9[/tex], and so is their difference:
[tex](2a-S(2a))-(a-S(a))=a-(S(2a)-S(a))=a-0=a[/tex]
Therefore [tex]a[/tex] is divisible by [tex]9[/tex].