My teacher just asked me a question like this but i do not know how to start and work it out at all. Can someone help me out with that?
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$\begingroup$Hint: The number is odd, so all of its prime factors are odd. Could they be all nasty (end in $1$ or $5$ or $9$? Examine products of nasty numbers. You will find they are all nasty.
Remark: Note that a nice number can have some nasty prime factors. For example, $77$ has the nasty prime factor $11$, but it also has the nice prime factor $7$. Note also that the product of nice numbers can be nasty, example $3\cdot 7=21$.
$\endgroup$ $\begingroup$Lets try elimination :
Numbers ending by 3 or 7 are odd.So they must consists of factors ending by {1, 3, 5, 7, 9}.(excluding even numbers)
Numbers having 5 as their factor will lead to numbers ending with 0 and 5 only. Hence the number must have factors ending with {1, 3, 7, 9}
Numbers ending with 1 might or might not be there in the factors list as multiplying 1 will result in the same number.Hence {3, 7, 9}.
Numbers ending with 9 alone cannot be there as the prime factors because 9 * 3 = 27 and 9 * 7 = 63 i.e 9 needs numbers ending with either 3 or 7 with it to satisfy the definition of nice number. Hence for the factors of a nice number ,numbers ending with either of {3, 7} are a must.
$\endgroup$ $\begingroup$Let $n$ denote a nice number.
Each prime factor of $n$ ends with $1$ or $3$ or $7$ or $9$.
Assume by contradiction that each prime factor of $n$ ends with $1$ or $9$.
Let $m$ denote the number of prime factors which end with $9$.
If $m$ is even then $n$ ends with $1$, since $9^m=9^{2k}=81^k\equiv1\pmod{10}$.
If $m$ is odd then $n$ ends with $9$, since $9^m=9^{2k+1}=81^k\cdot9\equiv9\pmod{10}$.
Hence one of the prime factors of $n$ must end with neither $1$ nor $9$.
Hence one of the prime factors of $n$ must end with either $3$ or $7$.
Hence one of the prime factors of $n$ is nice.
$\endgroup$ $\begingroup$Consider any two numbers $x,y$ , so that the product $xy$ ends in $3$, or in $7$ This means that the product of the rightmost digits $a,b$ respectivelymust end in $3$ or in $7$ Since $3,7$ are odd primes, their right most digits must be one of the pairs: $(1,3),(3,1)$ for those ending in $3$ , or $(1,7),(7,1)$ , for those ending in $7$. Notice this is also trivially-true for numbers ending in primes in {$0,1,..,9$}.
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