Division problems

Question

I came across these problems :

1) Find the lowest natural number $k$ that satisfies the condition :

$ 7 \mid A$ , where $A = 194^{19} + 125^{14} + k $

2) Find the different prime numbers $m, n, p$, that satisfy the condition :

$$2mnp - m - n - p - mn - mp - np \le 19$$

For the first one, I'm thinking at Fermat's little theorem..

Some hints would really help me.

Answer 1

(194)^19 = -2 (mod 7)

(125)^14 = 1 (mod 7)

194^19 + (125)^14 + K = 0 (mod 7)

-2 + 1 + k = 0 (mod 7)

the minimum natural number for k is 1

k = 1

Answer 2

For 2 :

One has $$\left((2n-1)\left(p-\frac 12\right)-\frac 32\right)m-n-p-np\le 19$$ $$\iff m\le \frac{19+n+p+np}{(2n-1)\left(p-\frac 12\right)-\frac 32}$$ Here, note that $(2n-1)\left(p-\frac 12\right)-\frac 32\gt 0$.

With $2\le m$, one has $$2\le\frac{19+n+p+np}{(2n-1)\left(p-\frac 12\right)-\frac 32}$$ $$\iff 2\left((2n-1)\left(p-\frac 12\right)-\frac 32\right)\le 19+n+p+np$$ $$\iff (n-1)(p-1)\le 8$$

Can you take it from here?

Answer 3

Another approach to problem 2) is to let $m=a+2$, $n=b+2$ and $p=c+2$. The desired inequality becomes

$$2abc+3(ab+bc+ca)+3(a+b+c)\le21$$

so in particular $a+b+c\le7$ with $0\le a,b,c$. Since these variables are each $2$ less than a prime, the allowed values are $a,b,c\in\{0,1,3,5\}$. We may as well assume $a\le b\le c$ (and then permute the results to get all solutions). The $(a,b,c)$ triples that satisfy the displayed inequality are $(0,0,0)$, $(0,0,1)$, $(0,0,3)$, $(0,0,5)$, $(0,1,1)$, $(0,1,3)$, and $(1,1,1)$.