NMTC Practice Part 10

Check $3$ cases.

• $\text{min}(x,y,z)=x$ . Divide both sides by $n^x$ .

$1+n^{y-x}=n^{z-x}$

The only natural number powers different by one are $2^0$ and $2^1$ , hence $n=2, x-y=0, z-x=1$ , which means equality holds if and only if the triplets are $(x,x,x+1)$ , where $0\le x\le 9$ (we have $z=x+1\le 10$ because of the restriction $2^z\le 2014$ ). There are $\boxed{10}$ such triplets.

• $\text{min}(x,y,z)=y$ leads to the same triplets.

• $\text{min}(x,y,z)=z$ . Divide both sides by $n^z$ .

$n^{x-z}+n^{y-z}=1$

But we have $x-z\ge 0, y-z\ge 0$ , hence $1=n^{x-z}+n^{y-z}\ge 2$ , impossible.

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All the cases add up to $\boxed{10}$ different triplets.

The triplets are ,

$2^{0}+2^{0}=2^{1}$

Triplet = (0,0,1)

$2^{1}+2^{1}=2^{2}$

Triplet = (1,1,2)

$2^{2}+2^{2}=2^{3}$

Triplet = (2,2,3)

$2^{3}+2^{3}=2^{4}$

Triplet = (3,3,4)

And so on till,

$2^{9}+2^{9}=2^{10}$

Triplet = (9,9,10)

Hence , total number of triplets is $\boxed{10}$ .