Condition : \(a,b,c,d\in N;a,b,c,d< 10;a\ne0\)

\(\left[b,c,d\right]\ge0\Rightarrow a^a\le100\Rightarrow a\in\left\{1;2;3\right\}\)

If a = 1,then \(\left[b,c,d\right]=100-1^1=99=3^2.11\)

So,there'll be at least 1 number that have the factor 11 in its/their form of prime factorization (asburd since b,c,d < 10)

If a = 2,then \(\left[b,c,d\right]=100-2^2=96=2^5.3\)

So,there'll be at least 1 number that have the factor 2⁵ in its/their form of prime factorization (asburd)

If a = 3,then \(\left[b,c,d\right]=100-3^3=73\)

So,there'll be at least 1 number that have the factor 73 in its/their form of prime factorization (asburd)

Hence,there's no satisfied number

We have a,b,c,d are four digits so \(0\le a,b,c,d\le9\)

Because a^a + [b,c,d] = 100

=> a^a \(\le100\)=> a \(\in\left\{1,2,3\right\}\)

If a = 1

=> [b,c,d] = 100 - 1 = 99 = 9.11

But 11 has two digits so b,c,d unsatisfactory 

If a = 2

=> [b,c,d] = 100 - 2² = 96 = 3.32

But 32 has two digits so b,c,d unsatisfactory 

If a = 3

=> [b,c,d] = 100 - 3³ = 73 

But 73 has two digits so b,c,d unsatisfactory 

Therefore, there are no numbers satisfying the question