Answer:
1) 82
2) 28
3) 18
4) 111
Explanation:
1) a two digit number can be written as:
N = b*10 + c*1
where b and c must be single digit numbers
in this case, c is a prime number, and b = 4*c
The prime numbers are:
2, 3, 5...
if c = 2, then:
b = 4*2 = 8
And the two digit number is: 82.
if c = 3, then:
b = 4*3 = 12, and this leads to a 3 digit number.
And for any large value of c, b will not be a single digit number, then the only option here is 82.
2) same as before:
N = b*10 + c
c = b^3.
c ≠ b
Let's give different values for b, and see what happens.
b can not be zero, because that will lead to a single digit number, then we start with one.
b = 1 ----> c = 1^3 = 1 ----> (c = b, then this is not our number)
b = 2 ----> c = 2^3 = 8, then:
N = 2*10 + 8 = 28.
if b = 3 ----> c = 3^3 = 28 (this is not a single digit number, then any value of b equal or larger than 3 does not work)
The only solution here is 28.
3) Same as before:
N = b*10 + c
c = b^2 + 7.
Same approach as before, let's give different values to b and see which one works:
b = 1 ----> c = 1^2 + 7 = 8
N = 1*10 + 8 = 18.
b = 2 ---> c = 2^2 + 7 = 4 + 7 = 11 (This is a two digit number, then any value of b equal or larger than 2 can be discarded)
The only option here is 18.
4) Now we have:
N = a*100 + b*10 + c.
where:
b = c^2
a = b^2.
Now let's give values to c. (here we can start with zero).
if c = 0 , then: b = 0^2 = 0, then: a = 0^2 = 0.
Then N = 0*100 + 0*10 + 0 = 0 (this is not a 3 digit number)
if c = 1, then: b = 1^2 = 1, then: a = 1^2
N = 1*100 + 1*10 + 1 = 111
If c = 2, then: b = 2^2 = 4, then: a= 4^2 = 16 (here a is not a single digit number, then values of c equal or larger than 2 can be discarded)
The only correct option here is 111.