Question

plz tell the process and answer

Answer 1

Answer:

Option 2):  a = 4

Step-by-step explanation:

The number with digits ab7 is 100a + 10b + 7, so the equation is

100a + 10b + 7 = a³ + b² + 7³   <=>  100a + 10b = a³ + b² + 7×48   ... (1)

Notice that ab7 is odd and 7³ is odd, so a³+b² is even.  Therefore, either both a and b are odd, or both a and b are even.

Case a and b are odd

Modulo 8, the square of an odd number is 1.  Hence equation (1) taken modulo 8 says

4a + 2b = a + 1  =>  3a = -2b + 1  =>  a = 2b + 3 (mod 8).

Since b is odd, this gives a = 1 (mod 4).  But none of the options satisfy this (the only options presented to us are 6, 4, 7 and 8).

Case a and b are even

Put a = 2k and b = 2m.  Then (1) becomes

50k + 5m = 2k³ + m² + 7×12

Modulo 5, this says

2k³ = -m² - 4  =>  k³ = 2m² - 2 = 2(m²-1) (mod 5).

As m² ∈ { 0, 1, 4 } (mod 5), we get 2(m²-1) ∈ { 3, 0, 1 } (mod 5).

So k³ ∈ { 0, 1, 3 } (mod 5) =>  k ∈ { 0, 1, 2 } (mod 5)

=> a = 2k ∈ { 0, 2, 4 } (mod 5).

This narrows it down to a = 4 or a = 7.

If a = 7, then b² - 10b = 100a - a³ - 7×48 = 7×(100-49-48) = 21

=> b(b-10) = 21.  But b is just a digit, so b-10 is negative, making this impossible.

That just leaves a = 4.

Finishing it off,

b² - 10b = 100a - a³ - 7×48 = 4×(100 - 16 - 84) = 0  => b = 0.

So

a³ + b² + 7³ = 4³ + 0² + 7³ =  407