$ Solve \mspace8pt (abc)^2$

Let p =(a,b,c)

STEP 1: $ \text{we can write abc} = a \times 10^2 + b \times 10 + c $

We have to square abc, Now we have $(a + b + c)^2 = a^2 +b^2 + c^2 +2ab +2bc +2ac$

STEP 2:$(a \times 10^2 + b \times 10 + c)^2 \\ = a^2 10^2 +2ab10^3 + 2ac10^2 +b^2 10^2 +2bc +c^2$

$ Let, \mspace8pt p^2 =(a \times 10^2 + b \times 10 + c)^2 \\ = a^2 10^2 +2ab10^3 + (2ac + b^2)10^2 +2bc +c^2$

STEP 3: $ p^2 =(a^2,2ab,2ac+b^2,2bc,c^2) $

In $p^2$ only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

Example 1: Find square of 231?

STEP 3: $(2^2,2\times 2\times 3,2\times 2\times 1 + 3^2,2\times 3\times 1,1^2) \\=(4,12,13,6,1)$

Only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

(4,12,13,6,1) =(4,13,3,6,1)=(5,3,3,6,1)=53361

Example 2: Find square of 475?

STEP 3: $(4^2,2\times 4\times 7,2\times 4\times 5 + 7^2,2\times 7\times 5,1^2)=(16,56,89,70,25)$

Only the leftmost part can have more than one digit. For the rest of the parts, we need to carry over the number preceding the units digit, to the immediate left part , and add it there respectively.

(16,56,89,70,25) =(16,56,89,72,5)=(16,56,96,2,5)=(16,65,6,2,5)=(22,5,6,2,5)=225625