# If O^3=DAD and (IM)^2=MOM, then what is MAID?

If $$O^3=DAD$$ and $$(IM)^2=MOM$$, then what is $$MAID$$?

Source: Taken from the book Neurone Abaro Onuronon by Muhammad Zafar Iqbal.

Start by constraining $$M$$:

$$M^2$$ must end in $$M$$. This leaves $$1,5,6$$ ($$0$$ can be ruled out because $$M$$ occurs as the highest digit of a multi-digit number.)

Next, $$I^2 \leq M$$ (otherwise $$\overline{I0}^2 \geq \overline{(M+1)00}$$) and $$(I+1)^2 \geq M$$ (otherwise $$\overline {IM}^2<\overline{M00}$$), giving $$M=1\implies I=1, M=5 \implies I=2, M=6 \implies I=2$$. Calculating the squares gives 1) $$11^2 = 121$$ so $$O=2$$ but the cube of $$2$$ does not have three digits. 2) $$25^2 = 625$$, not a palindrome. 3) $$26^2 = 676$$, so $$O=7$$ and $$7^3 = 343$$, so $$M=6,A=4,I=2,D=3$$.

No calculator required.

It can be seen that $$O^3 = DAD$$ can only happen with

$$7^3 = 343$$

and $$(IM)^2 = MOM$$ can only happen with

$$26^2 = 676$$

and hence the value of $$MAID$$ is

$$6423$$.

It is easy if you are familiar with squares and cubes of small numbers. Otherwise just take a calculator and calculate

cubes until $$10^3$$ and squares until $$32^2$$.

• Could you please elaborate on the "can only happen with" parts? Is this brute forcing logic or some analytical steps? May 7, 2021 at 8:30
• @GeorgeMenoutis It is brute forcing logic: just list out all possibilities. One can do it pretty fast in mind (if familiar with these numbers), or otherwise with the help of a calculator. The point is that there are not many cases to consider. May 7, 2021 at 11:19