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


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 $32^2$.

  • $\begingroup$ Could you please elaborate on the "can only happen with" parts? Is this brute forcing logic or some analytical steps? $\endgroup$ – George Menoutis May 7 at 8:30
  • $\begingroup$ @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. $\endgroup$ – WhatsUp May 7 at 11:19

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