# Solve this cryptarithm: PI * R^2 = AREA

PI × R^2 = AREA

If each letter in this expression (but not the exponent 2) is replaced with a corresponding digit, the resulting equation will be valid.

What are the values of the letters if

Case 1: no two letters can have the same value?

Case 2: there's no restriction to the value an letter can take? This means that if P = 6 then there can be other letter(s) which are also equal to 6.

P.S: Source contains the solution so will post it later.

• Seems easy with a direct search, but perhaps there is a more clever solution. Note: it's not clear from the rules whether $2$ is allowed as a value, though the solution found by search does not use an extra $2$.
– lulu
Commented Jul 31 at 19:47
• @lulu , an alphabet can take the value of 2 in both the cases. Commented Jul 31 at 19:54
• Didn't mean to do that. I'll delete those comments.
– lulu
Commented Jul 31 at 19:58
• For case 1, the solution is unique. Commented Aug 1 at 1:32
• @Lucenaposition , no leading 0s. Commented Aug 1 at 1:48

Solution for case 1:

Leading zeros in multi-digit integers are not permitted so $$A\ne0$$ so $$AREA>999$$. This means $$PI\times R^2>999$$. If $$R\le3$$ then $$R^2\le9$$ so $$PI\times R^2\le99\times9<999$$ which is too low. So $$R\notin\{0,1,2,3\}$$.

Next, if $$R=4$$ then $$AREA$$ is a multiple of $$16$$ so $$A$$ is even.
Leading zeros in multi-digit integers are not permitted so $$A\ne0$$. But that means $$2000>99\times4^2\ge PI\times R^2=AREA$$ so $$A=1$$. This is not possible so $$R\notin\{4\}$$.

Next, if $$R=5$$ then $$AREA$$ is a multiple of $$25$$ so $$A$$ is $$0$$ or $$5$$.
Leading zeros in multi-digit integers are not permitted so $$A=5$$. But that means $$99\times5^2\ge PI\times R^2=AREA\ge5000$$. This is not possible so $$R\notin\{5\}$$.

Next, if $$R=6$$ then $$AREA$$ is a multiple of $$36$$ so $$A$$ is even.
Leading zeros in multi-digit integers are not permitted so $$A\ne0$$. But that means $$4000>99\times4^2\ge PI\times R^2=AREA$$ so $$A<4$$. Therefore, $$A=2$$. So $$AREA=26E2$$. $$26E2$$ is a multiple of $$9$$ so $$2+6+E+2$$ is a multiple of $$9$$. This means $$E=8$$ but $$2682$$ is not a multiple of $$36$$. So $$R\notin\{6\}$$.

Next, if $$R=8$$ then $$AREA$$ is a multiple of $$64$$. The only multiple of $$64$$ that is of the form $$A8EA$$ is $$4864$$ but then $$PI=4864/64=76$$ so $$E=I$$. So $$R\notin\{8\}$$.

Next, if $$R=9$$ then $$AREA$$ is a multiple of $$81$$. The only multiple of $$81$$ that is of the form $$A9EA$$ is $$6966$$ but that has duplicate digits ($$E=A$$). So $$R\notin\{9\}$$.

Therefore, $$R=7$$. $$AREA$$ is a multiple of $$49$$. The only multiples of $$49$$ that are of the form $$A7EA$$ are $$3773$$ which has duplicate digits and $$4704$$.

Final answer: $$P=9,I=6,R=7,A=4,E=0$$ with $$96\times7^2=4704$$.

Solution for case 2:

The same arguments show $$R\notin\{0,1,2,3,4,5,6\}$$.

$$76\times8^2=4864$$
$$77\times7^2=3773$$
$$86\times9^2=6966$$