My First Cryptarithm!

Logic Level 3

$\begin{array}{ccccccc} & & & & J & E & E & P\\ \times & & & & J & E & E& P\\ \hline & B & E & E & B & E & E& P \end{array}$

The above shows a cryptarithm such that each letter represents a distinct single non-negative integer with $J, B$ being non-zero. Find the sum of all the possible values of the $4$ -digit number $\overline{JEEP}$ .

The answer is 2001.

2 solutions

Satyajit Mohanty
Jul 24, 2015

So we have $P^2 = P + 10k$ for some integer $k, \quad 0 \leq k \leq 8$ , we see that $P$ is one of $0,1,5$ or $6$ . Now by considering the last two digits of each factor and the product we have $(10E + P)^2 = 100n + 10E + P$ for some integer $n < 100$ .

This means that $20PE+P^2-10E-P = 10E(2P-1)+P(P-1)$ is a multiple of $100$ . Let us consider $P = 6$ . Then $110E+30$ is a multiple of $100$ , implying that $E = 7$ . This means that $776^2$ must end in the digits $776$ , but $776^2$ actually ends in the digits $176$ . Thus, $P \neq 6$ .

Next try $P = 5$ . Then $90E + 20$ is a multiple of $100$ , implying that $E = 2$ . This means that $225^2$ must end in the digits $225$ , but $225^2$ actually ends in the digits $625$ . Thus $P \neq 5$ . Therefore, $P = 0$ or $1$ .

In either case we have $P(P-1) = 0$ , which means that $10E(2P-1)$ is a multiple of $100$ . Since $2P - 1 = \pm 1$ , we conclude that $E=0$ implying that $P=1$ , since it must be different from $E$ . Thus we have $P=1, E=0$ . Then we have $B = J^2$ and $B = 2J$ , since $(J001)^2 = (J)^200(2J)001 = B00B001$ . Since $J^2 = 2J$ and $J \neq 0$ , we conclude that $J = 2$ , whence $\overline{JEEP} = \boxed{2001}$ .

If you've entered this problem via this wiki: Cryptogram , and if you're looking for a solution involving Chinese-Remainder Theorem , please check in the comments to this post for @PiHanGoh 's solution.

The multiplication shows the last three digits of a square of a number is the same as the last three digits of the very number itself. So we have $x^2 \equiv x \pmod{1000}$ or $x(x-1) \equiv 0 \pmod{1000}$ . Since $x$ and $x-1$ are consecutive integers, then they are coprime. Because $1000 = 2^3 \cdot 5^3$ , then one of $x,x-1$ is divisible by 8 and the other is divisible by 125. Solve via Chinese Remainder Theorem gives $x \equiv 001,376,625 \pmod{1000}$ but since the last three digits is of the form of $\overline{EEP}$ , then $x\equiv 001$ only. So $E=0, P = 1$ . And we are left to check for the squares of $1001,2001,3001,\ldots, 9001$ . But since $4001^2$ is already a 7-digit number, we are then reduce to check for the squares of $1001,2001,3001$ . Inspection shows that $J = 2$ only.

Pi Han Goh - 5 years, 10 months ago

Nicely Done! Why don't you post this as a solution instead of a comment?

Satyajit Mohanty - 5 years, 10 months ago

I promise myself I only post solutions on Mondays

Pi Han Goh - 5 years, 10 months ago

@Pi Han Goh – Superstitious! :D ... OK! You post this comment as a solution on Monday!

Satyajit Mohanty - 5 years, 10 months ago

@Satyajit Mohanty – SLACK SLACK SLACK SLACK NOW NOW NOW

Pi Han Goh - 5 years, 10 months ago

Well, there's just so many problems, all of the same type!

A similar problem ,

A similar problem

All have the same method, with a little bit of variation... Posting this, if somebody wants to solve problems of just 1 type, together!

Aditya Raut - 5 years, 10 months ago

Problems might be similar, but approaches to all would be different! :P

Satyajit Mohanty - 5 years, 10 months ago

That's right @Satyajit Mohanty , you follow the way in your solution, and I follow the brute force! Lol, the 'same way' for all is programming, right?

from itertools import permutations
li=[]
for i in permutations('012345678'):
    k=''.join(i)
    [J,E,P,B]=list(k)[:4]
    if int(J+E+E+P)*int(J+E+E+P)==int(B+E+E+B+E+E+P):
        li.append(int(J+E+E+P))

Though in your question a faster way than permutations is actually appending all the $10^4$ possible quadruples in a list before, but see? The above algorithm can be typed in, in all the similar problems ;) Btw very nice 'non-bash' solution of yours, really liked it.

Aditya Raut - 5 years, 10 months ago

@Aditya Raut – You're an Olympian Right! Programming never suits you. :P

Satyajit Mohanty - 5 years, 10 months ago

I found the solution of Pi Han Goh, and I wrote 6003. Am I wrong? and if I am right why I can't point?

Guillermo Templado - 5 years, 8 months ago

6003^2 is an 8 digit integer, not a 7-digit integer, so it didn't satisfy the criteria, so it's a not a solution.

Pi Han Goh - 5 years, 8 months ago

I meant I wrote 1001 + 2001 +3001=6003.

Guillermo Templado - 5 years, 8 months ago

@Guillermo Templado – Answer is 2001, not 6003. Check your arithmetic again.

Pi Han Goh - 5 years, 8 months ago

@Guillermo Templado – Ok, I saw my mistake

Guillermo Templado - 5 years, 8 months ago

Mike Sull
Aug 15, 2015

\sqrt{10000000} = around 3162 so Jeep is less than that because BEEBEEP is only 7 digits long. Unless P = 1, BEEBEEP won't make sense (5 and 6 would mess it up). E is obviously 0. Try 1001^2 in calculator, doesn't work. Try 2001^2 in calculator, works, Try 3001^2 in calculator, doesn't work. Answer is 2001.

My First Cryptarithm!

The answer is 2001.

2 solutions

If you've entered this problem via this wiki: Cryptogram , and if you're looking for a solution involving Chinese-Remainder Theorem , please check in the comments to this post for @PiHanGoh 's solution.

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