Question

T*(RBJBR)=VPLNT

Where the letters represent a digit in a number. "RBJBR" is five digits long.

Find possible letter values.

HOORAY FOR CRYPTO.

Answer 1

@FoolForMath @KingGeorge @Mr.Math :P Get in here, whether you like it or not. @TuringTest

Answer 2

any answer choices?

Answer 3

No, lol, multiple choice is for dorks.

Answer 4

lol what about fools?

Answer 5

lol

Answer 6

Insane question!!

Answer 7

I can solve it using a computer program, just traversing through all 5 digits palindromes.

Answer 8

Dang it, @FoolForMath , you are such a jerk. If I wanted I could get Mathematica to solve this for me.

Answer 9

6*(15251) = 91506
6*(15351) = 92106
6*(15451) = 92706
6*(15551) = 93306
6*(15651) = 93906
6*(15751) = 94506
6*(15851) = 95106
6*(15951) = 95706
6*(16061) = 96366
6*(16161) = 96966
6*(16261) = 97566
6*(16361) = 98166
6*(16461) = 98766
6*(16561) = 99366
6*(16661) = 99966
7*(10001) = 70007
7*(10101) = 70707
7*(10201) = 71407
7*(10301) = 72107
7*(10401) = 72807
7*(10501) = 73507
7*(10601) = 74207
7*(10701) = 74907

To name a few. Wrote a program.

Answer 10

Please be respectful. There are many solution to this problem so, the problem in this form is just not worth for a manual attempt.

Answer 11

>:| Maybe I should tell that to the math dept. then?

Answer 12

We are expect to find possible solutions by hand.

Answer 13

Not all of them, mind you.

Answer 14

0*(00000) = 00000

Answer 15

They have to be unique, unfortunately.

Answer 16

Forgot to mention. :(

Answer 17

The digits? No problem.

Answer 18

2*(15451) = 30902
2*(18481) = 36962
3*(15951) = 47853
3*(16261) = 48783
3*(16761) = 50283
3*(16961) = 50883
3*(18081) = 54243
3*(19091) = 57273
3*(19291) = 57873
4*(12721) = 50884
4*(15751) = 63004
4*(15951) = 63804
4*(17071) = 68284
4*(17271) = 69084
4*(18081) = 72324
4*(19091) = 76364
4*(19591) = 78364
5*(14641) = 73205
5*(16461) = 82305
5*(17271) = 86355
5*(18681) = 93405
6*(13431) = 80586
6*(14541) = 87246
6*(15451) = 92706
7*(12421) = 86947
7*(13631) = 95417
7*(14041) = 98287

Answer 19

6*(15451) = 92706 is a possible solution. There could be more!

Answer 20

Ah, I'm not looking for guessed numbers, or algorithmically determined ones. I'm looking for a formally written proof that logically deduces possible values.

Answer 21

Well, it's easy to find R. Since \(T\cdot R =T\mod 10\) it must be true that \(R=1\).

Answer 22

T * R = T mod 10
2 * 6 = 2 mod 10

Answer 23

Good point. T=5, R=3 also works.

Answer 24

Weird how 1 is the only of value of R that works though, huh.

Answer 25

(That is, unless I messed up writing my program, but the list with unique digits that I posted was the entire output)

Answer 26

It is still true that R=1 however. You just need to look at the other side. Since RBJBR is a palindrome, if R is greater than 1 such that \(TR=T\mod 10\), then \(TR\geq 10\)

Thus, if \(R\geq1\) The number VPLNT would have 6 digits. We conclude that \(R=1\)

Answer 27

Ahh, excellent point.

Answer 28

If I'm doing this correct, we also have that \(T\cdot B\equiv N \mod 10\)

Answer 29

Since these all values are distinct, this means that T and B are non-zero.

Answer 30

More strongly, \(B, T \geq 2\)

Answer 31

Another fact we know:

\(L= T\cdot J\mod 10\) plus the second digit of \(T\cdot B\) taken mod 10.

Answer 32

Also, \(V>T\) so \(T\leq 8\).

Answer 33

Let's list some pairs \((T, B)\) that are impossible with our current criteria. (Ignoring those less than 2 or those where T=B)

(2, 6)
(6, 2)
(4, 6)
(6, 4)
(6, 8)
(8, 6)
(5, 3)
(3, 5)
(5, 7)
(7, 5)
(5, 9)
(9, 5)
(3, 7)
(7, 3)

Answer 34

That leaves us with the possible pairs of (we can also reverse these)

(2, 3)
(2, 4)
(2, 5)
(2, 7)
(2, 8)
(2, 9)
(3, 4)
(3, 6)
(3, 8)
(3, 9)
(4, 5)
(4, 7)
(4, 8)
(4, 9)
(5, 6)
(5, 8)
(6, 7)
(6, 9)
(7, 8)
(7, 9)
(8, 9)

Let's see if we can narrow this down a bit. :/

Answer 35

I feel a bit like a jerk not participating in the problem solving process, but when you're on, I'm not, and vice versa. And now I need to go to sleep. :(

Answer 36

Good night. I'm sort of at a loss of where to go from here.

Answer 37

I'm perpetually at a loss.

Answer 38

I suppose we could also say that P equals the sum of N and the second digit of \(T\cdot J\).

I didn't show all steps to get that, but it's true.