OpenStudy (anonymous):

T*(RBJBR)=VPLNT Where the letters represent a digit in a number. "RBJBR" is five digits long. Find possible letter values. HOORAY FOR CRYPTO.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

No, lol, multiple choice is for dorks.

5 years ago
OpenStudy (anonymous):

5 years ago
OpenStudy (anonymous):

lol

5 years ago
OpenStudy (anonymous):

Insane question!!

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

We are expect to find possible solutions by hand.

5 years ago
OpenStudy (anonymous):

Not all of them, mind you.

5 years ago
OpenStudy (anonymous):

0*(00000) = 00000

5 years ago
OpenStudy (anonymous):

They have to be unique, unfortunately.

5 years ago
OpenStudy (anonymous):

Forgot to mention. :(

5 years ago
OpenStudy (anonymous):

The digits? No problem.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (kinggeorge):

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 $$R=1$$

5 years ago
OpenStudy (anonymous):

Ahh, excellent point.

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (kinggeorge):

More strongly, $$B, T \geq 2$$

5 years ago
OpenStudy (kinggeorge):

Another fact we know: $$L= T\cdot J\mod 10$$ plus the second digit of $$T\cdot B$$ taken mod 10.

5 years ago
OpenStudy (kinggeorge):

Also, $$V>T$$ so $$T\leq 8$$.

5 years ago
OpenStudy (kinggeorge):

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)

5 years ago
OpenStudy (kinggeorge):

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. :/

5 years ago
OpenStudy (anonymous):

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. :(

5 years ago
OpenStudy (kinggeorge):

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

5 years ago
OpenStudy (anonymous):

I'm perpetually at a loss.

5 years ago
OpenStudy (kinggeorge):

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.

5 years ago