In this multiplication problem, the 5 characters shown above represent 5 different numbers. Can you figure out what number corresponds to each character? There is no zero.

*Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!*

Hex| PUZZLE MASTER | Profile October 11th, 2010 - 4:30 am4973 x 8 = 39784

Jimmy Anders| PUZZLE MASTER | Profile October 11th, 2010 - 9:15 am4973

x 8

——

39784

I had to use the old “plug and chug” method to get it, though…

Shawn| PUZZLE GRANDMASTER | Profile October 11th, 2010 - 10:36 am4 = lady with earrings

9 = nutty professor

7 = hippy chick

3 = doc holliday

8 = mork’s uncle

4,973 x 8 = 39,784

Bobo The Bear| PUZZLE MASTER | Profile October 11th, 2010 - 12:14 pm4973 x 8 = 39784

suineg| PUZZLE MASTER | Profile October 11th, 2010 - 5:05 pmI think the answer is:

4973

X..3

39784

cool puzzle, I think it was key to find out the possible values of A that were very little and from that on, trial and error for me.

sad-b| Profile October 12th, 2010 - 2:51 amX~

Chris Edwards| Profile October 12th, 2010 - 6:31 am4973 x 8 = 39784

Thanks 4 this puzzle site! MATH RULES! Cool site, really. thanks again. Keep ’em coming, please!

Clearly, the 2 #’s in 1’s position couldn’t be 0, 1, or 5, and neither could be 6 if the other was an even #. And there were a few other general rules that became clear and allowed me to eliminate a lot of potential nos initially. But I eventually had to resort to trying potential combos of the 2 no’s in right-most position. I would have liked to come up with a system of equations to solve this — as I believe I ought to be able to do– rather than the ‘try to plug in possible solutions’ method. Any helpful hints on the most efficient methodology?

thx. Chris

Chris Edwards| Profile October 12th, 2010 - 11:30 am4973 times 8 = 39784. I’ve never used this site b/f 2day, so I’m checking 2 see whether the no of comments increases when I tell the system to post this response. Sorry 4 the redundancy, and thanks again for the cool site. Chris.

RK| Founder | Profile October 12th, 2010 - 1:07 pmHey there Chris, thanks for visiting and liking the site

As you describe, there are a few general rules that constrain some of the #’s, but I think there’s some plug & chug that’s necessary. However, am curious to see what Hex, Bobo, Shawn,Suineg etc… have to say

RK| Founder | Profile October 12th, 2010 - 1:13 pm@Shawn LOLs Mork’s uncle!

Used to watch that show too

suineg| PUZZLE MASTER | Profile October 12th, 2010 - 3:22 pmit was 8 in my comment as you see because 3×3 would be 9…, too much sleep disorder I guess.

@Chris Edwards: In this one I did not find any kind of equation system that helped me solved this problem right away, however, for me what really was helpful was translate this problem to a letter problem like this:

..ABCD

..X..E

=DBCEA

From this , for me what really reduce the search were the possible values of A

First A,B,C,D,E are differents numbers so:

A cannot be 1, because the only number that can be formed like that with 2 different digits is DXE=21 where D should be 7 or 3, but if A is 1, the maximun possible value of D would have been 1

A cannot be 3,5, the numbers are differents so it is impossible.

A cannot be 9 or 7, in the case of 9 because there is no multiplication between 2 different digits that gives you that unit digit, in the case of 7 , the only case is 9X3=27, D cannot be 9 for the reason exposed before so you assume E is 9, but EXA would be 63 so it gives you an incongruency with D

so A can be 2,4,6,8 with similar reasonings you get that the only possible value is 8 from that on it became very easy to find the others values.

suineg| PUZZLE MASTER | Profile October 12th, 2010 - 3:29 pmits 4 I meant, then you deduce the value of E and D and the rest is easy peasy

Chris Edwards| Profile October 12th, 2010 - 7:41 pm@ Suineg : thanks for your response. Interesting how we approach things from different angles!

I, too, used the same naming convention; ie, ABCD X E. But I focused on the possible values of D and E, noting that, of course, neither could be 0 nor 1, nor 5 because A would result in either a zero or another 5. Also, if either D or E is a 6 and the other is an even number, then A would be that same even number. And if E is even, the carry can’t be odd because E must be even, nor can c=1. I had been up too many hours before I started ‘playing’ with this puzzle to do my best reasoning. Next time, I’ll have to try it rested!

Thanks again 4 yer response!

Chris

Chris Edwards| Profile October 12th, 2010 - 7:44 pm@ RK : Thank you 4 the welcome and for your response . And thx 4 the site!

Bobo The Bear| PUZZLE MASTER | Profile October 13th, 2010 - 1:38 pmConfession time. I wrote a quick program on my TI-84 to run through all the possibilities.

Hex| PUZZLE MASTER | Profile October 13th, 2010 - 6:43 pmI started to look for conditions/equations which would help narrow the selection choices. However, being short of time, I quickly derived the resulting condition and threw up a small Pascal program that did the job in less than 1 second:

for a:= 1 to 9 do

for b:= 1 to 9 do

for c:= 1 to 9 do

for d:= 1 to 9 do

for e:= 1 to 9 do

if 10000*d + 1000*(b-a*e) + 100*(c-b*e) + 10*(e-c*e) + (a-d*e) = 0 then

writeln(a,b,c,d,e);

Chris Edwards| Profile November 13th, 2010 - 10:33 pmHahaha! I love computer and math peeps. My initial reaction was that it kind of ruins the point of doing a puzzle by having the computer find the solution 4 U, but writing a program is certainly the most efficient approach! And programming is fun! R U writing programs 4 the sake of your ‘stats’ or b/c when U C a math problem U must find the solution? I’m of the latter persuasion… Thanks 4 the input!

keerthi| Profile August 19th, 2013 - 3:40 am6412*3=19236

keerthi| Profile August 19th, 2013 - 3:42 am6412*3=19326