All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
Upon Reflection (Posted on 2013-11-22) Difficulty: 3 of 5
The Alphametics:
WED + TAB + PIN = X j k Y
and
DEW + BAT + NIP = Y k j X
..... when considered concurrently have 864 solutions.

In the totals X=1 and Y=2 or X=2 and Y=1, and j and k are distinct integers represented in the set of variables.

Why can X and Y never equate to W, T, P, D, B or N? The digits 0-9 are all available but there are no leading zeroes.

See The Solution Submitted by brianjn    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution My way | Comment 3 of 13 |

Observations: consider wed+tab+pin=1**2

If we add up both alphametics we must get a number ending by 3 (13,23,33,43).

It is quite easy to understand that this number is  either 23 or 33, since  a sum  of 6 biggest digits is at most 39  -less than 43 , and the sum of six smallest, zero included is 15, more than 13.

1.       W+D+T+B+P+N=23 OR 33

2.       23  subdivides  into 11+12  so both the 1st sum and the second will begin by 1

3.       So W+D+T+B+P+N= 33

4.       The subdivision must be  D+ B+N=22 and W+ T+ P=11, leaving A+E+I =10 or 11

5.       Trying 1hh2 , using  the same digit and trying leading zeros nulls immediately those options, - no need to disclaim them a priori.

6.       Same thing with VOWELS A+E+I=10 cannot produce a sum and its reversal.

7.       So be  D+B+N=22 and W+ T+ P=11, leaving A+E+I = 11

8.       Total of WEB PANDIT IS 44 i.e. DIGIT 1 NOT USED

It boils down to solving the basic set as follows:
wed+tab+pin=1232
dew+bat+nip=2321

W+D+T+B+P+N= 33, remember the subdivision

I=0;   ARBITRARY CHOICE
A+E =11.

9.       Trying 2+9, 3+8, 4+7, 5+6  for A+E we get generic solutions in no time presented in my previous post,-

Chose two generic solutions , and remember that each represents  864  permuted versions.

<o:p> </o:p>

So the answer to a modified version should refer to this number: 1728<o:p></o:p>

 I honestly apologize for the condensed explanation- I am sure that whoever

tried to solve this puzzle gets a coherent expose.
It is significantly easier to solve than to provide nicely structured explanation.

I will sum up, how in my eyes this puzzle should be rectified and

   I suggest you try to solve it from this text:
<begin
  <d3
A set of alphametics
WED + TAB + PIN = 1##2 or 2##1
DEW + BAT + NIP = 1##2 or 2##1,

  in which "#" may represent any digit,  has numerous solutions.
Find , how many distinct solutions are there.
<end>
<o:p></o:p>

Optional Rem: No need to solve the alphametics<o:p></o:p>

Another wording for puzzle's question"    Using each of  the distinct solutions of the above set - in how many different ways can you code " WEB PANDIT"?     <o:p></o:p>

 <o:p></o:p>

Ans: 1728<o:p></o:p>

WEBPANDIT :(SET1) &(SET2)- compare my prev post.<o:p></o:p>

Again , excuse my shorthand style, using both capital and regular letters for the same<o:p></o:p>

Item, not editing for punctuation  etc.<o:p></o:p>

<o:p> </o:p>

<o:p> Errrare  humanum  est,  perseverare  diabolicum.</o:p>



<o:p> </o:p>


  Posted by Ady TZIDON on 2013-11-23 15:21:18
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information