Make a list of distinct positive integers that are obtained by assigning a different base ten digit from 1 to 9 to each of the capital letters in this expression.
(A+B)*C + (D–E)/F + (GH)*I
What are the respective minimum and maximum positive palindromes from amongst the elements that correspond to the foregoing list?
As a bonus, what are the respective minimum and maximum positive
tautonymic numbers that are included in the list? How about the respective maximum and minimum prime numbers?
Overall the values extend from 24 to 939,524,151, with 13,253 different values altogether, and so too voluminous to list here.
For palindromes, the lowest is 33:
A B C D E F G H I
(2+6)*3+(9-4)/5+(1^7)*8 =33
(2+6)*3+(9-5)/4+(1^7)*8 =33
(4+5)*3+(2-8)/6+(1^9)*7 =33
(4+5)*3+(6-8)/2+(1^9)*7 =33
(4+9)*2+(8-3)/5+(1^7)*6 =33
(4+9)*2+(8-5)/3+(1^7)*6 =33
(5+4)*3+(2-8)/6+(1^9)*7 =33
(5+4)*3+(6-8)/2+(1^9)*7 =33
(5+7)*2+(9-3)/6+(1^4)*8 =33
(5+7)*2+(9-6)/3+(1^4)*8 =33
(5+8)*2+(7-3)/4+(1^9)*6 =33
(5+8)*2+(7-4)/3+(1^9)*6 =33
(5+9)*2+(3-7)/4+(1^8)*6 =33
(5+9)*2+(4-7)/3+(1^8)*6 =33
(6+2)*3+(9-4)/5+(1^7)*8 =33
(6+2)*3+(9-5)/4+(1^7)*8 =33
(6+7)*2+(4-9)/5+(1^3)*8 =33
(6+7)*2+(5-9)/4+(1^3)*8 =33
(6+9)*2+(3-8)/5+(1^7)*4 =33
(6+9)*2+(5-8)/3+(1^7)*4 =33
(7+5)*2+(9-3)/6+(1^4)*8 =33
(7+5)*2+(9-6)/3+(1^4)*8 =33
(7+6)*2+(4-9)/5+(1^3)*8 =33
(7+6)*2+(5-9)/4+(1^3)*8 =33
(7+8)*2+(3-9)/6+(1^5)*4 =33
(7+8)*2+(6-9)/3+(1^5)*4 =33
(8+5)*2+(7-3)/4+(1^9)*6 =33
(8+5)*2+(7-4)/3+(1^9)*6 =33
(8+7)*2+(3-9)/6+(1^5)*4 =33
(8+7)*2+(6-9)/3+(1^5)*4 =33
(9+4)*2+(8-3)/5+(1^7)*6 =33
(9+4)*2+(8-5)/3+(1^7)*6 =33
(9+5)*2+(3-7)/4+(1^8)*6 =33
(9+5)*2+(4-7)/3+(1^8)*6 =33
(9+6)*2+(3-8)/5+(1^7)*4 =33
(9+6)*2+(5-8)/3+(1^7)*4 =33
or if that's too simple a palindrome, then 101:
(those with A>B have been eliminated from this list, for brevity; just reverse them for the alternative.)
(2+6)*9+(8-3)/5+(4^1)*7 =101
(2+6)*9+(8-3)/5+(7^1)*4 =101
(2+6)*9+(8-5)/3+(4^1)*7 =101
(2+6)*9+(8-5)/3+(7^1)*4 =101
(3+5)*9+(8-2)/6+(4^1)*7 =101
(3+5)*9+(8-2)/6+(7^1)*4 =101
(3+5)*9+(8-6)/2+(4^1)*7 =101
(3+5)*9+(8-6)/2+(7^1)*4 =101
(3+8)*6+(2-7)/5+(4^1)*9 =101
(3+8)*6+(2-7)/5+(9^1)*4 =101
(3+8)*6+(5-7)/2+(4^1)*9 =101
(3+8)*6+(5-7)/2+(9^1)*4 =101
(3+9)*8+(2-7)/5+(1^4)*6 =101
(3+9)*8+(5-7)/2+(1^4)*6 =101
(3+9)*8+(7-2)/5+(1^6)*4 =101
(3+9)*8+(7-5)/2+(1^6)*4 =101
(4+5)*9+(2-8)/6+(3^1)*7 =101
(4+5)*9+(2-8)/6+(7^1)*3 =101
(4+5)*9+(6-8)/2+(3^1)*7 =101
(4+5)*9+(6-8)/2+(7^1)*3 =101
(4+6)*3+(2-7)/5+(8^1)*9 =101
(4+6)*3+(2-7)/5+(9^1)*8 =101
(4+6)*3+(5-7)/2+(8^1)*9 =101
(4+6)*3+(5-7)/2+(9^1)*8 =101
(4+7)*9+(2-8)/6+(1^5)*3 =101
(4+7)*9+(6-8)/2+(1^5)*3 =101
(4+9)*6+(2-7)/5+(3^1)*8 =101
(4+9)*6+(2-7)/5+(8^1)*3 =101
(4+9)*6+(5-7)/2+(3^1)*8 =101
(4+9)*6+(5-7)/2+(8^1)*3 =101
(5+6)*4+(9-1)/8+(2^3)*7 =101
(5+6)*4+(9-8)/1+(2^3)*7 =101
(5+6)*8+(9-2)/7+(3^1)*4 =101
(5+6)*8+(9-2)/7+(4^1)*3 =101
(5+6)*8+(9-7)/2+(3^1)*4 =101
(5+6)*8+(9-7)/2+(4^1)*3 =101
(5+6)*9+(2-8)/3+(1^7)*4 =101
(5+6)*9+(3-7)/2+(1^8)*4 =101
(5+7)*4+(1-9)/8+(3^2)*6 =101
(5+7)*4+(8-9)/1+(3^2)*6 =101
(5+7)*6+(3-9)/2+(4^1)*8 =101
(5+7)*6+(3-9)/2+(8^1)*4 =101
(5+7)*6+(8-4)/2+(3^1)*9 =101
(5+7)*6+(8-4)/2+(9^1)*3 =101
(5+7)*8+(9-3)/6+(1^2)*4 =101
(5+7)*8+(9-6)/3+(1^2)*4 =101
(5+8)*3+(2-6)/4+(7^1)*9 =101
(5+8)*3+(2-6)/4+(9^1)*7 =101
(5+8)*3+(4-6)/2+(7^1)*9 =101
(5+8)*3+(4-6)/2+(9^1)*7 =101
(5+8)*7+(6-2)/4+(1^3)*9 =101
(5+8)*7+(6-4)/2+(1^3)*9 =101
(5+9)*6+(7-3)/4+(2^1)*8 =101
(5+9)*6+(7-3)/4+(8^1)*2 =101
(5+9)*6+(7-4)/3+(2^1)*8 =101
(5+9)*6+(7-4)/3+(8^1)*2 =101
(5+9)*7+(2-8)/6+(1^3)*4 =101
(5+9)*7+(6-8)/2+(1^3)*4 =101
(6+8)*5+(9-3)/2+(4^1)*7 =101
(6+8)*5+(9-3)/2+(7^1)*4 =101
(6+8)*7+(2-5)/3+(1^9)*4 =101
(6+8)*7+(3-5)/2+(1^9)*4 =101
(6+8)*7+(9-4)/5+(1^3)*2 =101
(6+8)*7+(9-5)/4+(1^3)*2 =101
(6+9)*4+(8-1)/7+(2^3)*5 =101
(6+9)*4+(8-7)/1+(2^3)*5 =101
(6+9)*5+(2-8)/3+(4^1)*7 =101
(6+9)*5+(2-8)/3+(7^1)*4 =101
(7+8)*5+(2-6)/4+(3^1)*9 =101
(7+8)*5+(2-6)/4+(9^1)*3 =101
(7+8)*5+(4-6)/2+(3^1)*9 =101
(7+8)*5+(4-6)/2+(9^1)*3 =101
(7+9)*4+(8-3)/5+(6^2)*1 =101
(7+9)*4+(8-5)/3+(6^2)*1 =101
(7+9)*6+(5-2)/3+(1^8)*4 =101
(7+9)*6+(5-3)/2+(1^8)*4 =101
(7+9)*6+(8-3)/5+(1^2)*4 =101
(7+9)*6+(8-4)/2+(1^5)*3 =101
(7+9)*6+(8-5)/3+(1^2)*4 =101
The highest is 327723:
(6+7)*3+(9-1)/2+(4^8)*5 =327723
(7+6)*3+(9-1)/2+(4^8)*5 =327723
For tautonyms, the lowest is 1313:
(1+7)*4+(9-3)/6+(2^8)*5 =1313
(1+7)*4+(9-6)/3+(2^8)*5 =1313
(1+7)*8+(3-9)/6+(5^4)*2 =1313
(1+7)*8+(6-9)/3+(5^4)*2 =1313
(7+1)*4+(9-3)/6+(2^8)*5 =1313
(7+1)*4+(9-6)/3+(2^8)*5 =1313
(7+1)*8+(3-9)/6+(5^4)*2 =1313
(7+1)*8+(6-9)/3+(5^4)*2 =1313
and the highest tautonym is 19131913:
(1+6)*5+(8-2)/3+(9^7)*4 =19131913
(2+5)*6+(3-8)/1+(9^7)*4 =19131913
(5+2)*6+(3-8)/1+(9^7)*4 =19131913
(6+1)*5+(8-2)/3+(9^7)*4 =19131913
5 dim Used(9)
6 kill "paltaut3.txt":open "paltaut3.txt" for output as #2
10 for G=1 to 9
20 Used(G)=1
30 for H=1 to 9
40 if Used(H)=0 then
50 :Used(H)=1
60 :Gh=G^H
70 :for I=1 to 9
80 :if Used(I)=0 then
90 :Used(I)=1
100 :Ghi=Gh*I
110 :for A=1 to 9
120 :if Used(A)=0 then
130 :Used(A)=1
140 :for B=1 to 9
150 :if Used(B)=0 then
160 :Used(B)=1
170 :for C=1 to 9
180 :if Used(C)=0 then
190 :Used(C)=1
200 :for D=1 to 9
210 :if Used(D)=0 then
220 :Used(D)=1
230 :for E=1 to 9
240 :if Used(E)=0 then
250 :Used(E)=1
260 :for F=1 to 9
270 :if Used(F)=0 then
280 :Used(F)=1
290 :V=(A+B)*C+(D-E)//F+Ghi
300 :if V=int(V) then
310 :print #2,A;B;C;D;E;F;G;H;I;using(19,0),V
320 :print A;B;C;D;E;F;G;H;I;using(19,0),V
330 :endif
340 :Used(F)=0
350 :endif
360 :next
370 :Used(E)=0
380 :endif
390 :next
400 :Used(D)=0
410 :endif
420 :next
430 :Used(C)=0
440 :endif
450 :next
460 :Used(B)=0
470 :endif
480 :next
490 :Used(A)=0
500 :endif
510 :next
520 :Used(I)=0
530 :endif
540 :next
550 :Used(H)=0
560 :endif
570 next H
580 Used(G)=0
590 next G
600 close #2
The results were then sorted and then analysed with:
OPEN "paltaut3.txt" FOR INPUT AS #1
DO
LINE INPUT #1, l$
l$ = LTRIM$(RTRIM$(MID$(l$, 29)))
pal = 1
FOR i = 1 TO LEN(l$) / 2
IF MID$(l$, i, 1) <> MID$(l$, LEN(l$) + 1 - i, 1) THEN pal = 0: EXIT FOR
NEXT
IF pal THEN p$ = l$
IF LEN(l$) MOD 2 = 0 THEN
taut = 1
FOR i = 1 TO LEN(l$) / 2
IF MID$(l$, i, 1) <> MID$(l$, i + LEN(l$) / 2, 1) THEN taut = 0: EXIT FOR
NEXT
IF taut THEN
IF hadTaut = 0 AND pal = 0 THEN hadTaut = 1: PRINT l$
t$ = l$
END IF
END IF
LOOP UNTIL EOF(1)
PRINT p$
PRINT t$
Operation symbols were added manually.
BTW, the highest without consideration of palindrome/tautonym, 939,524,151, is (5+4)*6 + (3-2)/1 + (8^9)*7
Primes:
There are 1198 primes on the list and the minimum prime is 27 and the maximum prime is 939,524,147, from
list
1 Had=0
5 open "PALTAU~1.TXT" for input as #1
10 while not eof(1)
20 input #1,A:N=val(cutspc(A))
30 if prmdiv(N)=N then if Had=0 then print N:endif:Ct=Ct+1:Big=N:Had=1
40 if prmdiv(N)=0 then print "***";N:end
50 wend
60 print Big
70 print Ct
OK
run
29
939524147
1198
OK
(3+7)*2+(9-4)*5+(1^6)*8 =29
(3+7)*2+(9-5)*4+(1^6)*8 =29
(3+8)*2+(9-4)*5+(1^7)*6 =29
(3+8)*2+(9-5)*4+(1^7)*6 =29
(4+7)*2+(3-9)*6+(1^5)*8 =29
(4+7)*2+(6-9)*3+(1^5)*8 =29
(4+7)*2+(8-3)*5+(1^9)*6 =29
(4+7)*2+(8-5)*3+(1^9)*6 =29
(5+6)*2+(3-7)*4+(1^9)*8 =29
(5+6)*2+(4-7)*3+(1^9)*8 =29
(5+7)*2+(9-3)*6+(1^8)*4 =29
(5+7)*2+(9-6)*3+(1^8)*4 =29
(5+8)*2+(3-9)*6+(1^7)*4 =29
(5+8)*2+(6-9)*3+(1^7)*4 =29
(6+5)*2+(3-7)*4+(1^9)*8 =29
(6+5)*2+(4-7)*3+(1^9)*8 =29
(6+7)*2+(3-8)*5+(1^9)*4 =29
(6+7)*2+(5-8)*3+(1^9)*4 =29
(7+3)*2+(9-4)*5+(1^6)*8 =29
(7+3)*2+(9-5)*4+(1^6)*8 =29
(7+4)*2+(3-9)*6+(1^5)*8 =29
(7+4)*2+(6-9)*3+(1^5)*8 =29
(7+4)*2+(8-3)*5+(1^9)*6 =29
(7+4)*2+(8-5)*3+(1^9)*6 =29
(7+5)*2+(9-3)*6+(1^8)*4 =29
(7+5)*2+(9-6)*3+(1^8)*4 =29
(7+6)*2+(3-8)*5+(1^9)*4 =29
(7+6)*2+(5-8)*3+(1^9)*4 =29
(8+3)*2+(9-4)*5+(1^7)*6 =29
(8+3)*2+(9-5)*4+(1^7)*6 =29
(8+5)*2+(3-9)*6+(1^7)*4 =29
(8+5)*2+(6-9)*3+(1^7)*4 =29
(4+6)*5+(3-1)*2+(8^9)*7 =939524147
(4+6)*5+(3-2)*1+(8^9)*7 =939524147
(6+4)*5+(3-1)*2+(8^9)*7 =939524147
(6+4)*5+(3-2)*1+(8^9)*7 =939524147
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Posted by Charlie
on 2010-03-23 15:33:34 |
computer solution
