ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer that can take any value from 0 to 9.

full solution by jaypee

if you're multiplying by 4 A must be an even number

so If A = 2 then E = 3 or 8

if A = 4 then E = 1 or 6

if A = 6 then E = 4 or 9

if A = 8 then E = 2 or 7

the possibilities for ABCDE are

2_ _ _3
2_ _ _8
4_ _ _1
4_ _ _6
6_ _ _4
6_ _ _9
8_ _ _2
8_ _ _7

but A must be less than E so that narrows the possibilities down to

2_ _ _3
2_ _ _8
4_ _ _6
6_ _ _9

for 2_ _ _3 A = 2 but E would have to equal 8

so 2_ _ _8 is a possibility

for 4_ _ _6 A = 4 but E would have to equal either 1 or 2

for 6_ _ _9 A = 6 but E would have to equal either 2 or 3

so the only possibility is 2_ _ _8

regardless if D is even or odd B would always be odd because if you multiply by 4 any digit it will always come out even but if you add 3 to that the result will always be odd so B will be either
1, 3, 5, 7 or 9

now the sum of the middle three digits must equal 8 by elimination of 9's
for example: the the number 349 3 + 4 + 9 = 7

if B = 1 and D = 7 then C = 9
if B = 3 and D = 5 then C = 9
if B = 5 and D = 3 then C = 9
if B = 7 and D = 1 then C = 9
if B = 7 and D = 6 then C = 4
if B = 9 and D = 4 then C = 4 which is not possible

if B = 7, D = 6 and C = 4 in the number ABCDE
then C in the number EDCBA is equal to 8 but C must be equal in both numbers so the only possible number for C is 9

so far we have 2_9_8

so by eliminating 9 as a possible value for B
we now have 1, 3, 5 and 7 for possible value for B

the possible combinations are:

B = 1 and D = 7
B = 3 and D = 5
B = 5 and D = 3
B = 7 and D = 1
the only combination that will work is B = 1 and D = 7

so your number is 21978 for ABCDE

Edited on November 9, 2003, 9:09 pm

Edited on November 9, 2003, 9:15 pm

Edited on November 9, 2003, 9:17 pm
Edited on November 9, 2003, 9:21 pm

Posted by jaypee on 2003-11-09 21:03:31