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2178. do i need to xplain?

Yes, please do. All I seem to be able to prove is that A=A, B=B, C=C and D=D. I've approached it from a few different angles aside from brute force.

I erased a lot after I ran into the A=A, B=B, etc. However, I still had this remaining:

A*4*1000 + B*4*100 + C*4*10 + D*4*1 = D*1000 + C*100 + B*10 + A*1

4000A + 400B + 40C + 4D = 1000D + 100C + 10B + A

3999A = -390B + 60C + 996D

-390B = 3999A - 60C - 996D

60C = 3999A + 390B - 996D

996D = 3999A + 390B - 60C

Maybe isolate letters, reduce fractions?

A = -390B/3999 + 60C/3999 + 996D/3999

A = -13B/133.3 + C/66.65 + 16.6D/66.65

B = -3999A/390 + 60C/390 + 996D/390

B = -133.3B/13 + C/6.5 + 13D/33.2

C = 3999A/60 + 390B/60 - 996D/60

C = 66.65A + 6.5B - 16.6D

D = 3999A/996 + 390B/996 - 60C/996

D = 66.65A/16.6 + 33.2B/13 - C/16.6

Substitute for C

A = -13B/133.3 + 16.6D/66.65 + (66.65A + 6.5B - 16.6D)/66.65

A = -13B/133.3 + 16.6D/66.65 + A + 13B/133.3 - 16.6D/66.65

A = A

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I included the logic for solving it.

abcd = 2178

dcba = 8712

To solve, note that 4a must be 1 digit number. So a can only b 0,1,2.

A can not be 1 as 4d has to be even. If a = 0, then 4d must end in 0 and so d must be 5. 5 is too big, since 0bcd times 4 is less than 4000, nowhere near 5000 if d is 5. So a must be 2. 4d must end with 2, so d = 3 or d = 8. 3 is too small, since 4 times 2bc3 is going to be at least 8000. So d = 8.

Since 4*8 = 32, there is a 3 to carry over. so b = last digit of 4c+3. We also know that 4b is single digit, as there is no carry over to the thousands place. So b must be 0,1,2. b must also be odd, since it equals the last digit of 4c+3 which is odd. so b = 1 and 4c must end with an 8, which makes c either 7 or 2. Trying 7 gives 2178 * 4 = 8712 which works. Trying 2 for c does not work.

A four digit number ABCD when multiplied by 4 gives product DCBA. What is the number?

Edited by Nana7
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