Posted by Greg1959 on June 27, 2015 at 21:46:29 from (69.176.46.42):
In Reply to: Re: OT Math puzzle posted by Rich'sToys on June 27, 2015 at 21:34:06:
We're given a 5-digit number that becomes a new 5-digit number when multiplied by 4. Therefore the first number can't be greater than 24,999 because 25,000*4 = 100,000. This means A has to equal 1 or 2. "A" is also the last digit of the product, which has to be even if it's a multiple of 4. Therefore A=2. Since A=2, B can't be more than 4, so it must be 1, 3, or 4. Also, if "EDCBA" is a multiple of 4 then "BA" must be divisible by 4 too. Out of the possibilities for "BA" (12, 32, 42), only the first two work. So B is 1 or 3. Since A is 2 and B is greater than 0 but less than 4, then "ABCDE" is between 21000 and 24000. This means EDCBA is between 84000 and 96000. So E is 8 or 9. But since A is 2, E*4 must end in 2. So E=8. Notice that 23,000 * 4 = 92,000. This means that ABCDE must be less than 23,000 if E is to be 8. Thus B=1 and not 3.
Filling in the discovered values So far, we have: 21CD8 * 4 = 8DC12
If the product ends in "12", then 4D+3 must end in 1. (You can verify this by trying to multiply 21CD8 * 4 by hand; 8*4=32, so you carry the 3, etc.) So D=2 or 7. If D=2, then using the same reasoning, 4C+1 must end in the same digit as C. If D=7, then 4C+3 must end in the same digit as C. This only works for D=2 C=3 and D=7 C=9. Out of these pairs, only the second one gives positive results for ABCDE*4=EDCBA.
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