PARTS = 21978
I got it the hard way, by trial and error, after realizing that P had to be 2 and S had to be 8.
This kind of problem is not algebra; it is a riddle. There is no guarantee that there would be only one solution. It just turns out that way.
Replace each letter with a number to make the problem correct. Identicle letters are placed with identicle numbers. PARTS x 4 = STRAP
8 answers
PARTS x 4 = 87912
In looking at the numbers that I tried, without using the same number twice, only those with P = 2, A = 1, S = 8 and T = 7 would start out with 8 and end with 12 when multiplied by 4. That left only a few R numbers to try.
In looking at the numbers that I tried, without using the same number twice, only those with P = 2, A = 1, S = 8 and T = 7 would start out with 8 and end with 12 when multiplied by 4. That left only a few R numbers to try.
WIth different letters but the same result:
ABCDE x 4 = EDCBA
Lets take it one number at a time.
1--A must be either 1 or 2 or the answer would be a 7 digit number.
2--If A = 1, the 4 x E has to end with 1 which cannot be, so A = 2.
3--If A = 2, then E can only be 3 or 8 for the product to end in 2.
4--But since 4 x E = E, E must be 8 to satisfy both conditions.
5--4 x B can have no carryover to 4 x 2 so B must be 1 as A is already 2.
6--4 x D + the 3 carryover must equal 11, 21, or 31 since B = 1.
7--So D can only be 2 or 7 but since A = 2, D must be 7.
8--4 x 1 must have a carryover of 3 to become 7 so C must be 7, 8 or 9.
9--Since D = 7 and E = 8, C must be 9.
10--Therefore, the numbers are 21978 x 4 = 87912.
ABCDE x 4 = EDCBA
Lets take it one number at a time.
1--A must be either 1 or 2 or the answer would be a 7 digit number.
2--If A = 1, the 4 x E has to end with 1 which cannot be, so A = 2.
3--If A = 2, then E can only be 3 or 8 for the product to end in 2.
4--But since 4 x E = E, E must be 8 to satisfy both conditions.
5--4 x B can have no carryover to 4 x 2 so B must be 1 as A is already 2.
6--4 x D + the 3 carryover must equal 11, 21, or 31 since B = 1.
7--So D can only be 2 or 7 but since A = 2, D must be 7.
8--4 x 1 must have a carryover of 3 to become 7 so C must be 7, 8 or 9.
9--Since D = 7 and E = 8, C must be 9.
10--Therefore, the numbers are 21978 x 4 = 87912.
5--4 x B can have no carryover to 4 x 2 so B must be 1 as A is already 2.
why can't B be 0 at step 5? At this point, B can be 0,1,2, and you can eliminate 2 per your reason, but you can't eliminate 0 yet--which makes the rest of your steps invalid.
why can't B be 0 at step 5? At this point, B can be 0,1,2, and you can eliminate 2 per your reason, but you can't eliminate 0 yet--which makes the rest of your steps invalid.
4A72
B85C
92D6
Answer
E 7775
B85C
92D6
Answer
E 7775
thanks you are a big help
I NOT HAS MINE
IT IS HAS MINJES
1 answer