I'm trying to enter this so it shows up to what I need, but I need to know how to solve for the missing Letters:
ABCD
X 4
----
DCBA
It reads ABCD times 4 and the 4 is under the D.
i thank u have to rearange them
ABCD
x 4
DCBA
What does each letter stand for?
1--A must be 1 or 2
2--If A = 1, 4D = 1 cannot be so A = 2.
2BCD
.....x4
DCB2
3--D can be 3 or 8 to give an A = 2
4--If D = 3, 4A cannot be 3 so D = 8.
2BC8
.....x4
8CB2
5--C can be 1, 3, 5, 6, 7, or 9.
6--4C + 3 = 10 + B
....................C....1....3....5....6....7....9
....................B....7....5....3....7....1....9
....................R....0....1....2....2....3....3 R = remainder
4B + R = C....8....1....4....0....7....9
..........................X....X....X....X....Y....X
7--Only C = 7 results in the two C's being equal. (B = 9 from C = 9 would make 4B greater than 10 giving the answer 5 digits.)
8--Therefore, C = 7 and B = 1 for
2178
...x4
8712
To solve for the missing letters in the equation ABCD x 4 = DCBA, we can use basic multiplication principles.
1. Start by looking at the rightmost digit of the product, which is A. To find A, multiply the rightmost digit of the multiplier (4) with each digit in the multiplicand (ABCD) and check if it results in A. In this case, 4 multiplied by any single digit will not result in a number ending with 4. Therefore, A remains unknown.
2. Moving to the next digit, D. Again, multiply the next digit of the multiplier (4) with each digit in the multiplicand (ABCD) and check if it results in D. We know that 4 multiplied by B will give us a number that ends with 4, so B must be 6. So, we have:
ABCD
4
----
D6CA
3. Now, let's move on to the third digit, C. Multiply the remaining digits in the multiplicand (A, C, and D) by 4 and check if it results in C. We need to find a number that ends with 6 when multiplied by 4. The only possible digit for C is 1 because 4 multiplied by 1 equals 4 and ends with 6 when written in the product.
ABCD
4
----
D61A
4. Finally, we are left with the first digit, B. Multiply the remaining digits in the multiplicand (A and D) by 4 and check if it results in B. Since we already used the digit 6, the only possibility for B is 3 because 4 multiplied by 3 is 12.
ABCD
4
----
D613
Hence, the missing letters are A = unknown, B = 3, C = 1, and D = 6. The equation ABCD x 4 = DCBA can now be re-written as 3162 x 4 = 12648.