In the multiplication shown below, the digits have been replaced by
letters: different letters represent different digits, and the
same letters represent the same digits. What was the origianl
multiplication?
ABCDE
x 4
======
EDCBA
please help tried it many times can't get it. It is drivin me nuts.
Someone recently asked and someone else answered the similar question
ABCD x 4 = DCBA
In that case there was only one answer. It took quite a bit of reasoning to come up with the answer, 2178 x 4 = 8712
In your case, it is likely that there is no answer. The longer the string of numbers, the less likely it is that there is an answer. The problem has to be done by a "cut and try" method. If you try to do it with algebra, you end up with one equation and 4 unknowns.
ABCDE x 4 = EDCBA
21978 x 4 = 87912
You just use the fact that the last digit of a multiple of 4 is even. It is not zero, because A = 0 would mean that the number ABCDE is actually a four digit number. It cannot exceed two otherwise you get a six digit number when multiplying it by four. So:
A = 2
This means that the number will be EDCBA will be larger than 80,000 so E has to be 8 or 9. But E = 9 as the first digit of ABCDE
yields A = 6, so E must be 8. You then simply work your way through the multiplication. It is straightforward to come up with the answer (i.e., if it exists as in this case).
To solve this problem, we need to find the original multiplication where the digits have been replaced with letters. Let's go step by step:
1. Notice that the product ends with "EDCBA". This means that the unit's digit of the original multiplication is "A".
EABCDE
x 4
======
EDCBA
2. Multiplying a digit by 4 can either give a single digit or a two-digit number. The unit's digit of the multiplication must be A, so it means that E multiplied by 4 gives A and there is no carryover to the tens place.
E is a number such that 4E ends with A. The only value for E that satisfies this condition is 3.
3ABCDE
x 4
======
EDCBA
3. Now we have the following multiplication:
3ABCDE
x 4
======
EDCBA
4. The maximum value for the product of A and 4 is 9 (because 9 multiplied by 4 is 36). If there is any carryover from the previous step, the maximum value for the product of A and 4 is 8.
3ABCDE
x 4
======
EDCBA
Considering that A multiplied by 4 gives a one-digit number and there is no carryover, A must be 2.
3BCDE
x 4
======
EDCBA
5. Now we can solve for and determine the values of B, C, D, and E.
B multiplied by 4 gives C and there is no carryover from previous steps. B multiplied by 4 cannot give a single-digit number, so B must be greater than or equal to 3.
3CDE
x 4
======
EDCBA
Looking at the values of C and D, we can conclude that D multiplied by 4 must give a value greater than or equal to 10 and less than 20. The only value for D that satisfies this condition is 6.
3C6E
x 4
======
EDCBA
Lastly, let's solve for C. E multiplied by 4 gives A. The only value for C that satisfies this condition is 8.
386E
x 4
======
EDCBA
Therefore, the original multiplication is:
3864
x 4
=====
15456
To solve this problem, we will need to use a trial and error approach to find the values of the letters that satisfy the given conditions.
Let's analyze the multiplication step by step:
ABCDE
x 4
---------
EDCBA
From the equation, we can observe a few things:
1. The product is a 5-digit number, represented as EDCBA.
2. Multiplying a 5-digit number by 4 will result in another 5-digit number.
3. The leftmost digit, A, multiplied by 4, gives a one-digit result which starts the product.
Now, let's begin the trial and error process:
1. Start by assuming A = 1, as A * 4 should give a one-digit number at the beginning. Multiplying 1 by 4, we get the result 4.
BCDE
x 4
------
EDCB4
2. Since the rightmost digit of the product is 4, we can conclude that B must be either 1 or 6 (as only 1 or 6 multiplied by 4 can give a number with 4 at the rightmost digit). We begin by assuming B = 1.
CDE
x 4
------
EDC14
3. To find C, we need to multiply C by 4 and add any carry from the previous calculation. From the equation, we can see that C * 4 + E * 1 should end in E. Therefore, C should be an even number between 1 and 9. Let's start by assuming C = 2.
DE
x 4
------
E21DE
4. We can determine D in a similar manner. D * 4 + C * 1 + carry(2) should end in D. So, let's assume D = 3.
E
x 4
------
321E
5. Finally, we need to find E. E * 4 + D * 1 + carry(1) should end in E. By examining the equation, we can determine that E must be 9.
Thus, we have found the answer to the original multiplication:
9 8 7 6 3
x 4
--------------
3 2 1 9 6
Therefore, the original multiplication was 98763 x 4 = 32196.