Question

Another one, pretty tough D:

Answer 1

Is this from some sort of test earlier today?

Answer 2

Nope, some random exercises I found

Answer 3

Well, I know that B must be less than 5.

Answer 4

And not 1.

Answer 5

I believe the number should be ABCDE=84261

Answer 6

I'll try so solve it first then tell me how you solved it :) Thanks!

Answer 7

I believe B=4, A=8 and D=1.

Answer 8

I got my number wrong just a bit. It's actually

ABCDE=84269

Answer 9

I got it wrong before. I think B=4, A=8 and D=6.

Answer 10

Now, how I solved it. I started with looking at B. Obviously, B can't be 5 or greater. So I looked at B=1 first.

If B=1, then A=7. Hence, the second digit of 3C must be a 5. This can't happen. So B is not 1.

If B=2, then A=4, so C=1, 3. If C=1, the second digit of 4D must be 5, which is impossible, and if C=3, then the second digit of 4D is 9, which is also impossible. Thus B is not 2.

If B=3, A=1, and we run into similar problems with C and D.

Therefore, B=4.

Answer 11

Since B=4, A=8. From here we can continue solving.

This means that 3C is less than 10. So C=1, 2, 3. Of these, we can eliminate 1, since 4D would have a 5 for the second digit, and we can eliminate 3, since 5D would have to have a 9 in the second digit.

Thus, we have ABC=842

Answer 12

If B = 4, then A = 8, because B*7 = _8. So, there is a carry in of 2 to the second-to-last column.

Answer 13

Now we need to find a D such that the ones digit of 6D is 6, and the tens digit of 4D is 2. Using some guess and check, we get that D=6.

Answer 14

Because 6D+2=_8, 6D=_6, so D can be 1 or 6.

Answer 15

If we choose D = 6, then there is a carry in of 3 to the third-to-last column, so 5E + 3 = _8, therefore 5E = _5.

Answer 16

Now we just have to solve for E. We need a number such that the ones digit of 5E is 5, and the tens digit is 4. Hence, E=9.

Therefore, the number is 84269

Answer 17

If we have D=6, we know from the fourth-to-last column that 4D + some carry in = _8. Because D = 6, 4*6 + carry in = _8, 24 + carry in = _8, so carry in must be 4. So we conclude, from the fact that, 5E = _5 (which I wrote in a previous answer), that 5E = 45, therefore E = 9.

Answer 18

From the fact that D=6, we know we must have a carry in of 2 to the C column. So, 3C + 2 = _8, so 3C = _6, from which we conclude that C = 2. We can see then that the C column leaves no carry out, which makes sense.

Answer 19

Therefore, the number is 84269.

Answer 20

I got it! :D