1993 ABBC 2DDE, ABCD CDC ABC Each letter represents a number? Which master can answer? Thanks

If the letters represent the same number and are only considered in the range of positive integers from 0 to 9, from 1993+abbc=2dde, we can see that the value of a can only be 0 or 1, and from abcd-cdc=abc, we can know that d=2c or d+10=2c, and d must be even.

Let's look at the case of a=1 first, 1993+abbc=2dde becomes 1993+1bbc=2dde, b+9 can not produce carry, 3+c=e, so b can only =0, so d=9, and d is an even contradiction. Therefore a cannot be equal to 1

a=0, abcd-cdc=abc simplified to BCD-CDC=BC, 1993+abbc=2dde to obtain 1993+bbc=2dde

When d=2c, 10+c-d=b, b=c+1 can be obtained from bcd-cdc=bc, b=, and c=, which does not meet the requirements, so it is not valid.

Take d+10=2c, then b=c, c-d=b is obtained from bcd-cdc=bc, and d=0, c=b=5 is calculated, and substituting 1993+bbc=2dde is not valid.

All possibilities are ruled out, and there is no solution.

a=1, b=0, d=9 can be determined.

abcd-cdc=abc is definitely wrong.

abcd-cdc=aba

Then c=8You have a problem with the topic.

1993+bbc=2dde, bcd-cdc=bca=0 c= e=

474-747=47 d=

1993+100c=299e, 10c9-c9c=10c..My reasoning, the results are contradictory.

You inscribed the wrong abcd-cdc=abc is impossible.

Known a+ab+abc+abcd=1993

then there is a+10a+b+100a+10b+c+1000a+100b+10c+d=1993

1111a+111b+11c+d=1993 because of 1111a, so a can not be greater than 1, so a can only be equal to 1 because of 111b, so b can not be greater than 7, can only be equal to 7 because of 11c, c can only be equal to 9

d can only be equal to 6

Hence a=1, b=7, c=9, d=6

1) If you want to make the 3D pit is 8, then D=6, 3C=10-1, C=3, 2B=10-1, B=, impossible, this question is wrong. (2) Hypothesis: ABCD+BCD+CD+D=2008, then:

4d=8 ,d=2 ,c=0 ,b=0,a=2 .(3) Assum: abcd+bcd+cd=2009, then:

3d=9 ,d=3 ,c=0 ,b=0,a=2 .(4) Assumptions.

248 + 84 = 332, 257 + 75 = 332, etc.

9a+b+c=10d,b+c=10+a,147+74=221，174+47=221

2. The product of the four-digit ABCD and 9 is DCBA. where a, b, c, and d represent different Arabic numerals, and the original four digits are found. 3. There are four people, each of whom can play a musical instrument and a foreign language.

1. A can play the accordion; 2. French speakers can play the violin; 3. There is one person who can play the guitar, but not B; 4. Those who can speak German are not C; 5. Ding doesn't know the violin or English; 6. B does not know French and Russian; 7. C can't play the piano; 8. Guitar speakers don't speak Russian. Excuse me: What kind of musical instruments do A, B, C and D know?

What foreign language do you speak? 4. Seven students guessed what day of the week it was. The first one said:

Today, Wednesday. The second said, "Wednesday the day after tomorrow."

The third said, "Tomorrow Wednesday." The fourth said:

Today is not Monday, not Tuesday, not Wednesday. The fifth said, "Yesterday Thursday."

The sixth said, "Tomorrow Sunday." The seventh said:

Yesterday was not Saturday. Only one of their words was right, which one? Please analyze and determine what day of the week it is.

1089, and the simpler 1111x1 = 1111,,, hahahaha.

If these two equations are the same problem, then there is a problem with the problem itself, and if abcd is a different positive integer, then there is no solution.

First of all: abc 4 cda, abcd 4=dcba, then a must be greater than or equal to 4, otherwise abc 4 can not get three digits, for example, 312 4 = 78, abcd 4 = dcba is the same.

Again: if abc 4 cda, then the quotient of a 4 is either 1 or 2, but it can only be a unique number, and it cannot be in abc 4 = cda

is C, and in ABCD 4=DCBA it is D.

So, if these two formulas are the same problem, then there is no solution.

a=1, b=0, d=9 can be determined.

abcd-cdc=abc is definitely wrong.

abcd-cdc=aba

Then c=8You have a problem with the topic.

I hope I am helpful to you, and good luck! Try more questions like this yourself, and it will be next time!

I wish you progress in your studies! (*