Tuan Tuan Reunion Reunion Olympiad Problem Solving Urgent

If x, y, and z are all natural numbers below 10, there is no solution to this problem! The analysis is as follows:

1. According to the mantissa of multiplication, we know that there are: y*z=z+n*10(n=0,1,2,3......）

The displacement yields: (y-1)*z=10n(n=0,1,2,3......）

When n=0, (y-1)*z=0, it is further obtained that y=1 and 11*zz have the following equation:

z zz zz z

z （z+z）z

Obviously, (z+z)=2*z, and y=1, this number cannot be obtained by any means that the mantissa is 1, so n must not be equal to 0

When n=1, it is given by Eq. (y-1)*z=10n(n=0,1,2,3......Solution.

y-1）*z=10

y-1=2, z=5 or y-1=5, z=2

Substituting these two possibilities into the original formula, we get a four-digit number, and the original problem is a three-digit number, so n is not equal to 1

Therefore, when n=2 or more, it must be a larger number than four digits, so if x, y, and z are natural numbers below 10, there is no solution to this problem!

Therefore, I think that if there is a solution to this question, it must be that the questioner has missed some conditions, and finally, I hope that you will read the question clearly next time you ask a question, and don't waste everyone's time by making this kind of inconclusive question.

Original formula = y*11*z*11=(y*z)*121=xyz

1-5,9 is clearly off topic.

So, there is no solution at all.

Let these three numbers be 10 and a b n

10a+a)(10b+b)=100n+10a+b121*a*b=100n+10a+b

121*a*b-10a-b=100n

21*a*b-(10a+b)=0

21*a*b-10a=0 a(21b-10)=0 aNo solution.

In other words, there is no solution to this problem in the range of natural numbers.

Order group = a, circle = b, and = c, beauty = d (where a, b, c, d are all 1 9 numbers), then the above equation can be expressed as:

As can be seen from the above equation, 100c+d must be a multiple of 11, and the ten digit is 0, therefore, according to the law of multiplying two digits by 11 (the sum of the multiplier digits plus the tens of digits is equal to the number of tens of digits, and the full 10 digits into 1): we can derive this relation:

c-1)+d=10, that is, c+d=11 (2 formula), and ab = 10 (c-1) + d, (3 formula) synthesized 1 and 3 formulas, get:

100c+d=11(10c-10+d), (4 formulas), and then 2 and 4 are combined to obtain: c=8, d=3;

So ab=73, it is easy to get that a and b are 6 and 9 respectively; (Interchangeable).

So: Tuan Tuan = 66, Round = 99 (interchangeable); and and Meimei = 8833.

That's it! I hope you are satisfied, and if you don't understand, you can ask

aa*bb=ccdd

11a*11b=11*(100c+d)

11ab=100c+d=99c+c+d

C+D is a multiple of 11.

c=2, d=9, ab=209 11=19 no, c=3, d=8, ab=308 11=28 a=4, b=7 or a=7, b=4

c=4, d=7, ab=407 11=37 noc=5, d=6, ab=506 11=46 noc=6, d=5, ab=605 11=55 noc=7, d=4, ab=704 11=64 a=b=8c=8, d=3, ab=803 11=73 noc=9, d=2, ab=902 11=82 No.

aa*bb=ccdd

10a+a)(10b+b)=(1000c+100c+10d+d)121ab=(1100c+11d)

11*11ab=11(100c+d)

11ab=100c+d=99c+c+d

ab=9c+(c+d)/11

c+d is a multiple of 11 0 only when a=7, b=4, c=3, d=8 77*44=3388a=4,b=7,c=3, d=8 44*77=3388a=b=8, c=7, d=4, 88*88=7744 c=3, d=8, ab=308 11=28 a=4, b=7 or a=7, b=4

c=7, d=4, ab=704 11=64 a=b=8, the rest are not true, and the value of the integer a b is not obtained.

c=2, d=9, ab=209/11=19c=4, d=7, ab=407/11=37c=5, d=6, ab=506/11=46c=6, d=5, ab=605/11=55c=8, d=3, ab=803/11=73c=9, d=2, ab=902/11=82

You can't write the steps for this kind of question.

11 * group * round = 100 group + round.

100-11 round) group + circle = 0

Because 100-11 circle "100-11*9=1

So (100-11 round) Tuan + Circle "Tuan + Yuan" 0

So there is no solution to this problem.

There is no solution to this question.

In the vertical column, the reunion is divided by the group, and the first number on the quotient must be 1 in the hundred.

That is to say, the circle should be a number greater than or equal to 100, which obviously contradicts a two-digit number.

Tuan Tuan is clearly x times that of 11.

The circle is obviously y times that of 11.

If this is an Olympiad, the Olympiad will be hopeless.