Math Problem 57 What numbers do they represent?

142857 or 285714

Multiply the six-digit labcde by 3 to get the six-digit abcdel, so the single digit of 3*e is l, then l may be all the numbers in 1 9, and because of the possible carry on 3*l+, a

So l maybe

When l 1.

The number of digits of e*3 is 1, so e7

7*3=21, advance 2, that is, the number of digits of d*3+2 is 7, so there is only 5 multiplied by 3 and the last digit is 5, that is, d=5

Advance 1, the last digit of c*3+1 is 5, so c=8 into 2, the last digit of b*3+2 is 8, so b=2 does not carry, and the last digit of a*3 is 2, so a 4

Enter 1, the last digit of l*3+1 is =4, so l 1 meets the requirements.

So the first six digits are 142857

The same goes for when l 2.

The number of digits of e*3 is 2, so e4

4*3=12, advance 1, that is, the number of d*3+1 is 4, so d=1 is not carryed, and the last digit of c*3 is 1, so c=7

Enter 2, the last digit of B*3+2 is 7, so B=5 into 1, the last digit of A*3+1 is 5, so A 8 into 2, L*3+2 The last digit of the number is 8, so L 2 meets the requirements after cyclic verification.

So the first six digits are 285714

In the same way when l 3.

The number of digits of e*3 is 3, so e 1

1*3=3, no carrying, that is, the number of d*3 is 1, so d=7 into 2, the last digit of c*3+2 is 7, so c=5 into 1, the last digit of b*3+1 is 5, so b=8 into 2, the last digit of a*3+2 is 8, so a 2 is not carrying, and the last digit of l*3 is 2, so l 4

Round-robin verified that it does not meet the requirements.

So the first six digits are 142857 or 285714

1x7+1=8, 7x8+1=57, 8x57+1=457, you can fill in 457 - starting with the third number, each number is equal to the product of the first two numbers plus 1

One square plus seven equals eight, seven squared plus eight equals fifty-seven, and eight squared plus fifty-seven equals one hundred and twenty-one, so the answer is one hundred and twenty-one.

8=1*7+1

So, the next one should be 8*57+1=457

The original number is 2999998, or 1999999.

According to the inscription, the sum of the new seven-digit numbers obtained is 3, and the new seven-digit number contains only 0s, or 0s, or 0s.

And because the sum of the new numbers is 3, which is much smaller than the original 55, it means that continuous carry occurs when adding 2, and every time a carry occurs, the sum of the numbers is less than 10-1 = 9, therefore, a total of carry (55 + 2-3) (10-1) = 6 times, the single digit is at least 8, ten, hundred, thousand, thousand, and one hundred thousand must be 9.

interconnected

Because the sum of the new numbers is 3, which is much smaller than the original 55, it means that continuous carry occurs when adding 2, and every time a carry occurs, the sum of all numbers is less than 10-1 = 9, therefore, a total of carry (55 + 2-3) (10-1) = 6 times, the single digit is at least 8, ten, hundred, thousand, thousand, and 100,000 must be 9.

Therefore, the discussion and verification 2999998+2=3000000 is consistent, and the 1999999+2=2000001 is consistent, and the original number is 2999998 or 1999999, and the solution can be solved accordingly. There are several important properties of the addition method, it is commutative, which means that the order is not important, it is interrelated, which means that when more than two numbers are added, the order in which the addition is performed does not matter, the repeated addition of 1 is the same as the count, and the addition of 0 does not change the result.

Because the sum of the new numbers is 3, which is much smaller than the original 55, it means that when adding 2, a continuous carry occurs, and each time a carry occurs, the sum of the numbers is less than 10 - 1 = 9.

Thus, a total of (55 + 2 - 3) (10 -1) = 6 carries occur.

The single digit must be at least 8, and the ten, hundred, thousand, thousand, and 100,000 digits must be 9.

Hence the verification. 1) 2999998 + 2 = 3000000 (1) 1999999 + 2 = 2000001 The number that meets the original is 2999998, or 1999999

The answer is not the only one, there are 2 1999999 and 2999998 First of all, the method: 7 digits You are 55 plus 2 to get a new number, this new number The sum of the numbers on everyone is 3, the amount of reduction is 55-3 = 52, then there are a few digits that become 0 The first position cannot be 0, so the remaining bits become 0 or become smaller 6 * 9 = 54 You can know that the middle 5 digits are all 9, knowing that these last 5 9s are reduced by 45, the first digit is increased by 1, and the total is reduced by 44, so how is the remaining 8 reduced? Obviously, we get two cases: 9+2=11, 9-1=8 and 8+2=10, 8-0=8, so the answer is not unique, there are 2 1999999 and 2999998

It makes sense upstairs, 55-->3, the middle 5 digits must be 9, 5 * 9 = 45, so the 1st place + 7th digit = 10, plus 2 carry, the single digit is 8 or 9, the 7th digit is 2 or 1, check it, all right, the original number 1999999, 2999998

Because the sum of the digits on the new number is 3, which is much smaller than 55, there must be a lot of carry when adding 2, making 9 0.

This number is 1999999.

The four pants are grade mathematics, 458768 the meaning of the number in each table is modified is:

From the highest position on the left, 4 is on the 100,000 position, which means 4 100,000; 5 in the 10,000 position, which means 5 thousand; 8 in the thousands, which means 8 thousands; 7 in the hundreds, which means 7 hundreds; 6 in the ten place, which means 6 tens; 8 in the single digit means 8 ones. Hu Jianshan.

The sum of the original numbers 55 plus 2 is larger than the sum of the later numbers 3, and it can be concluded that there are several 0s in the later numbers, because 2 is added, it becomes 0, so the single digit number is 8, and the ten digit number is pushed above, and the ten digit number is 9, and in the same way, the hundred digit number is pushed.

Five 9s are 45, plus the single digit 8, which is 53, and it can be concluded that the first one is exactly 2. In line with the topic.

Therefore, the original seven-digit number is 2999998

The answer is not the only one, there are 2 1999999 and 2999998 First of all, the method: 7 digits You are 55 plus 2 to get a new number, this new number The sum of the numbers on everyone is 3, the amount of reduction is 55-3 = 52, then there are a few digits that become 0 The first position cannot be 0, so the remaining bits become 0 or become smaller 6 * 9 = 54 You can know that the middle 5 digits are all 9, knowing that these last 5 9s are reduced by 45, the first digit is increased by 1, and the total is reduced by 44, so how is the remaining 8 reduced? Obviously, we get two cases: 9+2=11, 9-1=8 and 8+2=10, 8-0=8, so the answer is not unique, there are 2 1999999 and 2999998

Because the sum of the new numbers is 3, which is much smaller than the original 55, it means that when adding 2, continuous carry occurs, and every time a rubber carry occurs, the sum of the words of the number of honors is less.

9。Therefore, carry occurs altogether.

The second digit must be at least 8, and the ten, hundred, thousand, thousand, and 100,000 digits must be 9. Therefore, verification (1) 2999998

Meets 1999999 (1).

The number that conforms to the original is 2999998, Qingruchong or 1999999

There is a seven-digit number, and the sum of the numbers on each digit is 55, after this number is added to the pants mountain and 2, it has to be a new number, and the sum of the numbers on each digit of this Hu Zhengxin number is 3, what is the original number?

The original number is 1999999

Because 1999999+2=2000001