What's the product of 35 plus the quotient of 14 divided by 34 and the sum multiplied by 57?

What's the product of 35 plus the quotient of 14 divided by 34 and the sum multiplied by 57?

(35 + 14 △ 34) × 57 = (35 + 13) × 57 = 1415 × 57 = 23

What's the product of 10 minus 2.5, divided by 0.3 and 2?

12.5

What is the result of dividing the sum of two 2.5 by the product of two 10? What is the number divided by 45, the quotient is 7, and the remainder is 3?
The number a is 430, which is 10 times more than 3 times of the number B. to find the number B, first move the decimal point of a decimal to the left by two places. After expanding by 10 times, the result is 0.78 less than 1.22. To find the decimal?
(formula calculation)

What is the result of dividing the sum of two 2.5 by the product of two 10? 10 x 10 ÷ (2.5 + 2.5) = 20
When a number is divided by 45, the quotient is 7 and the remainder is 3, what is the number? (45 - 3) △ 7 = 6
The number of a is 430, which is 10 times more than the number of B. find the number of B? (430 - 10) △ 3 = 140
Move the decimal point of a decimal to the left by two places, and the result is 0.78 less than 1.22 after 10 times of expansion?
（1.22 - 0.78 ）÷ 10 x 100 =4.4

What is the quotient of the sum of 44 25 divided by 11 times of 25? The expression should be ()
A. 25×44÷25×11B. （44+25）÷（25×11）C. （44×25）÷（25×11）

The quotient can be expressed as: (44 × 25) / (25 × 11)

How many digits should n be to minimize the sum of all the digits of a natural number n? What should be the first number?

First of all, the more n digits, the greater n. considering that each digit is 9, there are at most 33 9s. Because 300 = 33 × 9 + 43, the sum of 33 digit numbers is less than 300, so n is 34 digits. In order to make the minimum and highest order of 34 digits, the minimum and highest order should be 3 (300-297 = 3), so the answer is: 34; 4

For the natural number n, do the following operation: add each number to get another natural number. If the new natural number is one digit, then the operation will stop. If the new natural number is one digit, the operation will stop
Then continue the above operation for the new natural number until you get a one digit number 9 (2003 9) such operation, what is the final one digit?

2003×9+1=18028
1+8+0+2+8=19
1+9=10
1+0=1
Finally, 1

Find a natural number n, so that the sum of the first n natural numbers is a three digit number, and the three digits of the three digits are the same

According to the meaning of the question: the sum of the first n natural numbers has the same digit number, ten digit number and hundred digit number, so we can set the sum of the first n natural numbers equal to 111a, where a is one of the natural numbers 1-9; then there is 1 + 2 + 3 + +(n-1) + n = n (n + 1) △ 2 = 111a = 3 × 37 × a, so n (n + 1) = 37 × 6a, 1 ≤ a ≤ 9. After verification, when a = 6, the above formula is transformed into n (n + 1) = 36 × 37, so the natural number n = 36. Answer: when n = 36, the sum of the first n natural numbers is a three digit number, and the three digits of each, ten and hundred are the same

How many digits should n be to minimize the sum of all the digits of a natural number n? What should be the first number?

First of all, the more n digits, the greater n. considering that each digit is 9, there are at most 33 9s. Because 300 = 33 × 9 + 43, the sum of 33 digit numbers is less than 300, so n is 34 digits. In order to make the minimum and highest order of 34 digits, the minimum and highest order should be 3 (300-297 = 3), so the answer is: 34; 4

When n is a natural number, try to prove that the number of N + 4 times of 3 is the same as that of N times of 3

3 ^ (n + 4) = 81 * 3 ^ n = 80 * 3 ^ n + 3 ^ n because 80 * 3 ^ n has zero single digits, so 3 ^ (n + 4) and 3 ^ n have the same single digits

Write down the natural number starting from 1 until 201 bits. The remainder of this number divided by 3 is ()

Sum up the law
123 is divisible by three
456 can also be divided by three
123456 can also be divided by 3
So, the number of times of 3 can be divided by 3,
Therefore, write down the natural number starting from 1 until 201 bits, which is also a multiple of 3;
The remainder of this number divided by 3 is (0)