40 times of a number minus 1 can be divided by 97, so the minimum natural number is ()

40 times of a number minus 1 can be divided by 97, so the minimum natural number is ()

The original problem is to find the minimum natural number solution of the following equation: 40 * m - 1 = 97 * n, 40m - 1 = 97n40m = 97n + 1. If the number of 40 m on the left side of the equal sign is 0, then the number of 97 n on the right side of the equal sign should be 9, then the number of n should be 7

What is the sum of all odd numbers that cannot be divisible by 9 in the 100 natural numbers 1-100?

（1+3+5+… +99）-（9+27+… +81 + 99) = (1 + 99) × 50 / 2 - (9 + 99) × 6 / 2 = 2500-324, = 2176

In the 100 natural numbers 1-100, what is the odd sum of all the numbers that cannot be divisible by 11

2005
(1+99)*50/2-(11+99)*9/2=2005

What is the 100th odd number in the natural number 1-2008 that cannot be divisible by 3 or 5?

The 100th odd number is 373. It is equivalent to 15 odd numbers in every 30 natural numbers as a group: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 2931, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59 Cross out multiples of 3, 3, 9, 15, 2

What is the sum of all the numbers from 1 to 100 that cannot be divisible by 5 or 9?

The sum is: 1 + 2 + +100 = 5050, the sum of the numbers divided by 5 is 5 × (1 + 2 +...) +20) The sum of the numbers divided by 9 is: 9 × (1 + 2 +...) 11) The sum of the numbers divisible by 45 is: 45 + 90 = 135, so all the numbers from 1 to 100 cannot be divisible by 5 and 9

In the 100 natural numbers from 1 to 100, the sum of all the numbers that cannot be divisible by 3 or 5 is_____ .

15n+1、15n+2、15n+4、15n+7、15n+8、15n+11、15n+13、15n+14
There are eight groups, each of which is an arithmetic sequence
1+16+31+46+…… +91=（1+91）×6÷2＝276
Please count the rest by yourself!

How many natural numbers can't be divisible by 2357 in 1-100
I know there are 22,

First of all, we can make sure that all prime numbers are not divisible by 2,3,5,7. All prime numbers are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97. A total of 25 numbers are removed, 2,3,5,7, and there are 21 left, plus 1 (because 1 cannot be divisible by 2,3,5,7), so 25-4 + 1 = 22 (numbers) cannot be divisible by 2,3,5,7
A: there are 22 numbers that cannot be divided by 2, 3, 5, 7

In the 100 natural numbers 1-100, the numbers that can be divided by 2 or 3 have the same number______ One

There are 50 numbers that can be divided by 2, 100 △ 2 = 50, 100 △ 3 = 33 1, there are 33 numbers divisible by 3; 100 △ 6 = 16 There are 16 numbers that can be divided by 2 and 3. 50 + 33-16 = 67

How many natural numbers from 1 to 100 can be divided by 2 or 3 or 5?

It's all in the prime table

In natural numbers 1 to 100, the numbers that can be divided by 3 or 5 are

We all give the program, I think what you want is not programming! You can think like this: 1, from 1 to 15, 16 to 30 The rule of 15 consecutive numbers in each group is the same, that is, 5 of 15 numbers in each group can be divided by 3, 3 by 5, and 1 by 3 and 5 at the same time