Introduction:
In this problem, we need to find out the number of integers from 1 to 100 which are not divisible by 3, 5 or 7.

Method:
We can use the principle of inclusion and exclusion to solve this problem.

Step 1: Count the numbers divisible by 3, 5 or 7:
To count the numbers that are divisible by 3, we can divide 100 by 3 and round it down to the nearest integer. The result is 33. Similarly, there are 20 numbers divisible by 5 and 14 numbers divisible by 7.

Step 2: Count the numbers divisible by both 3 and 5:
To count the numbers that are divisible by both 3 and 5, we can divide 100 by their least common multiple (LCM). The LCM of 3 and 5 is 15. Therefore, there are 6 numbers divisible by both 3 and 5.

Step 3: Count the numbers divisible by both 3 and 7:
To count the numbers that are divisible by both 3 and 7, we can divide 100 by their LCM, which is 21. Therefore, there are 4 numbers divisible by both 3 and 7.

Step 4: Count the numbers divisible by both 5 and 7:
To count the numbers that are divisible by both 5 and 7, we can divide 100 by their LCM, which is 35. Therefore, there are 2 numbers divisible by both 5 and 7.

Step 5: Count the numbers divisible by 3, 5 and 7:
To count the numbers that are divisible by 3, 5 and 7, we can divide 100 by their LCM, which is 105. Therefore, there is only 1 number divisible by 3, 5 and 7, which is 105 itself.

Step 6: Apply the principle of inclusion and exclusion:
To count the numbers that are not divisible by 3, 5 or 7, we need to subtract the numbers that are divisible by 3, 5 or 7 from the total number of integers from 1 to 100, and then add back the numbers that are divisible by both 3 and 5, both 3 and 7, or both 5 and 7, and subtract the number that is divisible by 3, 5 and 7.

Therefore, the number of integers from 1 to 100 that are not divisible by 3, 5 or 7 is:

100 - 33 - 20 - 14 + 6 + 4 + 2 - 1 = 26.

Conclusion:
There are 26 integers from 1 to 100 that are neither divisible by 3 or by 5 nor by 7.