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There are 12 numbers between 100 and 199 that are divisible by eight: 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184 and 192.

Q: What numbers between 100 and 199 are divisible by 8?

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No, 199 is not divisible by 9

There are 21 prime numbers between 100 and 199 inclusive. These are as follows: 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199

101 to 199 give 50 odd numbers. Their sum is 7,500.[I'm not too certain that the question meant 'the average total'? But, the odd numbers average total would be 7,500 ÷ 50 = 150]

In math, y is divisible by x if the division y/x produces no remainder. All whole numbers from 1 to 99, if divided by 100, would produce a remainder. That pattern repeats from 101 to 199 and similarly with all whole numbers of the form ...x01 to ...x99. Therefore: That only leaves whole numbers in the form ...x00, all of which are divisible by 100. So, to state it more simply, the set of all whole numbers divisible by 100 is characterized by endings of 00, such as 100, 200, 3100, and 18100. Any whole number ending in 00 is divisible by 100. The only tricky part is determining if zero and negative integers are included, which depends on the definition given for whole numbers. Sometimes zero is included in the definition, sometimes not. Sometimes negative integers are included, sometimes not. If they are included, the same reasoning as above would apply. Note that zero could be written 00.

199 31+37+41+43+47=199

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50

200

No, 199 is not divisible by 9

There are 21 prime numbers between 100 and 199 inclusive. These are as follows: 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199

81: 8 + 1 = 9 which is divisible by 9, so 81 is divisible by 9 162: 1 + 6 + 2 = 9 which is divisible by 9, so 162 is divisible by 9 199: 1 + 9 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1 which is not divisible by 9, so 199 is not divisible by 9. 1125: 1 + 1 + 2 + 5 = 9 which is divisible by 9, so 1125 is divisible by 9. So 199 is the only one not divisible by 9.

19 divisible by 5, 14 divisible by 7. Of these, 3 divisible by both.

101 to 199 give 50 odd numbers. Their sum is 7,500.[I'm not too certain that the question meant 'the average total'? But, the odd numbers average total would be 7,500 ÷ 50 = 150]

200

-99

Testing the largest whole number less than 200, 199, we find that it is not evenly divisible by 2, 3, 5, 7, 11 or 13. Since the square root of 199 is between 14 and 15, there's no need to test any prime numbers higher than 14. Therefore, the answer is 198, since it is an even number, and all even numbers are composite numbers because they are divisible by 2. A composite number is the opposite of a prime number.

In math, y is divisible by x if the division y/x produces no remainder. All whole numbers from 1 to 99, if divided by 100, would produce a remainder. That pattern repeats from 101 to 199 and similarly with all whole numbers of the form ...x01 to ...x99. Therefore: That only leaves whole numbers in the form ...x00, all of which are divisible by 100. So, to state it more simply, the set of all whole numbers divisible by 100 is characterized by endings of 00, such as 100, 200, 3100, and 18100. Any whole number ending in 00 is divisible by 100. The only tricky part is determining if zero and negative integers are included, which depends on the definition given for whole numbers. Sometimes zero is included in the definition, sometimes not. Sometimes negative integers are included, sometimes not. If they are included, the same reasoning as above would apply. Note that zero could be written 00.

101103107109113127131137139149151157163167173179181191193197199