Not evenly. The answer would be 18 remainder 1.
No, 199 is not divisible by 9
No, it would be divisible by ten.
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.
To find the natural numbers less than 200 that are divisible by 6, 9, or both, we can use the principle of inclusion-exclusion. First, the count of numbers divisible by 6 is ( \left\lfloor \frac{199}{6} \right\rfloor = 33 ), and for 9, it is ( \left\lfloor \frac{199}{9} \right\rfloor = 22 ). The count of numbers divisible by both 6 and 9 (i.e., 18) is ( \left\lfloor \frac{199}{18} \right\rfloor = 11 ). Applying inclusion-exclusion, we get ( 33 + 22 - 11 = 44 ). Thus, there are 44 different natural numbers less than 200 that are exactly divisible by either 6, 9, or both.
11 is divisible by no whole numbers except 1 and 11.
No, 199 is not divisible by 9
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.
1, 3, 199, 597.
No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.No. It is divisible by 11.
19 divisible by 5, 14 divisible by 7. Of these, 3 divisible by both.
Here's a mathematical trick. Any number wholly divisible by 3 has the sum of its digits equal to 3 or a multiple of 3. 199 : 1 + 9 + 9 = 19....NOT divisible by 3 209 : 2 + 0 + 9 = 11....NOT divisible by 3 390 : 3 + 9 + 0 = 12....YES, this is divisible by 3 499 : 4 + 9 + 9 = 22....NOT divisible by 3 Note : 390 ÷ 3 = 130
No, it would be divisible by ten.
Yes because 597/3 = 199
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.
To find the natural numbers less than 200 that are divisible by 6, 9, or both, we can use the principle of inclusion-exclusion. First, the count of numbers divisible by 6 is ( \left\lfloor \frac{199}{6} \right\rfloor = 33 ), and for 9, it is ( \left\lfloor \frac{199}{9} \right\rfloor = 22 ). The count of numbers divisible by both 6 and 9 (i.e., 18) is ( \left\lfloor \frac{199}{18} \right\rfloor = 11 ). Applying inclusion-exclusion, we get ( 33 + 22 - 11 = 44 ). Thus, there are 44 different natural numbers less than 200 that are exactly divisible by either 6, 9, or both.
No, it is not, but 121 and 132 are divisible by 11.
Yes. 11 is divisible by 1 and itself.