No.
All multiples of 8 are even, but 141 is odd, so 141 is not divisible by 8.
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To test if a number is divisible by 8, add 4 times the hundreds digit to 2 times the tens digit to the ones digit; if this sum is divisible by 8, then so is the original number. The test can be repeated on the sum, so continue the summing process until a single digit remains - only if this single digit is an 8 is the original number divisible by 8.
141 → 4 × 1 + 2 × 4 + 1 × 1 = 13
13→ 4 × 0 + 2 × 1 + 3 = 5
5 is not 8, so 141 is not divisible by 8
141 ÷ 8 has a remainder of 5.
141 can be divisible by 3. As you can see 141/3 = 47.
No. 141 is not evenly divisible by four.
it is divisible by 3 which is 141
No. 141 is not evenly divisible by nine.
No.
Yes.
No. If the last two numbers are not divisible by 4, then the number is not divisible by 4.
Any real number is divisible by any other real number
17.625
282 is divisible by: 1, 2, 3, 6, 47, 94, 141 and 282.
I'm not quite sure about that theory, but if all the numbers in the number add up to a number that's dividable by 3, then the number itself would then be divisible by 3. Ex. 141 1+4+1= 6 6 is divisible by 3, so 141 is too. 141/3= 47.
If the last 3 digits are divisible by 8, the number is divisible by 8.