1, 2, 5, 7, 10, 14, 35, 49, 70, 98, 245, 490.
No, it's not.
No, it is divisible by 1, 2, 5, 7, 10, 14, 35, 49, 70, 98, 245, 490
Not exactly because it will have a remainder
no. its divisible by 2 and 10
Any multiple of 490.
Composite; any number ending with zero (besides zero itself) is divisible by 10. 490 = 49 * 10 = 7 * 7 * 5 * 2
490
90% of 490 = 90% * 490 = 0.9 * 490 = 441
25% of 490= 25% * 490= 0.25 * 490= 122.5
60% of 490= 60% * 490= 0.6 * 490= 29460% of 490.00 is 294.00
use the tens place to multiply by 2, then use the number to subtract the ones place.The answer can be divided by 7, if not it will be 0.use 490 as an example.9x2=18.18-4=14.14 divided by 7 is 2. so 49 can be divide by 7. -- Above is falsely formulated, but the idea is correct. Also there are different solutions. Correct: Subtract 2 times the last digit from the truncated number. Ignore negative sign. Repeat until 0 or 7, then original number is divisible by 7. Example: 28 -> 2.8 - 2 = 14 -> 2.4 - 1 = 7, so 28 is divisible by 7. Example: 490 -> 2.0 - 49 = -49 -> 49 -> 2.9-4 = 14 -> 2.4 - 1 = 7, so 490 is divisible by 7. Take the last digit and subtract it from 4 times the truncated number. Repeat until 0 or 7, then original number is divisible by 7. Example: 28 -> 4.2 - 8 = 0, hence 28 is divisible by 7. Example: 490 -> 4.49 - 0 = 196 -> 4.19 - 6 = 76 - 6 = 70 -> 4.7 - 0 = 28 -> 0 (see previous example), hence 490 is divisible by 7. Such computations are not that much easier than dividing by 7 and finding a remainder of 0, but it uses smaller factors.
490% = 490/100 = 49/10