See the following: http://www.artofproblemsolving.com/Wiki/index.php/Divisibility_rules/Rule_for_3_and_9_proof
No - octal numbers use only the digits 0-7.
It appears that only single digit numbers work (0 thru 9)
That usually refers to a floating-point number that is stored in 8 bytes, and has (in decimal) about 15 significant digits. In contrast, single-precision is stored in 4 bytes, and has only 6-7 significant digits.
a neon number is that wheresum of digits of square of the number is equal to the number ...for example --sample input => 9sample output =>9*9 = 81,8+1=9therefore,9 is a neon numberPRACTICALLY, THERE IS NO OTHER SUCH NUMBER...ITS ONLY A THEORETICAL CONCEPT OF NEON NUMBERS !!
Only 1 exists, and it is "999"
There is no number, no matter the number of digits, that is only divisible by 2.
If a number is divisible by 4, the last two digits are divisible by 4. For example, the number 314 is not divisible by 4, but 316 is divisible by 4, because 16 is divisible by 4.
2,4,6,8,and 0 are divisible by 2 in the ones digit. Zero is only divisible in a number with 2 digits or more. 0 itself is not divisible by 2.
A number is divisible by 4 if and only if its last two digits are divisible by 4. For example, 53112 is divisible by 4, because the last two digits (12) are divisible by 4.
30
That the number is divisible by 4* and the sum of its digits is a multiple of 3. *If the number has three of more digits then it is only necessary to look at the tens and units to determine if it is divisible by 4, as 4 is a factor of 100 and therefore of any multiple of 100. Examples : 75 : is not divisible by 4 although its digits total 12 which is a multiple of 3. 132 : is divisible by 4 as 32 is divisible by 4, and its digits total 6 which is divisible by 3, then 132 is divisible by 12.
A prime number is only divisible by 1 and itself. If all three digits have to be used then any number formed from them is divisible by 9 as the sum of the digits = 9. Thus no prime number can be formed. If only two of the three digits are used then it cannot end in 0 or 8 as it would be divisible by 2. This only leaves 81 where the digits again total 9 and thus is divisible by 9. None of the single digits 0,1,8 are prime. 8 is an even number. 1 is classed as unity, 0 is zero.
only 180- if the last 2 digits of a number is divisible by 4, then it is
For a number to be divisible by 105 it must be divisible by 3, by 5 and by 7. So, divisibility by 3 requires all three of the following to be satisfied:Sum the digits together. Repeat if necessary. If the answer is 0, 3, 6 or 9 the original number is divisible by 3.If the final digit of the number is 0 or 5, the original number is divisible by 5.Take the number formed by all but the last digit. From it subtract double the last digit. Keep going until there is only one digit left. If it is 0 or 7 then the original number is divisible by 7.
Because any number to left of the last two digits is, by definition a multiple of 100, and as that (100) is of course divisible by 4, then they can be disregarded when checking to see if the whole number id divisible by 4
The foolproof way is to divide 4 into it. If the answer is a whole number, it's divisible by 4. You can also tell by looking at it. If the last two digits are a multiple of 4, like 24 or 80, the whole number is divisible by 4.
A number is divisible by 3 only if the sum of its digits is. 1+1+2=4, which is not divisible by 3, therefore 112 is not.