AND it to 1 and test for zero.
All even numbers have 0 in the last bit.
If it is even
There are none because two consecutive even integers would add up to an even number and the number given of 217 is an odd number.
No, thanks.
Composite. It is an even number after all, so the factor 2 is a given.
Type your answer here... i think we should first enter 1 number then check it
No. A given number need not even be divisible by a given prime.
The only even prime number is 2.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
Assuming you know that your number is a perfect square, the square root of an even number is even, and the square root of an odd number is odd.
If it is even
yes it can be tested with a cdi tester!!! that even as a part number
The lowest priority interrupt in the 8085 microprocessor is INTR, unless you also consider the software interrupts, RST 0 through RST 7, which are even lower.
The MOV A,A instruction in the 8085 does nothing, not even change flags. It only consumes time, specifically four clock cycles plus applicable wait states.
16 and The eighth even number
even number can be divided by 2 to give a natural number undecimal whereas the odd number can't be divided by 2 or else given a decimal number
Test the least significant bit of the number. If it is 1, the number must be odd. If it 0 it must be even. To achieve this, place the value in B then use the ANI instruction: ANI 01H Thus if B is 01H, B is odd, otherwise it is even. Works for both positive and negative integers.