The code of five digits is of the form:
A B C D F
Where A (the first digit) may be any of 9 digits (because 0 is excluded)
B, C, and D, may be each any of 10 digits (0 - 9)
F (the last digit) may be 0, 2, 4, 6 or 8, i.e. 5 different digits.
So the number of different available codes is:
9 * 10 * 10 * 10 * 5 = 45,000.
Answer: 45,000
Answer:
5,10; 6,12; 7,14
Explanation:
We will demonstrate the iteration of the loop:
First iteration: Number = 7, Count = 5 at the beginning. We will check if Count <= Number? Since it is correct, prints 5,10. Increment the Count by 1.
Second iteration: Number = 7, Count = 6. We will check if Count <= Number? Since it is correct, prints 6,12. Increment the Count by 1.
Third iteration: Number = 7, Count = 7. We will check if Count <= Number? Since it is correct, prints 7,14. Increment the Count by 1.
Forth iteration: Number = 7, Count = 8. We will check if Count <= Number? Since it is not correct, the loop stops.
The single instruction that can inverts bits 5 and 6 in the bl register is xor bl,1100000b.
<h3>What is single instruction?</h3>
Single Instruction is a term that connote all the data streams are said to be processed though the use of the same compute logic.
Note that in the case above, the single instruction that can inverts bits 5 and 6 in the bl register is xor bl,1100000b.
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