I am less than 100 so the range is 01 - 99, but as I am divisible by 2 then I am even. As my tens digit and ones digit are the same then I am a 2 digit number so the range is now 10 - 98.
The sum of my digits is 8, my tens digit and my ones digit are the same . . so the only solution is 44.
The question cannot be answered. All that I know about your number is that is has at least 4 digits. Your number could be a hundred digits long - I have no way of knowing.
Mental math like if 12 into 72 it will not be a double digit if you divide 100 by 10 it will be double digits. Most times if you have a tripple digit number and divide it by double digit the answer will be a double digit.
4
Oh, dude, dividing two digits by three digits is like, totally easy. You just, like, do long division. You put the three-digit number outside the division house, the two-digit number inside, and then you, like, divide and stuff. It's not rocket science, man. Just remember to carry the one if you have to, or whatever.
It's basically the algorithm to break a number down to digits. Then each digit would be represented as a natural binary number on its own.To break it down to digits, get a remainder after division with 10 (that's the last digit), then divide the number by 10 to remove that digit. Repeat until there are digits, and store each one into an array (or print them to screen without storing).
The question cannot be answered. All that I know about your number is that is has at least 4 digits. Your number could be a hundred digits long - I have no way of knowing.
Mental math like if 12 into 72 it will not be a double digit if you divide 100 by 10 it will be double digits. Most times if you have a tripple digit number and divide it by double digit the answer will be a double digit.
by making them half
4
The number 2464 fulfils the requirements.
Oh, dude, dividing two digits by three digits is like, totally easy. You just, like, do long division. You put the three-digit number outside the division house, the two-digit number inside, and then you, like, divide and stuff. It's not rocket science, man. Just remember to carry the one if you have to, or whatever.
Unless you are using remainders, no because the divisor may not divide evenly into the dividend you idiots.
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9 excluding the two previous digits
Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number
Since there are only five different digits, a 6-digit number can only be generated if a digit can be repeated. If digits can be repeated, the smallest 6-digit number is 111111.
It's basically the algorithm to break a number down to digits. Then each digit would be represented as a natural binary number on its own.To break it down to digits, get a remainder after division with 10 (that's the last digit), then divide the number by 10 to remove that digit. Repeat until there are digits, and store each one into an array (or print them to screen without storing).
987654321 is the highest nine-digit number with no repeated digits.