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
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).
To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; i
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
by making them half
The number 2464 fulfils the requirements.
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
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).
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
You can't tell anything about the quotient until you know whatthe divisor is going to be.-- If I divide your 4,796 by 4, the quotient is 1,199 . . . 4 digits.-- And if I divide it by 2,398, the quotient is 2 . . . . only 1 digit.
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.
To count the number of times a digit occurs in an integer, start by initializing an array of ten counts of digits, such as int digits[10];Then, in a loop while the number is non zero, increment the element in the digits array that corresponds to the units digit, and then divide the number by ten, such as digits[number%10]++ and number/=10;int digits[10];int i;int number = some number;for (i=0; i