The number uses every number 0-9 the numbers are used only once the fourth digit is 4 the first digit in the billions place is 3 the 5 is next to the last digit the sixth digit is 7 the digit after 3 is 9 the digit before 5 is
I have no idea what this problem means: For how many two-digit numbers if the ones digit larger than the tens-digit? Can you find a systematic way to arrive at the number? Ok, every 2-digit number looks like xy, where x is the
1. Set I contains six consecutive integers. Set J contains all integers that result from adding 3 to each of the integers in set I and also contains all integers that result from subtracting 3 from each of the integers in set I.
1. First of all, pick the number of times a week that you would like to have chocolate (try for more than once but less than 10) 2. Multiply this number by 2 3. Add 5 4. Multiply it by 50 5. If you have already had your birthday
Find the sum of all 3-digit positive numbers N that satisfy the condition that the digit sum of N is 3 times the digit sum of N+3. Details and assumptions The digit sum of a number is the sum of all its digits. For example, 1123
You have 2 digits, 2 numbers, reverse digits and 54 if the difference and the sum of all is 10 I'm not clear what you're asking, could you clarify what the conditions are for us please? My grandson came home with the problem that
From the digits 1, 2, 3, and 4, how many positive integers are less than 100,000? Consider the possibilities for 5-digit, 4-digit, 3-digit, 2-digit, and 1-digit numbers and repetition of digits. 1,364 1,024 256 A telephone dial