Want this question answered?
Six of them.
123, three number 45, two number
6 possible 3 digit combonations
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
A 5-digit number that is divisible by 4 can be any number in the form of ( 10,000n ), where ( n ) is an integer. This means the last two digits of the number must form a number divisible by 4. Therefore, the possible last two digits could be 00, 04, 08, 12, 16, ..., 96.
Six of them.
123, three number 45, two number
6 possible 3 digit combonations
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
Assuming a seven digit whole number: 1,111,111. But if you have to use at least one (2) digit, then 1,111,112.
Any two digit number in which: (a) the units digit is not 0, and (b) the two digits are different will form a new 2-digit number when the digits are interchanged.
By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
The number 4.If you want to use all the digits, proceed from left to right, using the SMALLEST possible digit in each case. Thus, the first digit would have to be 3, the next one a 5 (since you need to keep the only even digit for the end), the next one 9, and finally the 4.
Every square of a 4 digit number has more than four digits.
001234557889 is the answer.