There are 5 choices for the first digit, as any of the 5 given digits can be used. After selecting the first digit, there are 4 remaining choices for the second digit, and 3 choices for the third digit. Therefore, the total number of ways to write a 3-digit positive integer using the given digits without repetition is 5 x 4 x 3 = 60 ways.
A number with more than one digits: that is, an integer greater than 9.
This is a question of permutations; the answer is equal to the factorial of 5 (number of digits) divided by the factorial of 3 (number used in each selection), written 5! / 3!. This equals 120 / 6, or 20 ways.
A 17-digit number is typically referred to as a "seventeen-digit number." In mathematics, it would be classified as a whole number or an integer with 17 digits. In computing, a 17-digit number may be represented as a data type such as a long integer or a double-precision floating-point number.
A three digit number can be chosen in 9*10*10 = 900ways No. of ways to choose a three digit number without 7 is 8*9*9 = 648 ways probability of picking a three digit number that includes atleast one digit that is 7 is 1- (648/900) = 252/900 = 7/25
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
The smallest positive integer with four different digits is 1023. This number uses the smallest digits available (1, 0, 2, and 3) while ensuring that the first digit is not zero, thus maintaining its status as a positive integer.
It is -987. The smallest positive 3-digit integer with unique digits is 102.
The largest three-digit positive integer is 999. It is the highest number that can be represented with three digits, where each digit can range from 0 to 9, with the first digit being non-zero.
81
The answer is 9*9!/9*109 = 0.0003629 approx.
The answer is a negative or positive integer with one or two digits.
The smallest positive five-digit integer with all different digits that is divisible by each of its non-zero digits is 10234. Each of its digits (1, 0, 2, 3, 4) is unique, and it is divisible by 1, 2, 3, and 4. Note that 0 is not considered in divisibility but does not repeat any digits.
12
Let the three-digit integer be ( x ) and the two-digit integer be ( y ). According to the problem, ( x - y = 288 ), which implies ( x = y + 288 ). The digits used are 7, 8, 2, 3, and 0. After testing possible combinations, the solution is ( x = 782 ) and ( y = 494 ). Therefore, the sum of the three-digit integer ( x ) and the two-digit integer ( y ) is ( 782 + 494 = 1276 ).
The least possible integer is -98765432. The least possible positive integer is 10234567.
First, separate the negative and positive integers (put them into two separate groups). If there is a zero, you can put it in its own group - or put it into the same group with the positive integers. Negative integers come first, then zero, then positive integers.For positive integers:An integer with less digits comes before an integer with more digits.For integers with the same number of digits, look at the first digit. The integer with the smaller digit in this position comes first.If the first digit is the same, look at the second digit. If those are equal, look at the third digit, etc.For negative integers, it is the other way round - for example, an integer with MORE digits comes first.
9*9*9 = 729 ways.