What is the number of 4 digit natural numbers such that the sum of digits is even?
But the even digit can be in the hundreds, tens or units place which gives us 3 possibilities. Hence, total no. of required numbers= 625+1500+500+1875=4500.
How many four digit numbers are such that the sum of the digits is 9?
So there are 165 four-digit numbers with a digit-sum of 9.
How many four digit numbers are such that the sum of the digits is 10?
There are 219 4-digit numbers whose sum of digits is 10.
How many four digit numbers are there with distinct digits are such that the sum?
Hence 4536 is the number of possible arrangements of four distinct digit numbers.
How many 4 digit numbers with distinct digits are there such that sum of the digits of each of these numbers is an odd natural number?
Therefore, there are 4536 four-digit numbers with distinct digits.
How many four digit numbers with non zero digits have the sum of their digits is 12?
Therefore, 48 positive 4-digit integers of four different non-zero digits have a sum of digits of 12.
How many four digit numbers are such that the sum of the digits is 12?
The number of 4 digit natural numbers such that the product of their digits is 12 is. Hence, total numbers are 36.
How many 4 digit numbers with distinct digits are there such that sum of digits of each of these numbers is an odd natural number?
How many 4 digit numbers exist with all digits even?
But the 2nd,3rd & 4th positions can have any of the 5 digits. So even number can be chosen in 4*5*5*5=500 ways. So total 625+500=1125 4-digit numbers have all their digits either even or odd.
How many four digit numbers are always odd numbers?
Same count holds for (7)ooeo (8)oooe. Thus in this later count we have got 500 + 625 + 625 + 625 = 2375 numbers. Thus in total we have all 2125 + 2375 = 4500 four digit numbers whose digits’ total sum is odd number always.
What is the sum of four digit numbers in the units?
Four digit numbers $ = 4 \\cdot 3 \\cdot 2 \\cdot 1 = 24$ ways we can form a four digit number. Since it’s a 4 digit number, each digit will appear $6= 24/4$ times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is $6(1+2+5+6)=84$.
How many 4-digit numbers can be formed using the given digits?
Since we form a 4 digit number, let us create 4 places. So, totally we may form 5 (24) = 120 numbers using the given digits. From the above steps, we may understand that we may see the number 1 in 24 times in the unit place. Find the sum of all 4-digit numbers that can be formed using digits 0, 2, 5, 7, 8 without repetition?
How many ways can you choose the first digit of a number?
The first digit can be chosen in 9 ways (since it can’t be a 0), the second in 10 ways, the third in 10 ways, the fourth in 10 ways, and the fifth in five ways (the last digit must be one of the five even digits if the first four digits add up to an even number, and odd otherwise, so that the total sum is even).