 How many 3-digit numbers satisfy the property that the digit in the middle is the average of the first and the third digit?
Answer: 45 Step by Step Explanation:- Average of any two numbers a and b is equal to . This signifies that, for their average to be a natural number, the sum of a and b has to be an even number, which further means that both a and b have to be either even or odd.
- The number of such pairs will be:
Let the first box represents the hundreds digit and the second box represents the units digit.
Let us first consider the case when both of them are odd. The first and the second box both have five possibilities: 1, 3, 5, 7, 9
This means that there will be 5 × 5 = 25 such pairs. - Now let us consider the case when both of them are even. The possible number of digits at the hundreds place are 2, 4, 6, 8 i.e. 4 digits (the hundreds digit of a three digit number cannot be 0).
The possible digits for units place will be equal to 0, 2, 4, 6, 8, i.e. 5 digits.
This means that there will be 5 × 4 = 20 such pairs. - Therefore, the total number of such possible numbers is equal to 25 + 20 = 45.
Hint: There are total numbers from 1 to 100. Use the symmetricity of the numbers from 101 to 200, 201 to 300 and so on. And three digit numbers are the numbers which have only three digits in them.Complete Step-by-Step solution:
As we know any number is said to be a three digit number if it consists of only three digit. So, as we know that the smallest three digit number is 100. And the maximum three digit number is 999. So, we need to count the number from 100 to 999. Now, we know that there are 100 numbers from 1 to 100 as we can count them like one, two, three …..hundred.
So, if we follow the same pattern then there should be exactly 100 numbers from 101 to 200 and similarly, there will be 100 numbers from 201 to 300. So, we can get the series as
Numbers from 101 to 200 $=100$
Numbers from 201 to 300 $=100$
Numbers from 301 to 400 $=100$
Numbers from 401 to 500 $=100$
Numbers from 501 to 600 $=100$
Numbers from 601 to 700 $=100$
Numbers from 701 to 800 $=100$
Numbers from 801 to 900 $=100$
Numbers from 901 to 1000 $=99+1 $( 1 for 100)
$\overline{900}$
Now, we can get a total of three digit numbers by counting the numbers from 100 to 999. There are 900 numbers from 101 to 1000 (including both), since 1000 is a 4-digit number, so we cannot count it for a three digit number. Hence, there are $900-1=899$ numbers from 101 to 999. And since 100 was not included in counting, we can add one more number $\left( 100 \right)$ to the total number of three digit numbers. Hence, there are 899 + 1 $=900$ numbers of three digit.
Hence, the answer is 900. Note: One may use the other approach that is based on the chapter Arithmetic progression. As we know A.P.(Arithmetic progression) is a series with same successive difference and general term ( ${{n}^{th}}$ term) of A.P. is given as
${{T}_{n}}=a+\left( n-1 \right)d$
So, let the series be $\left( 100,101,102,103......999 \right)$
Put ${{T}_{n}}=999$, $a=$ first term $=100$ and $d=$ common difference $=1$. So, we get
$999=100+\left( n-1 \right)\left( 1 \right)$
$999-100+1=n$
$n=900$
Hence, 999 is the ${{900}^{th}}$ term of the sequence i.e. there are 900 three digit numbers.
One may go wrong if he/she tries to find the total numbers from 100 to 999 by just subtracting 100 and 999. He/she will get an answer as $999-100=899$, which is wrong. As we get there are 900 numbers in the situation. So, one may go wrong with this approach.
So, one may generalised the relation that there are $\left( A-B+1 \right)$ numbers between A and B (including A and B) where A > B. How many 4 digit numbers can be formed by using the digits 1 to 9, if repetition of digits is not allowed? Improve Article Save Article Like Article - Last Updated : 02 Dec, 2021
ReadDiscussCoursesPracticeVideoImprove Article Save Article Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter. In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered. Permutation Formula In permutation r things are picked from a group of n things without any replacement. In this order of picking matter. nPr = (n!)/(n – r)! Here, n = group size, the total number of things in the group r = subset size, the number of things to be selected from the group
Combination A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used. Combination Formula In combination r things are picked from a set of n things and where the order of picking does not matter. nCr =n!⁄((n-r)! r!) Here, n = Number of items in set r = Number of things picked from the group
Answer: Repetition of the digit is not allowed. So, for the first digit we have 9 option for second digit we have 8 option for third digit we have 7 option and for fourth digit we have 6 option There are total 9 digit from which we have to select 4, repetition is not allowed Total no. of ways = 9P4 = 9!/(9-4)! = 9!/5! = 3024
Similar QuestionsQuestion 1: How many 5 digit numbers can be formed by using the digit 1 to 9. If repetition of digits is not allowed? Answer: Repetition of the digit is not allowed. So, for the first digit we have 9 option for second digit we have 8 option for third digit we have 7 option for fourth digit we have 6 option and for fifth digit we have 5 option There are total 9 digit from which we have to select 5, repetition is not allowed Total no. of ways = 9P5 = 9!/(9-5)! = 9!/4! = 15,120
Question 2: How many 3 digit numbers can be formed by using the digit 0,1,2,3. If repetition of digits is allowed? Answer: Repetition of digit is allowed. So, for the ones place we have 4 option i.e., 0,1,2,3 similarly for tens place we have again 4 option i.e., 0,1,2,3 and for the hundredth place we have 3 option i.e., 1,2,3 we can’t take 0 at hundredth place because if 0 will be filled at hundredth place it will not become 3 digit number it will be taken as two digit number. Total no. of three digit number = 3 × 4 × 4 = 48
Question 3: How many 5 digit numbers can be formed by using the digit 0,1,2,3,4. If repetition of digits is allowed? Answer: Repetition of digit is allowed. So, for the ones place we have 5 option i.e., 0,1,2,3,4 similarly for tens place we have again 5 option i.e., 0,1,2,3,4 for the hundredth place we have 5 option i.e., 0,1,2,3,4for the thousandth place we have 5 option i.e., 0,1,2,3,4 and for the ten thousandth place we have 4 option i.e., 1,2,3,4 we can’t take 0 at ten thousandth place because if 0 will be filled at ten thousandth place it will not become 5 digit number it will be taken as 4 digit number. Total no. of five digit number = 4 × 5 × 5 × 5 × 5 = 2500
Question 4: How many 4 – digit even numbers can be formed using the digits (3,5,7,9,1,0) if repetition of digits is not permitted? Answer: For even number unit digit must be 0, Now the remaining digits are 5 i.e., 3,5,7,9,1 now for the thousand place we have 5 option for the hundredth place we have 4 option and for the tens place we have 3 option
How many 3 digits number are there for which the product of their digits is more than 2 but less than 7?
Solution: The product of the digits of the three-digit numbers should be more than 2 and less than 7 . Hence the possible numbers are as follows. Hence there are a total of 21 possibilities.
How many 3 digit numbers are there for which the sum of their digits is more than 3?
Hence the answer is "20"
How many 3 digit numbers are there in which all 3 digits are different?
Interpreting the result
There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed. Note: We can also use the multiplication principle to answer this question.
How many 3 digit numbers are there in which the sum of digits is 7?
There are 28 numbers.
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