You are given three digits 7, 2, and 9.
A. How many 3-digit numbers can be
formed without repeating any of them?
B. How many of the numbers formed are
divisible by 3? Why?
C. How many of the numbers formed are
divisible by 9? Why?
D. How many of the numbers formed are
divisible by 6? Why?
Answers
Answer:
A. 6 numbers are formed.
B. 6 numbers.
C. 6 numbers.
D. 2 numbers.
Explanation:
(A)
Total digits = 3 ( 7, 2 and 9 )
Number of digits formed = 3!
3! = 3 × 2 × 1 = 6
6 digits will be formed.
The digits are:
279, 297, 729, 792, 927 and 972
( B )
Divisibility by 3: Sum of all the digits of the number should be divisible by 3.
As we know all the 6 numbers are made up of the same digits. So, sum of all the digits of the numbers is same.
Sum of all the digits of the numbers :
7 + 2 + 9 = 18
18 is divisible by 3
So, all the 6 numbers are divisible by 3.
( C )
Divisibility by 9: Sum of all the digits of the number should be divisible by 9.
Sum of all the digits of all the numbers is same i.e. 18
18 is divisible by 3
So, all the 6 numbers are divisible by 3.
( D )
Divisibility by 6 : Sum of all the digits should be divisible by 3 and the unit place should be even.
as we've checked earlier, all the number are divisible by 3, so we just need to check whose unit place is even.
The numbers: 279, 297, 729, 792, 927 and 972
The numbers with even unit place: 972 and 792
So, two numbers are divisible by 6
Answer:
Step-by-step explanation: