Suppose a number is divisible by both 3 and 5. Which of the following digit(s) can be at unit’s place of this number? WITH EXPLANATION .GIVING OF 50 POINTS A GOOD CHANCE .THE BEST ANSWER WILL BE MARKED BRAINLIEST .TELLTHE ANSWER
Answers
Answer:
by 3 is to be formed using the digits 0,1,2,3,4 and 5, without repetition. The total number of ways this can be done, is
Since a five-digit number is formed by using digits 0,1,2,3,4 and 5, divisible by 3 i.e., only possible when the sum of digits is multiple of three which gives two cases.
Case I: Using digits 0,1,2,4,5 the number of ways =4×4×3×2×1=96.
Case II: Using digits 1,2,3,4,5 the number of ways =5×4×3×2×1=120
Therefore, total number formed =120+96
Step-by-step explanation:
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Concept:
Divisibility by 3: The number itself is divisible by 3 if the sum of its digits is also divisible by 3.
Divisibility by 5: The number is divisible by 5 if it ends in 0 or 5, meaning that its unit digit is either 0 or 5.
To find: A number to be divisible by both 3 and 5 the unit place of the number must contain the digit.
Given: The number is divisible by both 3 and 5.
Explanation:
Many numbers can be divided by the numbers 5 and 3.
The LCM of 5 and 3 is 15, therefore find it first.
The numbers 5 and 3 can now be used to divide all multiples of 15.
15n is typically divisible by 5 and 3 for every n=1, 2, 3,4,5,6,7,8 and 9.
So the unit's place must contain 0 or 5 and the sum of all digits of the number should be in the form 3n for every n=1,2,3...........,9.
the example of such numbers is- 225,715 etc.
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