N is the smallest natural number such that in decimal representation it ends with 6 and if we move last digit to the front of the number we get a number 4 times larger than the given number. Find the sum of digits of N.
Answers
Sum of the digits of N = 27
Step-by-step explanation:
Given: N is the smallest natural number such that in decimal representation it ends with 6 and if we move last digit to the front of the number we get a number 4 times larger than the given number.
Find: Find the sum of digits of N.
Solution:
- Given that the new number starts with a 6 and the old number is 1/4 of the new number, the old number must start with a 1.
- As the new number now starts with 61, the old number must start with 61/4 = 15.
- Progress in this way to find all the 6 digits in the number N. We can terminate the repeated process when we arrive at the new number 615 , 384 and the old number N = 153846
Sum of the digits of N = 1 + 5 + 3 + 8 + 4 + 6 = 27
Given : N is the smallest natural number such that it ends with 6 and if we move last digit to the front of the number we get a number 4 times larger than the given number
To find : smallest natural number
Solution:
Let say Number is
N = ABCD...........X6
As we do no know How many digits are in the Number
we move last digit to the front of the number
Hence
Number Becomes
6ABCD...........X
ABCD...........X6 * 4 = 6ABCD...........X
6 * 4 = 24 ( 2 is carried over)
Hence X must be 4
4 * 4 = 16 + 2 = 18
Hence Digit before X (4) = 8 and carried over = 1
4 * 8 = 32 + 1 = 33
Digit before 8 = 3 & Carried over = 3
4 * 3 = 12 + 3 = 15
Digit before 3 is 5 & 1 is Carried
4 * 5 = 20 + 1 = 21
Digit before 5 is 1 and 2 is Carried over
4 * 1= 4 + 2 = 6 ( Digit we need at left Most )
Hence we get 615384 ( as Multiplication )
and number before was 153846
Sum of Digit = 1 + 5 + 3 + 8 + 4 + 6 = 27
Sum of Digit of N = 27
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