A natural number when divided by 12, 15, 18 leave remainder 5, 8, 11 respectively. Find sum of digits of least such number? (A) 9 (B) 10 (C) 11 (D) 12
Answers
Given : A natural number when divided by 12, 15, 18 leave remainder 5, 8, 11 respectively.
To Find : sum of digits of least such number
(A) 9 (B) 10 (C) 11 (D) 12
Solution:
Let say Number = N
N = 12A + 5 => N = 12A + 12 - 7 => N + 7 = 12(A + 1)
N = 15B + 8 => N = 15B + 15 - 7 => N + 7 = 15(B + 1)
N = 18C + 11 => N = 18C + 18 - 7 => N + 7 = 18(C + 1)
LCM - Least common multiplier of given numbers is the least number which is perfectly divisible by given numbers.
LCM = product of each factor of highest power
N + 7 = LCM ( 12 , 15 , 18)
12 = 2 * 2 * 3
15 = 3 * 5
18 = 2 * 3 * 3
LCM = 2 * 2 * 3 * 3 * 5 = 180
N + 7 = 180
=> N = 173
173 = 12 * 14 + 5
173 = 15 * 11 + 8
173 = 18 * 9 + 11
Sum of digits of 173
1 + 7 + 3 = 11
Hence correct option is C) 11
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Answer:
Option C is the correct answer.