Find the smallest number which, when divided by 5, 6 and 7 leaves remainder 3 in each case but the number is exactly divisible by 9?
Answer
Correct Answer : c ) 423
Explanation :According to the question
The number exactly divisible by 5,6 and 7 = K times of LCM of (5, 6 and 7)
Where, K is any integer
Taking the LCM of (5, 6, 7) K = 5 * 6 * 7 * K
⇒ 210K
According to the question,
Remainder in each case = 3
⇒ Thus, the required number becomes = 210 * k + 3
The number is exactly divisible by 9,
So, \(210K+3 \over 9\) should give zero remainder
⇒Remainder \(({207K \over 9})\)+Remainder\(({{3K+3}\over 9})\)=0
As, 207 is completely divisible by 9 so,
Remainder\(({{3K+3} \over 9})\)=0
Let the value of K be 2
Then, 3K + 3 = 3 * 2 + 3 = 9
⇒Re\({9 \over 9}\)=0
Thus, Required number = 210 * 2 + 3 = 423
Hence, (c) is the correct answer.
Check the divisibility rule of 9.
Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, UPSC CDS, UPSC NDA, etc.
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