Question of The Day19-04-2021

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.

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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