Answer

Correct Answer : a ) 497

Consider the given expression

4 × 5 × 6 × …………… × 998 × 999

Or,

⇒\(1×2×3×4×5×6× …………… ×998×999 \over 1×2×3\)

⇒\(999! \over1×2×3\)

To calculate the number of ‘3s’ in a factorial of the number we will divide the number first by 3 and write the quotient, then we will divide the number by 9 and write the quotient again and we will continue doing so until we get ‘0’ as the quotient.

So, highest power of 3 that can exactly divide 999! will be

⇒\(Q({999 \over 3})+Q({999 \over 9})+Q({999 \over 27})+Q({999 \over 81})+Q({999 \over 243})+Q({999 \over 729})\)

⇒ 333 + 111 + 37 + 12 + 4 +1

⇒ 498

Therefore, number of 3s in 999! are ‘498’ thus we can say that

Number of '3s' in \(999! \over1×2×3\) will be 497

Therefore, 3497 is the largest power of ‘3’ in the expression 4 × 5 × 6 × …………… × 998 × 999

Hence, (a) 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, UP SI etc.

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