Consider the following number : $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$
What is the number of consecutive zeros at the end of the number given above ?
[amp_mcq option1=”50″ option2=”46″ option3=”37″ option4=”35″ correct=”option3″]
This question was previously asked in
UPSC CISF-AC-EXE – 2020
– The given number is $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$.
– Prime factorize each term:
– $3^5 = 3^5$
– $5^5 = 5^5$
– $6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$
– $10^6 = (2 \times 5)^6 = 2^6 \times 5^6$
– $15^{12} = (3 \times 5)^{12} = 3^{12} \times 5^{12}$
– $12^{15} = (2^2 \times 3)^{15} = (2^2)^{15} \times 3^{15} = 2^{30} \times 3^{15}$
– $25^7 = (5^2)^7 = 5^{14}$
– Combine the prime factors:
– Factors of 2: $2^{10} \times 2^6 \times 2^{30} = 2^{10+6+30} = 2^{46}$. Total power of 2 is 46.
– Factors of 5: $5^5 \times 5^6 \times 5^{12} \times 5^{14} = 5^{5+6+12+14} = 5^{37}$. Total power of 5 is 37.
– (Factors of 3 are $3^5 \times 3^{10} \times 3^{12} \times 3^{15}$, but these do not contribute to zeros).
– The number of factors of 2 is 46. The number of factors of 5 is 37.
– The number of pairs of (2, 5) is $\min(46, 37) = 37$.
– Therefore, there are 37 consecutive zeros at the end of the number.