N -7 -3 - How many factors of N are multiples of 10 -A- 736B- 1008C- 1356D- 894

The correct option is B 1008Prime factorization of (7!)^3 = (2^4 × 3^2 × 5 × 7)^3 = 2^12× 3^6 × 5^3 × 7^3 Now, for a multiple of 10, there should be at least ...

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Divisibility Rule for Powers of 2 & 5

N =7 !3 . How...

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$N = 7!^{3} .$ How many factors of N are multiples of 10?

736

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1008

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1356

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894

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Solution

The correct option is B 1008
Prime factorization of $(7!)^{3} = (2^{4} \times 3^{2} \times 5 \times 7)^{3} = 2^{12} \times 3^{6} \times 5^{3} \times 7^{3}$
Now, for a multiple of 10, there should be at least one 5 and at least one 2 present in the number.
Therefore, the number can be like $= 2^{1 - 12} \times 3^{0 - 6} \times 5^{1 - 3} \times 7^{0 - 3}$
Hence, number of factors $= 12 \times 7 \times 3 \times 4 = 1008$

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N=7!3. How many factors of N are multiples of 10?

The correct option is B 1008Prime factorization of (7!)3=(24×32×5×7)3=212×36×53×73 Now, for a multiple of 10, there should be at least one 5 and at least one 2 present in the number. Therefore, the number can be like =21−12×30−6×51−3×70−3 Hence, number of factors =12×7×3×4=1008

$N = 7!^{3} .$ How many factors of N are multiples of 10?