If n=9!-6^4, which of the following is the greatest integer k such that 3^k...

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If n=9!-6^4, which of the following is the greatest integer k such that 3^k is a factor of n ?

GMAT Panda · Accepted Answer

6 We first find how many times 3 appears in 9! by counting its factors in the numbers 3, 6, and 9. 9! = 123456789 = 3^1(23^1)3^2(other factors not divisible by 3) = 3^4(integer not divisible by 3) We then break down 6^4 into its prime factors. 6^4 = (23)^4 = 2^43^4 We now factor out the common 3^4 from both terms in the expression for n. n = 9! - 6^4 = 3^4(9!/3^4 - 6^4/3^4) We simplify the first fraction by canceling four factors of 3 from 9!. 9! = 12345678(33) = 12345(23)78(33) 9!/3^4 = 1245278 = 4480 We simplify the second fraction by dividing out all factors of 3 from 6^4. 6^4/3^4 = (23)^4/3^4 = 2^4 = 16 We then subtract to find the integer inside the parentheses. 4480 - 16 = 4464 We next find how many times 3 divides into 4464 by testing divisibility by 9 and then by 3. 4 + 4 + 6 + 4 = 18 Since 18 is divisible by 9, 4464 is divisible by 9 4464/9 = 496 4 + 9 + 6 = 19 Since 19 is not divisible by 3, 496 is not divisible by 3 We now add the exponent of 3 from the remainder to the exponent we factored out to get the total exponent. n = 3^4(3^2496) = 3^4+2496 = 3^6496

Explanation	Calculations	Help
$We first find how many times 3 appears in 9! by counting its factors in the numbers 3, 6, and 9.$	$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 = 3^{1} \times (2 \times 3^{1}) \times 3^{2} \times (other factors not divisible by 3) = 3^{4} \times (integer not divisible by 3)$
$We then break down 6^{4} into its prime factors.$	$6^{4} = (2 \times 3)^{4} = 2^{4} \times 3^{4}$
$We now factor out the common 3^{4} from both terms in the expression for n .$	$n = 9! - 6^{4} = 3^{4} \times (\frac{9 !}{3 ^{4}} - \frac{6 ^{4}}{3 ^{4}})$
$We simplify the first fraction by canceling four factors of 3 from .$	$9! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times (3 \times 3) = 1 \times 2 \times 3 \times 4 \times 5 \times (2 \times 3) \times 7 \times 8 \times (3 \times 3) \frac{9 !}{3 ^{4}} = 1 \times 2 \times 4 \times 5 \times 2 \times 7 \times 8 = 4480$
$We simplify the second fraction by dividing out all factors of 3 from .$	$\frac{6 ^{4}}{3 ^{4}} = \frac{( 2 \times 3 ) ^{4}}{3 ^{4}} = 2^{4} = 16$
$We then subtract to find the integer inside the parentheses.$	$4480 - 16 = 4464$
$We next find how many times 3 divides into 4464 by testing divisibility by 9 and then by 3.$	$4 + 4 + 6 + 4 = 18 Since 18 is divisible by 9, 4464 is divisible by 9 \frac{4464}{9} = 496 4 + 9 + 6 = 19 Since 19 is not divisible by 3, 496 is not divisible by 3$
$We now add the exponent of 3 from the remainder to the exponent we factored out to get the total exponent.$	$n = 3^{4} \times (3^{2} \times 496) = 3^{4 + 2} \times 496 = 3^{6} \times 496$

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