What is the greatest value of the positive integer p such that 2^p is

Question

What is the greatest value of the positive integer p such that 2^p is a factor of 80^50?

(A) 4
(B) 10
(C) 50
(D) 100
(E) 200

kunalcvrce · Answer

80^50=(2^200*5^50)

so greatest value is 200

option E

KSBGC · Answer

we are asked to find the greatest value of the positive integer p. In order to find out the greatest value of p as a exponent of base 2 we have to factorize 80^50.

(2*2*2*2*5)^{50} 
= {(2^4)^{50}*5^{50} 
= (2^{200} )*5^{50}

so, 2^{200} shows that the greatest value of p. The correct answer is E.

generis · Answer

(ab)^{n} = a^{n}b^{n} (a^{m})^{n}=a^{(m*n)} Find prime factors of 80, raise to 50th power 80 = (2*2*2*2*5) = (2^4*5) (2^4*5)^{50}= (2^4)^{50}*5^{50}= 2^{200}5^{50} 5 to any power ends in 5. NEVER divisible by 2. The other factor of (80)^{50}=2^{200} (80)^{50}= (2^{200}*5^{50})=(2^{p}*5^{50}) 2^{p}=2^{200} p=200 There are 200 powers of 2 in (80)^{50} Answer E selim , I think there is a small typo in your answer: 5^{200}

ScottTargetTestPrep · Answer

Solution:

Since 80^50 = (16 x 5)^50 = (2^4 x 5)^50 = 2^200 x 5^50, the largest value of p such that 2^p is a factor of 80^50 is 200.

Answer: E

kaylaquijas · Answer

I got to the answer a little differently. Let's call 80^50 8^50. We know that the pattern of the units digits for 8 to any power is 8,4,2,6. Therefore, 8^50 will have a units digit of 4, meaning it is divisible by 4 (and 2, of course). The largest option available that is divisible by 4 is 200, thus 200 is the answer. Does this work?

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What is the greatest value of the positive integer p such that 2^p is

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What is the greatest value of the positive integer p such that 2^p is

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