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An integer p is chosen at random from the set of first 100 nonnegative

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PyjamaScientist
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An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   19 Nov 2021, 04:07
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An integer \(p\) is chosen at random from the set of first \(100\) nonnegative integers. What is the probability that \(p(p+2)\) is divisible by 4?

A. \(\frac{1}{4}\)

B. \(\frac{49}{100}\)

C. \(\frac{1}{2}\)

D. \(\frac{51}{100}\)

E. \(\frac{2}{3}\)
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Re: An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   19 Nov 2021, 08:04
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PyjamaScientist wrote:
An integer \(p\) is chosen at random from the set of first \(100\) nonnegative integers. What is the probability that \(p(p+2)\) is divisible by 4?

A. \(\frac{1}{4}\)

B. \(\frac{49}{100}\)

C. \(\frac{1}{2}\)

D. \(\frac{51}{100}\)

E. \(\frac{2}{3}\)



There are two case:
1) p is odd: There are 50 such case in each case the answer is no, as the product p(p+2) will be odd
2) p is even: Again 50 cases. p(p+2) means product of two consecutive even numbers. The product of two consecutive even numbers is not only multiple of 4 but of 8.
Let p=2k, so p(p+2) =2k(2k+2)=4k(k+1). As k(k+1) is also even, p(p+2) becomes product of 4*2.

So half of the cases give us answer YES. => P=1/2


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An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   19 Nov 2021, 08:15
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First 100 non negative numbers would be 0 to 99.

Now p(p+2) pairs divisible by 4 would be (0,2), (2,4), (4,6), (6,8)........(98,100)

Hence we would have 50 such pairs and hence 50 values of p

IMO the answer should be 50/100 = 1/2

Option C
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Re: An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   19 Nov 2021, 08:18
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If p is even, p(p+2) will be the product of two consecutive even integers, so not only will it be divisible by 4, it will actually be divisible by at least 8 (since if you look at any two consecutive even integers, one of them is always a multiple of 4). If p is odd, clearly p(p+2) isn't divisible even by 2, so the question is just: what is the probability p will be even if we pick p randomly from a set of 100 consecutive integers, and of course in any such set 50 numbers will be even, and 50 will be odd, so the answer is 1/2.
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Re: An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   25 Feb 2023, 01:28
total range of numbers is 0 to 99
we have 50 each of odd and even set
for \(p(p+2)\) to be divisible by 4
p has to be even which is 50
50/100 ; 1/2
option C


PyjamaScientist wrote:
An integer \(p\) is chosen at random from the set of first \(100\) nonnegative integers. What is the probability that \(p(p+2)\) is divisible by 4?

A. \(\frac{1}{4}\)

B. \(\frac{49}{100}\)

C. \(\frac{1}{2}\)

D. \(\frac{51}{100}\)

E. \(\frac{2}{3}\)
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Re: An integer p is chosen at random from the set of first 100 nonnegative [#permalink] An integer p is chosen at random from the set of first 100 nonnegative   19 May 2024, 07:28
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Re: An integer p is chosen at random from the set of first 100 nonnegative [#permalink]
19 May 2024, 07:28
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