Five consecutive positive integers are chosen at random

stmt1: third of the five integers is a prime number.
If the middle number is 7 then the largest number 9 which gives reminder 1
If the middle number is 5 then the largest number 7 which gives reminder 1

Insufficient

stmt2: the second of the integers is the square of an integer. so it must be divisible by 4 bcause the 2nd number is even number as per the question.
so the largest number will be 4*2 + 3.

Sufficient.

Ans :B

I think u did some typo hen explaining statement 1.
If the middle number is 5 then the largest number 7 which gives reminder 3

Can u pls elaborate stmt 2 . I dint get that

Hi Bunuel

could you please explain how we get 4k+3 here i am unable to understandho we get 3 there.

The second of the five consecutive integers is 4k;
The third of the five consecutive integers is 4k+1;
The fourth of the five consecutive integers is 4k+2;
The fifth of the five consecutive integers is 4k+3.

Hope it's clear.

Hi. What am I missing to assume in the 2nd Statement that the 2nd integer could also be odd: 9, 25, 49, etc.?

Hi,

Here are my two cents for this question

We are told that the average of 5 consecutive number say A,B,C,D,E is odd, which means C is odd . So our Sequence begins with Odd Number , ( O E O E O) this implies A, C ,E are odd and B &D are even

And any Odd number divided by 4 will give remainder as 1 or 3. So we need to know in which for is E is it 4k+1 or 4k+3

Statement 1 tells us that C is prime number , again this means C can be any prime number >2. But again we have E can be any odd number
So Not Sufficient

Statement 2: Tells us that its a square of a integer. from question stem We know that B is even and is square of even number which means B is in the form of 4k.
Since C,D,E are consecutive numbers we have C=4k+1, D=4k+2, E=4k+3

This statement tells us in which form E would be and hence we can answer the question .

Hence Statement 2 is sufficient.

Probus

Dear Bunuel

It is given in question stem
That the average of five integers is odd. We are only meeting this criteria when we are starting this first number with 0

Kindly clarify

Regards

Posted from my mobile device

There are two examples given in the solution you quote in which the average is odd and neither of them starts with 0. If the first integer in the sequence of five consecutive integers is 0, then the average is 2, so even: {0, 1, 2, 3, 4}.

The average of five consecutive integers to be odd the sequence must start with odd number. For example, {7, 8, 9, 10, 11}. The average = 9 = odd.

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