99 consecutive composite natural numbers

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Zelda22

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99 consecutive composite natural numbers.

a. What is the smaller number on this set?

99!+1
99!+2
100!+1
100!+2
101!+1
101!+2

b. What is the largest number on this set?

99!+99
100!+100
100!+101
101!+101
101!+102

I think the smaller is 100!+2 and the largest is 100!+100
(n+1)!+2,(n+1)!+3,⋯,(n+1)!+n+1.

Is this correct? Thanks
 
Steven G said:
Why do you think those are the correct answers?
Click to expand...
(n+1)!+2 is divisible by 2
(n+1)!+3 is divisible by 3
(n+1)!+4 is divisible by 4 and so on.

(99+1)!+2,(99+1)!+3,⋯,(99+1)!+99+1
100!+2 smaller
100!+100 largest

Is it wrong?
 
I would think that the smallest composite number is 4.
The 99th composite number will surely be less than 200 (there are 99 even composite number), which rules out 100! + 100.

If you think that my answers are wrong then please state the question differently.
 
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Steven G said:
I would think that the smallest composite number is 3.
The 99th composite number will surely be less than 200 (there are 99 even composite number), which rules out 100! + 100.

If you think that my answers are wrong then please state the question differently.
Click to expand...
I need to select one answer from the options provided, and the 99 composite numbers need to be consecutive.
 
4, 6, 8,10,12,14,15,16,18, 20, 21, 22, 24, 25 ,....

You are to pick one of the choices if in fact one of them is true.

I started with the smallest composite number, which is 4. Are you really going to debate this?
Now continue with the list I started until you get to the the 99th number.

Are you sure that you stated the exact problem?
 
@Steven G

I think that we have another badly worded question. I suspect that the intended question is along the lines of

What is the smallest number possible in a set of 99 consecutive integers, all of which are composite?

Moreover, it does not look as though the original question was written in English so asking for an image of the question may not help.
 
JeffM said:
@Steven G

I think that we have another badly worded question. I suspect that the intended question is along the lines of

What is the smallest number possible in a set of 99 consecutive integers, all of which are composite?
Click to expand...
@JeffM
I know what the problem should be. I just want the OP to realize that it is poorly written and that just because there are choices doesn't always mean that one of them is correct.
I do appreciate you trying to clarify the problem for me.
 
@Cubist and @Zelda22

My problem is that even if I understand the problem correctly, it seems insanely difficult.

I can see that the numbers running from 99! + 2 up through 99! + 99 are all composite and that there are only 98 such numbers, but it is not obvious to me that there is no integer > 99 that cannot divide 99! + 1. There are quite a few candidates available.

I can also see that the numbers running from 100! + 2 through 100! + 100 are all composite and that there are indeed 99 of them.

But I do not see how to answer the question without determining whether 99! + 1 is composite.

Furthermore, it is not obvious, at least not to me, that no set of 99 consecutive composite integers contains a number less than 99! + 1.

So I think Zelda has done the best she could to deal with an atrocious problem. The set of consecutive integers starting with 100! + 2 and continuing through 100! + 100 does contain 99 members that are all composite numbers. Whether there is a set of such numbers that includes a number less than 99! + 2 seems hard.
 
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Cubist said:
The problem is poorly worded. But since we know the intent...
What does n represent? (What if n=2)
Click to expand...
n clearly represents the number of required integers in the set

JeffM said:
@Cubist and @Zelda22

My problem is that even if I understand the problem correctly, it seems insanely difficult.

I can see that the numbers running from 99! + 2 up through 99! + 99 are all composite and that there are only 98 such numbers, but it is not obvious to me that there is no integer > 99 that cannot divide 99! + 1. There are quite a few candidates available.

I can also see that the numbers running from 100! + 2 through 100! + 100 are all composite and that there are indeed 99 of them.

But I do not see how to answer the question without determining whether 99! + 1 is composite.

Furthermore, it is not obvious, at least not to me, that no set of 99 consecutive composite integers contains a number less than 99! + 1.

So I think Zelda has done the best she could to deal with an atrocious problem. The set of consecutive integers starting with 100! + 2 and continuing through 100! + 100 does contain 99 members that are all composite numbers. Whether there is a set of such numbers that includes a number less than 99! + 2 seems hard.
Click to expand...
I think this method just guarantees a set of consecutive integers will all be composite. But it doesn't say this is the lowest in value. It seems very likely that much smaller solutions will exist. I think we have to choose from the available options only.

In fact 99!+100 (click) is divisible by 100 therefore 99!+2 starts a lower valid set with 99 elements, but I suspect this isn't what the question's author intended.

EDIT: 99!+1 is composite too (click)
 
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Cubist said:
I think this method just guarantees a set of consecutive integers will all be composite. But it doesn't say this is the lowest in value. It seems very likely that much smaller solutions will exist. I think we have to choose from the available options only.

In fact 99!+100 (click) is divisible by 100 therefore 99!+2 starts a lower valid set with 99 elements, but I suspect this isn't what the question's author intended.

EDIT: 99!+1 is composite too (click)
Click to expand...
ROFL

What a terrible problem.

And I cannot believe that I missed that 99! + 100 is a composite number. Clearly, 4 and 25 are factors of both 99! amd 100.
 
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