For every positive even integer n, the function h(n)

[RonPurewal]

Hi ,

I agree with the analogy. But greater than 40 also includes numbers that are not greater than 50 and those numbers will not satisfy the condition.
Hence the question regarding choice E.

It is saying that if eqn X satisfies x> 15 it also satisfies x>10. I think that is incorrect assumption.

Please let me know your thoughts

irrelevant.

here's another example:
if N is a positive integer greater than 1, then N! is
(a) even
(b) odd

the answer here is (a), because all of these factorials are even.
your objection, in this case, would be like saying "But (a) is wrong because there are even numbers that aren't factorials!" that's not the point -- those would be irrelevant. the problem is just saying that all the factorials are even, not the other way around.

more generally, "all X's are Y's" does not imply, nor does it require, that all Y's also be X's.

[chandrahasreddy]

Mgmat staff - Very good explanation. Thank you.

Can we generalize (sort off a takeaway ..):

For all positive integers of N, (N!+1) will have factors"  > N.

The factors"  of (N! + 1) will be above N

"  - factors other than 1

[RonPurewal]

Mgmat staff - Very good explanation. Thank you.

Can we generalize (sort off a takeaway ..):

For all positive integers of N, (N!+1) will have factors"  > N.

The factors"  of (N! + 1) will be above N

"  - factors other than 1

yes, except "factors" shouldn't be plural. it should be "at least one factor".
in some cases there will be only one such factor, e.g., 3! + 1 = 7, which only has one factor greater than 3 (namely, 7 itself).

[rkafc81]

Hi ,

I’m having a lot of trouble understand the bit about + 1 and that yielding a remainder of 1? I understand everything else, just not that (in bold...):

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Thanks!

[RonPurewal]

Hi ,

I’m having a lot of trouble understand the bit about + 1 and that yielding a remainder of 1? I understand everything else, just not that (in bold...):

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Thanks!

the best way to figure this kind of thing out is to experiment with it yourself.

* think of an integer (2 or greater).
* think of a random multiple of that integer (or, even better, several such multiples).
* add 1 to that multiple (or to all of them, if you picked more than one).
* now, divide by your original integer, and find the remainder. it will always be 1.

e.g., let's say i think of 5.
here are some multiples of 5: 15, 55, 100, 290.
adding 1 to these gives 16, 66, 101, 291.
if i divide these by 5, each one gives a remainder of 1.

that should help you understand the comment that you quoted.

[rkafc81]

ok that's great, thanks Ron!

[tim]

:)

[pooj306]

How about the numbers 14 and 15? The greatest prime of 14 is 7, but the least prime of 15 is 3 which is less than 7.
I understand the concept here, but I don't understand why it doesn't work for all consecutive numbers, and not just factorial+1 numbers

[RonPurewal]

How about the numbers 14 and 15? The greatest prime of 14 is 7, but the least prime of 15 is 3 which is less than 7.
I understand the concept here, but I don't understand why it doesn't work for all consecutive numbers, and not just factorial+1 numbers

What concept/rule/principle are you discussing? I honestly can't tell. Please clarify, thanks.

[pooj306]

The concept explained for this problem is:

For any number n!+1, the greatest prime of n! is always less than the least prime of n!+1

Shouldn't the above rule be applicable to ANY two consecutive numbers? When I tried applying the rule to the consecutive numbers, 14 and 15, I realized that the greatest prime of 14 (which is 7) is GREATER than the least prime of 15 (which is 5).

Is there something I have missed here? Does the rule only work for n!+1 numbers and not really for consecutive numbers?

[RonPurewal]

The concept explained for this problem is:

For any number n!+1, the greatest prime of n! is always less than the least prime of n!+1

oh... hmm
if you are trying to memorize "rules" that look like this, then, yeah, this exam is going to seem a lot harder than it really is.

this is like trying to memorize the following: "if you take step n with the left foot, then you'll take step (n + k) with the left foot if k is even, and with the right if k is odd".
i mean, yeah, that's true, and, sure, you could memorize it -- but for heaven's sake why would you want to? if you just have the basic concepts of even and odd numbers (and a basic understanding of how walking works), then it's clearly pointless to try to remember such an arcane (and overly specific) formulation.

same thing here... this fact is a fairly direct consequence of how factorials work.
namely,
* n! contains all the numbers from 2 through n. therefore, n + 1 can't contain any of those numbers.
* all of the prime factors of n! are between 2 and n (since you can break it into that product).
* none of those primes are in n! + 1, so the smallest prime factor of that number must be bigger.

Shouldn't the above rule be applicable to ANY two consecutive numbers?

no, definitely not.
think about it for a sec: if this "rule" applied to every possible pair of numbers, then no two different numbers could share any prime factor!

[pooj306]

Thanks for explaining Ron.

I will keep in mind to not apply "rules" to every situation!

[jlucero]

Glad it helped.

[chetan86]

Hi Ron,

I am trying to understand how these concepts will be applicable.

In the question, if we had h(40)+1 rather than h(100)+1, then calculation would have been as follows?

h(40)+1 = 2^40*40!+1

As 2^40*40! and 2^40*40!+1 are consecutive number, they are co-prime and will no share any prime factor.

So the smallest prime factor for h(40)+1 would be greater than 40.
Am I correct?

Thanks!!!

[RonPurewal]

Hi Ron,

I am trying to understand how these concepts will be applicable.

In the question, if we had h(40)+1 rather than h(100)+1, then calculation would have been as follows?

h(40)+1 = 2^40*40!+1

As 2^40*40! and 2^40*40!+1 are consecutive number, they are co-prime and will no share any prime factor.

So the smallest prime factor for h(40)+1 would be greater than 40.
Am I correct?

Thanks!!!

not quite right.

don't skip steps!
write stuff out!

using the definition given here, h(40) is
2 • 4 • 6 • ... • 38 • 40
which is
(2•1) • (2•2) • (2•3) • ... • (2•19) • (2•20)

so, as you can see, this is actually 20! • 2^20. your 40's should actually be 20's.

(incidentally, that last sentence is also pretty good advice for how to live once you turn 40.)