If n is a positive integer and the product of all integers..

[guest]

If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, wht is the least possible value of n?

The answer is 11 but I cant figure it out --- can someone please explain? THanks

[tmmyc]

The question states the product of 1 to n inclusive is a multiple of 990. This means this product will be 990 multiplied by "other stuff".

First, find the prime factors of 990: 2 3 3 5 11
From here, we know that n must be at least 11.

Let's assume n is 11.
Product of all integers from 1 to n where n = 11: 1*2*3*4*5*6*7*8*9*10*11

Here we see that 990's prime factors are all covered in the product above: 2, 3, 5, and 11 are present. The extra 3 can be taken from the 6. The rest of the product is the "other stuff".
Hence 11 is the least possible value of n.

[StaceyKoprince]

Very nice explanation, tmmyc!

[LP1]

Iam not quite sure I understand. is there an alternate explanation/approach? Also, why should n be atleast 11?

thanx much

[sinhavis]

any number will be multiple of 990 if it is multiple of 11 and 10 and 9 (11*10*9=990).
Given that the number in question is product of 1...n. So 'n' has to be atleast 11.

[RonPurewal]

Iam not quite sure I understand. is there an alternate explanation/approach?

not really one that's practicable, no.

if you are INSANELY LIGHTNING FAST at arithmetic, you may be able to actually multiply out these numbers and perform long division.
1 x 2 x 3 x ... x 10 x 11 = 39,916,800
divided by 990 = 40,320
you could do this within the two-minute time limit, if your initial guess was close enough to eleven.

--

barring this approach, there is no way, none at all, to get around finding the prime factorization.

and in any case, you SHOULD NOT be looking for ways around prime factorizations!
the prime factorization is the single most fundamental concept in Number Properties. if you're trying to avoid using it, you're going to have some serious problems.

Also, why should n be atleast 11?

11 is the biggest prime factor of 990. you'll see this when you break 990 down into primes (= 3 x 3 x 11 x 5 x 2).

therefore, you have to have an '11' somewhere in your product. since 11 is prime, you're not going to get it from a product of numbers that are smaller than 11.

[LP1]

thank u guys..now i get it.

[JonathanSchneider]

: )

[gkhan]

Iam not quite sure I understand. is there an alternate explanation/approach? Also, why should n be atleast 11?

thanx much

Notice that any positive integer factorial greater than 4! will yield divisibility of 10.

Working from top down, assume n=990 divide by 11

Irrelevant to know what n actually is since each multiple of any number, n, has the same number of distinct prime factors).

[RonPurewal]

Working from top down, assume n=990 divide by 11

i'm not sure what this is supposed to mean. could you explain further?

--

Irrelevant to know what n actually is since each multiple of any number, n, has the same number of distinct prime factors).

absolutely not true.

for instance, n = 10 has two distinct prime factors, but 30 (which is a multiple of 10) has three distinct prime factors. and 210 has four distinct prime factors. etc. etc.

what does "distinct prime factors" have to do with this problem, anyway? are you sure you're posting in the correct thread?

[gkhan]

Working from top down, assume n=990 divide by 11

i'm not sure what this is supposed to mean. could you explain further?

Irrelevant to know what n actually is since each multiple of any number, n, has the same number of distinct prime factors).

absolutely not true.

For instance, n = 10 has two distinct prime factors, but 30 (which is a multiple of 10) has three distinct prime factors. and 210 has four distinct prime factors. etc. etc.

what does "distinct prime factors" have to do with this problem, anyway? are you sure you're posting in the correct thread?

To clarify...

The stem asks for the least so checking the divisibility of the first answer choice, 10 seems smart to try because 5! and up are divisible by 10, BUT since the product of all integers from 1 to n has to be divisible by 990, 1980, 2,970 one would realize that it must one of the prime factors of 990.

I meant original distinct integers. I understand your instance given,

n=10 2*5 (the original distinct primes:2,5)
n=30 2*5*3 (remain 2,5)
n=210 2*5*3*7 (2,5)

Ron you are very right, its very strange I was given a problem very close to this one with a slight twist. My fault. I tried to edit my post so I don't confuse but it wouldn't allow such.
--

[RonPurewal]

\To clarify...

The stem asks for the least so checking the divisibility of the first answer choice, 10 seems smart to try because 5! and up are divisible by 10, BUT since the product of all integers from 1 to n has to be divisible by 990, 1980, 2,970 one would realize that it must one of the prime factors of 990.

ok.

I meant original distinct integers. I understand your instance given,

n=10 2*5 (the original distinct primes:2,5)
n=30 2*5*3 (remain 2,5)
n=210 2*5*3*7 (2,5)

i'm still not following this part -- there's no such thing as "original distinct integers".
it looks as though you're basically saying:
1) multiply by a number n
2) then, completely ignore n, and just find the number of primes in the number you started with!

i think i must be missing something here -- but there is definitely no distinction between "original" distinct integers and other distinct integers.

[madhan.naidu]

The L.C.M of 990 is

2 L990
5 L495
11 L99
9

:. the least possible number is 11.

[jnelson0612]

madhan, that's certainly one way to look at it. Thanks!

[metman82]

990 = 9 * 110 = 9 * 2 * 55 = 9 * 2 * 5 * 11
No other primefactors.