A Nice Factorial Math Olympiad| Factorial Simplification|

This video is about factorial math . Here we need to find the value of n . The value of n is 4 . This is a real value . Math factorial problems
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Nice Factorial Math Olympiad| Factorial Simplification| nice factorial problems
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A Nice factorial problems
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Maths Olympiads are held all around the world to recognize students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications,
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[@user-ig1gu9dj6t]

After dividing we have (n+2)(n+3)(n+4)(n+5)=3024 four consecutive numbers. None of said numbers can be multiple of 5 as 3024 doesn't end in 5 or 0. Method 1 is simply estimate the 4th root of 3024. Since 3024 is less than 3600 the square root is going to be less than 60 and that means the 4th root is less than 10. since 1*2*3*4 is too small, that leaves 6*7*8*9 method 2 is using fundiamental theorem of arithmatic 3024=2*2*2*2*3*3*7 giving the 4 consecutive numbers of 6*7*8*9 n+2=6 n=4.

[@diamantnt]

i cant see nothing wrong, very good video

[@Nrlogic1]

Thanks stay connected

[@xyzmanizator]

Step 1: rewrite the equation as (n+2)(n+3)(n+4)(n+5)=3024 Step 2: by definition, n is positive integer, therefore all the brackets are positive integers. Step 3: it's easy to prove that for every positive x and y=x+1, (x+2)(x+3)(x+4)(x+5) < (y+2)(y+3)(y+4)(y+5), therefore f(n)=(n+2)(n+3)(n+4)(n+5) monotonously increases for all positive N, meaning there is at most one solution for the equation Step 4: At this point simply showing that n=4 is a valid solution for (n+2)(n+3)(n+4)(n+5)=3024 means that it's the only valid solution and the original problem is done. But it's easy to show that since 3024 isn't divisible by 5, its factors (in brackets) shouldn't be divisibly by 5 either, thus all valid n would look like n=5k-1, k being natural number. So 4 would be the first potential valid solution.

[@Nrlogic1]

Good work

[@57thorns]

I was going the factorization route: 3024 = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 7 Fourth rote of 3024 is about 7. So one of (n+2)(n+3)(n+4)(n+5) is 7. We also know that none of them is 5 or 10 as there is no factor 5 in the 3024. So the only possible sequence would be 6*7*8*9 which is indeed a factorization of 3024. so n=4

[@Nrlogic1]

Good

[@ceejay0137]

Yes, much easier than the algebraic method! Simply inspecting the prime factors of 3024 showed that you can make 6, 7, 8 and 9 by combining groups of them (obviously the 7 is on its own). Therefore the (n+2) term has to be the 6, thus n = 4.