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Mathematics
General Math
Factorials and Exponent Challenge
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[QUOTE="kaliprasad, post: 6780397, member: 704062"] [SPOILER="my attempted solution"] Without loss of generalty we can have $a>=b>=c$. Now we get a multiple of power of 2 only when we add multiples of same power of 2 So $a!$ and $b!+c!$ should be muiltiple of same power of 2. and when we add the 2 we shall get multiple of power of 2 say of the form $m2^x$. if m is power of 2 then we are done. Now b and c should be multiple of same power of 2 and when we add the same we get a multiple of higher power and further this should be same as multiple of power of 2 of a. So we have 2 cases to check $b = c$ and $a = 2$ or 3 (for reason please see below )- And $ b = c + 1$ then any a. As c devides a!+b!+c! so c can not be greater than 2 as sum is power of 2 Now Put the values b= 1, c = 1 giving a = 2 or 3 as a =4 gives a! divsible by 4 but b!+c! is not a =2 gives c = 2 a =3 gives c = 3 c= 2, b= 3 gives b! + c! = 8 so we need to check for a = 4 and 5 only as a = 6 or above a! is divisible by 16 so it is not possible a = 4 gives 32 power of 2 so n = 5 so solution (4,3,2,7) a = 5 gives 128 power of 2 so n = 7 so solution (5,3,2,7) so solution set $(2,1,1,2), (3,1,1,3), (4,3,2,7),(4,3,2,7)$ and any permutation of 1st 3 numbers is each set [/SPOILER] [/QUOTE]
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Factorials and Exponent Challenge
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