33) i think it must be 13 after that 11. :)
2^15 * 3^6 * 5^3 * 7^2 *11 * 13.
this is 16!.
the rest two involve some modular algebra??
\hspace{-16}(1)\;\; $If $\mathbf{19!=1216451\underline{a}0408832000}$. Then $\mathbf{a=}$\\\\ (2)\;\; If $\mathbf{34!=95232799\underline{c}\;\underline{d}96041408476186096435\underline{a}\;\underline{b}000000}$\\\\ Then $\mathbf{(a,b,c,d)}$ is\\\\ (3)\;\; If $\mathbf{n!=2^{15}.3^{6}.5^3.7^2.11.3},$ Then $\mathbf{n=}$
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7 Answers
1057i think 1st one can be done like this....
since 19! is divisible by 11...sum of digits at odd places shud be equal or differ by a mutiple of 11 to the sum of digits at even places...
thus i got a as 0....
3
2621) 19! is divisible by eleven.
thus (sum of even placed digits ) - (sum of odd placed digits) =11k (k=integer)
thus (28+a) -(17) =11k
a=11(k-1)
since 0≤a≤9
thus a=0
1057for the 3rd one....
since the power of 5 is 3 the number is between 15-19...
but since 17 is not a factor it has to be 15 or 16....but since power of 2 is 15 the number is 16....
1057for the 2nd one..
since 34! has seven zeroes at the end....b is 0...
calculating the last digit of prime factors and multiplying a=2
1057the value of 34! is wrong.....u missed a 2 at the beginning.... :P
3if a = 2. c = 0. d = 3. since 34! is divisible by 9 and 11.
it is a british olympiad question. :) i solved it in PEN.. 95232799039604140847618609643520000000 this is the value.
