3for the second qn u have to find the no of zeros at the end of 1000!
Q1)Show that there can be infinite number of sets of four distinct real numbers in AP which are such that the square of the last term is equal to the sum of squares of the first three terms.
Q2)1000! is divisible by 10n.Find the largest positive integral value of n.
-
UP 0 DOWN 0 0 7
7 Answers
11) assume a, a+d, a+2d, a+3d, add up the squares of d 1st 3, equate to dat of the last to see there exist 2 d's for every real a, so infinite set of numbers.
2) we need exponent of 10 in 1000!
exponent of 2 in 1000!= [1000/2]+[1000/4]+[1000/8]+...
dat of 5 in 1000!=[1000/5]+[1000/25]+[1000/125]+[1000/625]=249
exp of 2 in 1000!>exp of 5 in 1000!, so exp of 10 in 1000! is 249
cheers!!
62a2 + (a+d)2 + (a+2d)2 = (a+3d)2
Thus, a2=2d2
Thus, for any value of a we have a value of d given by d=a/√2 for which the AP is formed satisfying the criterion above.
We can chose d in infinitely many ways. (Every real number)
Hence we get an AP for each value of d
62[url=http://targetiit.com/iit_jee_forum/posts/9th_january_2009_1549.html]Highest power of 3 in 100![/url]
Read this link above for the 2nd part and try to understand and find out..