consider the set = {1,2,3......1000}. How many AP can be formed from the element of S that start with 1 andend with 1000 and have at least 3 elements. a) 3 b) 4 c)6 d)7 Q2. Let M be the set of all the distinct factors of the number N=6^5*5^2*10, which are perfect squares.find the product of the elements contained in the set M Q3. There are two clocks .One of them gains 2min in 12 hrs and the another loses 2 min in 36 hrs .both are set right at 12 noon on tuesday .What will be the correct time when both of them show the same time for the next time? |
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Number of factors of 999 = (2 + 1) x (1 + 1) = 6 = Possible number of common differences
But I don’t think that is the answer as it includes the factor 1 (1, 3 , 9, 111, 333, 999)
And in that case the number of terms in the AP ‘ll be 2.
So the answer should be 5 but unfortunately its not among the available options.
1000 – 1 = 999
999 = 9 x 111
= 3^3 x 37
So no. of factors are (3+1)(1+1)= 8
but ignoring factor '1'...
so there will be 7 such APs
regards,
Shaheen
oops!!!
Such a blunder
I wrote 111 but was thinking it as 101 a prime number
Thanks buddy
thx Anita...try the pbms i hv posted in another thread....
regards,
Shaheen
kindly xplain ur funda guys
pls guys,
explain how u got "(3+1) (1+1)"
can u pls explain
yes..plzz explain
there ir no point repeatin th same thing again n again ....if ny1 has an idea they will obviously contribute when they come online...b patient
nyways i got some idea
they r calculatin total no. of common diff. possible.
1000-1=999
=3^3*37^1
total no. of factors =4*2=8
Hey Thanks for the solution...
Tell me is this the best way to solve these type of probs???
and wats the logic behind taking out factors for an AP problem???
Hi guys, Posting a detailed approach......
Let's start small in this case. In place of 1 000 numbers, let me take the first 10 natural numbers. Notice that two consecutive numbers have a common difference of 1 between them. If you take this common difference as an interval, you have 9 such intervals.
Now I choose 3 or more numbers in AP with first and last terms being 1 and 10, respectively, by putting partitions, as shown. The partitions will have to be put such that there are equal numbers of intervals between any two consecutive partitions (since the numbers are to be in AP, equal number of intervals mean same common difference).
In essence we are trying to find the number of ways of dividing 9 intervals in equal parts. This will be nothing but the number of divisors of 9 i.e. 1, 3, and 9. Since we cannot take 1, we can only take 3 and 9.
Similarly, we are trying to find the number ways of dividing 999 intervals equally, or the number of divisors of 999. Now 999 = 33 × 37 therefore number of divisors = 4 × 2 = 8. And not counting 1, the number of ways = 7.
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