71Ohh.. Yes!
1. 1+22+333+4444+......... upto n terms
2. Find the sum of Products of 1st n natural numbers taken 2 at a time.
3. Show that 49,4489,44489,........, the no's obtained by inserting 48 in the middle of the subsequent terms are square of an ODD Integer.
4. Prove that (666....... n terms)2 + (888......n terms) = (444...... 2n terms)
5. MISCELLANEOUS:
A.) Find the remainder when 750 is divided by 342.
B.) a^{\sqrt{log_{a}b}}- b^{\sqrt{log_{b}a}}
C.) Find the no. Of solution sin4x+cos2x = 1
D.) No of solutions of (x-2)2 + (x-5)2 + (x+3)2 = 0
E.) How many naughts are present at the end of product of all whole numbers from 1 to 100.
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19 Answers
71@ Joydoot 24 is the correct answer. But please Elaborate! and also All the sums are correct.
i mean isnt the the power of 7 a multiple of 3..
But how is it so?
@Nishant Sir
2nd one is simply [(1+2+3....+n)^2-(1^2+2^2+....+n^2)]/2
Yes Sir! Solved and got! But the main problem is that how to think of such things. I'm really troubled in that. People just solve by newer tricks and ides but how to develop them. Someone suggest.
One SUm More:
If a+ \frac{1}{b} = b+\frac{1}{c}=c+\frac{1}{a} = x.
Prove that abc = x
11If you dont like learning method, then make your own, it will be helpful for you. But in order to derive something new you ought to learn the old methods to know the basics
Well I think there would be some method for that Vivek, you do one thing start to find out those methods or if you can't then make yours and share with us [1]
71May be. I don't know that. That's what I was thinking too. But there must be something.Learning method's I don't like.
11E.) How many naughts are present at the end of product of all whole numbers from 1 to 100.
the product of (even number) and 5 results in "0" of unit's
place.
no. of 5's from 1 to 100 = 100/5 = 20
but in 25, 50, 75 and 100, there is one more 5
25 = 5*5*1
50 = 5*5*2
75 = 5*5*3
100= 5*5*4
four 5's extra
therefore no. 0f naughts = 20+4 = 24
Or
Any Number N you want to find the number of zeros in
product expand it in Standard form say for example
(22*25*20*11)=(2*11)*(5*5)*(2*5*2)*(11)
No of 2's= 3
No of 5's= 3 as 2*5=10 so there are 3 combinations of 2*5
so totaly 3 zeros.
For Fafctorial just use this 100! = [100/5]+[100/25]=20+4=24
115) E) The no of zeros can be determined by the no of no. of 5s and the no of 2s as 5*2 gives 10.
there are 2s in every second no.
like 2,4,6,8....
but there are 5s after every 5 no....
therefore fr getting the no of zeros, we have to only find the no of 5 in 100! ie, the product frm 1 to 100.
for the no of 5 from 1 to 100... it is 100/5=20
but 25 = 52 again 50 = 2*52 again 75 = 3*52 and 100 = 4*52
thus the no of 25s in 100!... it is 100/25 = 4
the total no. of fives = 20 + 4 = 24.
the total no. of zeros is thus again 24.
341I get x =-abc. You can check my working
We have x = \frac{ab+1}{b} = \frac{ac+1}{a} = \frac{a(b-c)}{b-a}.......1
Also a + \frac{1}{b} = b + \frac{1}{c} \Rightarrow (a-b) = \frac{b-c}{bc}
\Rightarrow abc = \frac{a(b-c)}{a-b}......2
From 1 and 2 we get x = -abc
edit: this is a prob from INMO '97. with required conditions a≠b≠c http://www.artofproblemsolving.com/Forum/viewtopic.php?p=341948&sid=5962917334680e42e0f3bc3f7ba814c9#p341948
231+22+333+4444+....=1+2(11)+3(111)+4(1111)+...n(11111..n)
=19(9+2(99)+3(999)+4(9999)+.....n(9999....n) )
=19(10-1+2(100-1)+3(1000-1)+4(10000-1)+.....)
=19(10+2(102)+3(103)...n(10n) -1- 2 - 3 - 4 ...-n )
=19 (-1-2-3-4-5.... - n) + 19(10+2(102)+3(103)...n(10n))
do first part urself
2ndpart:
S=10+ 2(102)+3(103)+4(104)+...n(10n)
10S= 102 +2(103)+3(104).....n(10n+1)
subtract S - 10S
so -9S = 10 +102+103+104+....10n - n(10n+1)
so S = 19[n(10n+1) - 10(10n - 1 )9]
11vivek, is the sum 5 a) correct...
i mean isnt the the power of 7 a multiple of 3..
if it was so... we could have written it as
73n = 343n = (342+1)n... where n = a positive imteger.
and then applying expansion we can easily do it...
115.
e) does naughts means zeros...
if it is so...
is the answer 24..?
115.
d)
as the sum of the squares are given to be zero...
the individual squared terms should be zero...
because a squared term gives a minimum value of 0... and can take no negetive values...
therefore...
(x-2)2 =0
and (x-5)2 =0
and (x+3)2 =0
then x=2,5 and -3.
234th is also almost same
36(11111..n)2+8(111..n) = 4(111..2n)
3681(999...n)2+89(999...n)=49(999...2n)
3681(10n-1)2+89(10n-1)=49(102n-1)
i.e 36(a-1)2+72(a-1)=36(a2-1)
wich is true
for Q5 C ,
sin4x-sin2x=0
so sin2x(sin2x-1) = 0
now u can solve
for 5.D there will be no real solutions since square of a no. is ≥ 0
so here all terms will have to be zero simultaneously i.e x = 2, x=3 , x=-5 , wich cant be true simultaneously
62Q 5b)
a^{\sqrt{log_ab}}=b^{log_ba\times\sqrt{log_ab}}=b^{\sqrt{log_ba}}
Hence that one is zero..
622nd one is simply [(1+2+3....+n)^2-(1^2+2^2+....+n^2)]/2
Think of the expansion of (a+b+c)^2 and that you have to find ab+ac+bc