[hi.shivani]

Originally Posted by subu_222412

Can please elaborate

I don't have any idea what the correct answer is but anyways lemme elaborate what I have done.

1^(1/2)=1
and (100)^(1/2)=10


Square root of 1=1
Square root of 2,3 i and 4 s 1.414 and 1.73 and 2 .So,{}=2 .
{}square root of no. from 5 to 9 will be 3
From 10 to 16 will be 4
17-25=5
26-36=6
37-49=7
50-65=8
65-81=9
90-100=10
So total sum=1+2*3+3*5+4*7+9*5+11*6+13*7+15*8+17*9+19*10=71 5

[sourabhforu]

Hi,
I came across an interesting problem while surfing the internet:-

Let a,b and c be the positive real numbers. Determine the minimum value of (a+3c)/(a+2b+c) + 4b/(a+b+2c) -8c/(a+b+3c)
a)12√2-17
b)6√2-17
c)12√2+17
d)6√2+17


I saw the answer to it as:-

Let x = a +2b+c, y = a+b+2c and z=a+b+3c
z-y = c and x - y = b - c => x -y = b-(z-y),
or b = x+z-2y
We note that a+3c=2y-x. By the AM-GM Inequality, it follows that
(a+3c)/(a+2b+c) + 4b/(a+b+2c) -8c/(a+b+3c) = (2y-x)/x + 4(x+z-2y)/y - 8(z-y)/z
=-17 + 2y/x + 4x/y + 4z/y + 8y/z
≥ -17 + 2√8+2√32 = 12√2-17
therefore the minumum value is 12√2 -17
Hence option a is correct



Here, what I did not understand is that how
-17 + 2y/x + 4x/y + 4z/y + 8y/z ≥ -17 + 2√8+2√32

[shreyans_nitt]

Originally Posted by hi.shivani

I don't have any idea what the correct answer is but anyways lemme elaborate what I have done.

1^(1/2)=1
and (100)^(1/2)=10


Square root of 1=1
Square root of 2,3 i and 4 s 1.414 and 1.73 and 2 .So,{}=2 .

7+9*5+11*6+13*7+15*8+17*9+19*10=715

correct!!.........

[shreyans_nitt]

Originally Posted by sourabhforu

Hi,
I came across an interesting problem while surfing the internet:-

Let a,b and c be the positive real numbers. Determine the minimum value of (a+3c)/(a+2b+c) + 4b/(a+b+2c) -8c/(a+b+3c)
a)12√2-17
b)6√2-17
c)12√2+17
d)6√2+17


I saw the answer to it as:-

Let x = a +2b+c, y = a+b+2c and z=a+b+3c
z-y = c and x - y = b - c => x -y = b-(z-y),
or b = x+z-2y
We note that a+3c=2y-x. By the AM-GM Inequality, it follows that
(a+3c)/(a+2b+c) + 4b/(a+b+2c) -8c/(a+b+3c) = (2y-x)/x + 4(x+z-2y)/y - 8(z-y)/z
=-17 + 2y/x + 4x/y + 4z/y + 8y/z
≥ -17 + 2√8+2√32 = 12√2-17
therefore the minumum value is 12√2 -17
Hence option a is correct



Here, what I did not understand is that how
-17 + 2y/x + 4x/y + 4z/y + 8y/z ≥ -17 + 2√8+2√32

By AM>GM,

2y/x + 4x/y >= 2 sqrt8

4z/y + 8y/z >= 2 sqrt(32).....

[abhishek.smiles]

Let S be the set {1, 2, 3, . . . , n} consisting of the first n positive integers.
What is the maximum value of n for which every 100-element subset of
S contains two integers which differ by 25 ?

(a) 149 (b) 150 (c) 174 (d) 175 (e) none of these

My answer comes as 174.Is it fine?

[shreyans_nitt]

Originally Posted by abhishek.smiles

Let S be the set {1, 2, 3, . . . , n} consisting of the first n positive integers.
What is the maximum value of n for which every 100-element subset of
S contains two integers which differ by 25 ?

(a) 149 (b) 150 (c) 174 (d) 175 (e) none of these

My answer comes as 174.Is it fine?

ya 174 is correct......

choose first 25....leave next 25...choose next 25...leave next 25....choose next 25...leave next 25..choose next 25.. (total 175 elements shud be there for single subset )

this way we can get one subset which will not have integer difference as 25....

so inorder to avaoid that, we reduce the number by 1...if total elements are only 174, we will have subsets with integer difference 25....

[hi.shivani]

Originally Posted by shreyans_nitt

ya 174 is correct......

choose first 25....leave next 25...choose next 25...leave next 25....choose next 25...leave next 25..choose next 25.. (total 175 elements shud be there for single subset )

....

But we need to take subset of 100 element. Isn't it??

[phattuSingh]

Originally Posted by subu_222412

Thank you very much MaskedMenace
Hey..Just needed one more clarification...
Very Silly though...
The question was..to find the sum of the digits in the multiplication...
So there should have been an option with answer 7...
why did choose the answer having the digit sum 7 .i.e 160 in this case..What if there was one more option having digit sum as 7...So just wanted to know. if we can directly arrive at a particular value..like 160 in this case....

Subu...the single digit sum method is the quickest way..But by chance if you get a catch by muntiple choices that have 7 as the sum..May be you can visualise the question a bit more.

You can see 202 * 20002, The zeroes are such placed that with every single muntiplication there will not be any carry because of 2 rows of muntiplication. eg.

202
*
20002
___________
0404
404xxxx
___________

So There wont be any carry as such for the first 3 muntiplications.

they will all have (0)808080808080808
In this way..All digits have a zero in the next place to accomodate the carry they get for the 4th and 5th muntiplications.

in the end the final result will have 3232323232323232....some 64 times..

So ,you can look for an option that is a muntiple of (2+3) = 5.

****
The same method shall be valid in case there are access options if there were one more muntiplicant as 2----(63 zeroes)---2.

[sourabhforu]

Here is another problem:

If x and y are natural numbers and 19x + 97y = 1997 then the smallest value of x+y is
a)21
b)23
c)38
d)41

The solution says:-
We can rewrite the given equation as 97y - 97 = 1900 -19x.
taking out common factors this becomes 97(y-1) = 19(100 -x)
Now Both 19 and 97 are prime numbers, therefore 97 must be a multiple of 100-x
Thus x = 100 or x =3
If x = 100 then y = 1 => x+y = 101
If x = 3 then y= 20 => x+y=23
Hence option b is correct.


However, I think that it is not necessary 97 must be multiple of 100-x because the above equality can be written as:-
97/(100-x)=19(y-1)
Now for this equality to hold, it is not necessary that 97 and 100-x be exactly divisible..

What do u guys think??

[thebornattitude]

Originally Posted by sourabhforu

Here is another problem:

If x and y are natural numbers and 19x + 97y = 1997 then the smallest value of x+y is
a)21
b)23
c)38
d)41

The solution says:-
We can rewrite the given equation as 97y - 97 = 1900 -19x.
taking out common factors this becomes 97(y-1) = 19(100 -x)
Now Both 19 and 97 are prime numbers, therefore 97 must be a multiple of 100-x
Thus x = 100 or x =3
If x = 100 then y = 1 => x+y = 101
If x = 3 then y= 20 => x+y=23
Hence option b is correct.


However, I think that it is not necessary 97 must be multiple of 100-x because the above equality can be written as:-
97/(100-x)=19(y-1)
Now for this equality to hold, it is not necessary that 97 and 100-x be exactly divisible..

What do u guys think??

Coming to my fav thread after a long while ........
neways, the solution is right Saurabh, check that the que says x,y are natural numbers, even if say that

y= (100-x)19/97,,,,so for y to be natural no, 100-x has to be a multiple of 97 for sure, ie x either 3 or 100......

here comes a qu from my side......this was a qu that came in Maths olympiad , also in QQAD last yr , also one thing to be noted that olympiad questions have a loving relationship with CAT .....

Two footballs are coming together along two perpendicular lines with same speed but they are at 10 m and 20 m away from the point of intersection of the lines along which they are moving. At this instant, one of the football is deflected perpendicular to its direction of motion with speed 5m/s (its net motion being the superposition of the motion in two directions) and they collide. The time of collision (in seconds) of the footballs after one of them is deflected is

a) 2 (b) 2√2 (c) 4 (d) 5 (e) none of the foregoing