[nbangalorekar]

Originally Posted by SUPER XERO

dude........ agree with ur second answer......
but in the first one it shud be a 2 digit number.........

please post the explanation for 1st answer

[SUPER XERO]

okkkk........ it goes like this....

f(f(f(f(k)))=1
since the sum is 1 ...... and k has least possible digits....
so,
f(f(f(k))= 10 {1+0=1}
noww...... we want the sum to be 10..... and with k having least possible digit so the expression can be 19 {1+9=10}
so....
f(f(k))=19
now...... we want the sum to be 19 so ...... {1+9+9=19}
so.....
f(k)=199
following the same method we have .........
22 9's and one 1 (22x9+1x1=199)
so , k =23 (22 9's and one 1)
i hope i m correct .................

[vicky113]

Originally Posted by SUPER XERO

okkkk........ it goes like this....

f(f(f(f(k)))=1
since the sum is 1 ...... and k has least possible digits....
so,
f(f(f(k))= 10 {1+0=1}
noww...... we want the sum to be 10..... and with k having least possible digit so the expression can be 19 {1+9=10}
so....
f(f(k))=19
now...... we want the sum to be 19 so ...... {1+9+9=19}
so.....
f(k)=199
following the same method we have .........
22 9's and one 1 (22x9+1x1=199)
so , k =23 (22 9's and one 1)
i hope i m correct .................

correct solution

[nbangalorekar]

Originally Posted by SUPER XERO

okkkk........ it goes like this....

f(f(f(f(k)))=1
since the sum is 1 ...... and k has least possible digits....
so,
f(f(f(k))= 10 {1+0=1}
noww...... we want the sum to be 10..... and with k having least possible digit so the expression can be 19 {1+9=10}
so....
f(f(k))=19
now...... we want the sum to be 19 so ...... {1+9+9=19}
so.....
f(k)=199
following the same method we have .........
22 9's and one 1 (22x9+1x1=199)
so , k =23 (22 9's and one 1)
i hope i m correct .................

dude i used the same method but c once we get 10 then no matter how many times u put it in the function loop the sum of digits will be 1.
for eg: u have f(k)=1 then u will have f(f(k))=1 and so on...
so the least possible number of digits wud be 3..
this is my reasoning.... i may be wrong

[SUPER XERO]

Originally Posted by nbangalorekar

dude i used the same method but c once we get 10 then no matter how many times u put it in the function loop the sum of digits will be 1.
for eg: u have f(k)=1 then u will have f(f(k))=1 and so on...
so the least possible number of digits wud be 3..
this is my reasoning.... i may be wrong

we also have a condition
k>f(f(k)).f(f(f(k)))>1
so in above case (highlighted) k wud be less than the product of f(f(k).f(f(f(k)))

[nbangalorekar]

Originally Posted by SUPER XERO

we also have a condition
k>f(f(k)).f(f(f(k)))>1
so in above case (highlighted) k wud be less than the product of f(f(k).f(f(f(k)))

suppose k=199
f(f(k)=10, f(f(f(k)))=1
so k>f(f(k)).f(f(f(k))) {which is = 10}

[SUPER XERO]

Originally Posted by nbangalorekar

suppose k=199
f(f(k)=10, f(f(f(k)))=1
so k>f(f(k)).f(f(f(k))) {which is = 10}

YES DUDE ANSWER SHUD BE 3..... great work.......

[Wynand]

This qs has been posted before. So sorry to break the flow of this thread, but I will really appreciate it if you could provide help again. There is obviously a big hole in my concept

The no of integral solns of |x| + |y| <= 4

My soln -> As mod of x = x when x>0, and -x when x<0.
So |x| + |y|<=4 in the 1st quadrant (x>0,y>0) will have 15 solns
In the 2nd quadrant (x<0,y>0) it becomes -x+y<=4, which shall have infitnite solns as
y-x can be <=4 for infinite sets of (x,y) values :tellme:.

The incumbent Quant Gods went about it calculating the solns in 1st quadrant as 15, abd in 4 quads as 15 x 4 = 60. And subtracting values common to all quadrants like (x=0,y=0) and axis points, arrived at 41.

Please Help!

[nbangalorekar]

Originally Posted by Wynand

This qs has been posted before. So sorry to break the flow of this thread, but I will really appreciate it if you could provide help again. There is obviously a big hole in my concept

The no of integral solns of |x| + |y| <= 4

My soln -> As mod of x = x when x>0, and -x when x<0.
So |x| + |y|<=4 in the 1st quadrant (x>0,y>0) will have 15 solns
In the 2nd quadrant (x<0,y>0) it becomes -x+y<=4, which shall have infitnite solns as
y-x can be <=4 for infinite sets of (x,y) values :tellme:.

The incumbent Quant Gods went about it calculating the solns in 1st quadrant as 15, abd in 4 quads as 15 x 4 = 60. And subtracting values common to all quadrants like (x=0,y=0) and axis points, arrived at 41.

Please Help!

dude ur right that y-x<=4 has infinite solns but question says l x l & l y l
so max x & y can be 4
work with permutation comb principle its easier...
sumtimes working with the graphical soln is sumtimes confusing n also u cant use it when question has many variables

[SUPER XERO]

Originally Posted by nbangalorekar

dude ur right that y-x<=4 has infinite solns but question says l x l & l y l
so max x & y can be 4
work with permutation comb principle its easier...
sumtimes working with the graphical soln is sumtimes confusing n also u cant use it when question has many variables

hey........ can u explain how you would use PnC to solve this question????