official quant thread for cat08

[astute10]

Originally Posted by Vishal2704

True...both the formulas are same.....

No. of squares = m*(m+1)*(3n-m+1)/6

if n=m no. of squares become m*(m+1)*(2m+1)/6

No. of rectangles = (m+1)C2 * (n+1)C2

I hope u get it.....

Hey puys pl note:


This formula (m+1)C2 x (n+1)C2 gives you no of rectangles and squares both ...............think of it using this formula i will also include cases when I will select twp adjacent horizontal lines and two adj vert also ......... so no: of rectangles with different L and W will be (m+1)C2x(n+1)C2 - No of squares ...................Yes if you define a rectangle such that L >=W then the above is valid.


cheers

[ahaprashant]

Find the last TWO digits of:
87^474
79^83
583^512
78^379
Hope many of you would have already come across these in some forum, still try.

[naga25french]

Originally Posted by ahaprashant

Find the last TWO digits of:

87^474

is the answer 89 ?

79^83

is the answer 39 ?

583^512

is the answer 61 ?

78^379

is the answer 92 ?

[ahaprashant]

Originally Posted by naga25french

is the answer 89 ?



is the answer 39 ?




is the answer 61 ?




is the answer 19 ?

YES^3. Post your quick approach too (if u using some special funda)
Like if You're able to finish off in <a minute, then must share the quickapproach!

[Vishal2704]

Originally Posted by ahaprashant

find the last two digits of:
87^474
79^83
583^512
78^379
hope many of you would have already come across these in some forum, still try.

1. 89
2. 39
3. 61
4. 92

[ahaprashant]

-----------edited-----------

[naga25french]

Originally Posted by ahaprashant

YES^4. Post your quick approach too (if u using some special funda)
Like if You're able to finish off in <a minute, then must share the quickapproach!

i jus did it by dividing by 100

perhaps if u need to solve in very quick time,u need to solve by cyclicity rule

basically these problems r sitters if u know cyclicity of tenth digits...

for any number , we can have last two digits repeating every 20 powers

eg 7^23 last two digit is same as 7^3

so if we know the pattern , we can guess it easily

anyways the best is to solve by dividing by 100... or if choice are quite obvious , answer can be got by elimination of choices

eg. last two digit of 4567^2303

options
a) 49 b) 07 c) 01 d) 69 e) 63

from the given question its quite obvious that last digit is 3(hope every one agrees with it)

so we can easily select e)option

if the choices are

a) 43 b) 03 c) 23 d) 83 e) 63

if the options r like this , for sure u have to division by 100 method...

hope i make sense...

[Vishal2704]

Originally Posted by ahaprashant

Check the 2nd one again.

yar...edit kar dia....4th post karta hoon..within 2 mins

[ahaprashant]

Originally Posted by naga25french

for any number , we can have last two digits repeating every 20 powers

eg 7^23 last two digit is same as 7^3

so if we know the pattern , we can guess it easily

anyways the best is to solve by dividing by 100... or if choice are quite

Couple of small doubts, are you sure for ANY number (7 / 8 / 459 / 2463 etc. ), we have last two digits repeating every 20 powers!??? I knew it was true for some numbers, didn't realize this universality!
Secondly, by saying "dividing by 100", you mean divide the power with 100, then simplify right ? (which is nothing but making us eof the fact that last two digs repeat after 20 power) ? RIGHT ?

[Vishal2704]

Originally Posted by ahaprashant

Couple of small doubts, are you sure for ANY number (7 / 8 / 459 / 2463 etc. ), we have last two digits repeating every 20 powers!??? I knew it was true for some numbers, didn't realize this universality!
Secondly, by saying "dividing by 100", you mean divide the power with 100, then simplify right ? (which is nothing but making us eof the fact that last two digs repeat after 20 power) ? RIGHT ?

Is the 4th one 92 ??...already edited d post