@albiesriram said:Q 13,14 nd 15
last one is E 20
for second one alpha =4 and Beta=8 i am getting both
for first one a/b=2/3
then how can any of the options be true?
@Subhashdec2 said:last one is E 20for second one alpha =4 and Beta=8 i am getting bothfor first one a/b=2/3 then how can any of the options be true?
Bhai actually it was like answer all the correct options kind of question . Just now clarified it. So u r bang on OA :-).
@Ani1308 said:thats not even in the option..oa--6, 4 ,3, 9, 2
@Bigshu said:@chillfactor yar ye test funda me aaj ka question haii.......aur '0' option h nahi hai............
OAs can be wrong sometimes... But chill sir? Aur galat? 
....not happening...
....not happening...@chillfactor said:10^40000 = Q(10^400 + 40) + R, where Q is quotient and R is remainder when 10^40000 is divided by 10^400 + 40Remainder when 10^40000 is divided by (10^400 + 40) is (-40)^100 or 40^100=> 10^40000 = Q(10^400 + 40) + 40^100Clearly unit digit of Q has to be 0
____/\____
http://www.wolframalpha.com/input/?i=%28Greatest+integer+of+[10^40000%2F%2810^400%2B40%29]%29+mod+10
regards
scrabbler
albiesriram
1. all true
2. a, d
3. b - 7 numbers
oa kya hai
@scrabbler said:OAs can be wrong sometimes... But chill sir? Aur galat? ....not happening...____/\____ http://www.wolframalpha.com/input/?i=%28Greatest+integer+of+[10^40000%2F%2810^400%2B40%29]%29+mod+10regardsscrabbler
it can be...but open testfunda.. and see question of the day...
a guy posted solution there with unit digit 9.. it
seems logical and consistent..
@albiesriram said:Q 13,14 nd 15
for 1 a is 2 b is 3 all options r right
for 2 alpha =4 beta =8
it is as a/b=c/d=e/f
2/2alpha=7/28=3/alpha+beta
solve for alpha and beta
for 3rd it is 20
case 1 when 0 occurs
210 ----2*2!=4
510 ----2*2!=4
when no 0 occurs
213 - 3!=6
531- 3!=6
4+4+6+6=120
@chillfactor said:10^40000 = Q(10^400 + 40) + R, where Q is quotient and R is remainder when 10^40000 is divided by 10^400 + 40Remainder when 10^40000 is divided by (10^400 + 40) is (-40)^100 or 40^100=> 10^40000 = Q(10^400 + 40) + 40^100Clearly unit digit of Q has to be 0
hemant sir m a nobody here but cud u please tell me 1 thing
lhs has 0 unit digit
10^400+40 has 0 unit digit
40^100 has 0 unit digit
if q has ne number as its unit digit other then 0 still it will satisfy
suppose qs digit is 8
8*0+0=0??????
how did u deduce q unit is 0????
@abhishek.2011 said:hemant sir m a nobody to argue with u but cud u please tell me 1 thinglhs has 0 unit digit10^400+40 has 0 unit digit40^100 has 0 unit digitif q has ne number as its unit digit other then 0 still it will satisfysuppose qs digit is 88*0+0=0??????how did u deduce q unit is 0????
It is not just one zero in RHS here, there are more than 1, in fact 100 zeros !
So, (10^400 + 40)*Q = (10^40000) - R
(10^400 + 40)*Q = 10^40000 - (40)^100
10*Q*(10^399 + 4) = (10^100)*(10^400 - 4^100)
So, you see there are 100 zeros in RHS, and only one in LHS.. so, where will the rest of the zeros come from ? :wow:
So, not only units, even tens, hundreds, thousands, and what not, are 0 for Q π .
It is like Q*10*(something that doesn't end in zero)= (very big number with ends in 100 zeros)
Now, you tell me if Q ends in zero or not π ?
@DivineSeeker said:plz puys.... help this !!!!what is the digit in the unit's place of the quotient of 10^40000/(10^400+40).
For cross check : (Wolframalpha)
Ani1308 Wrong options..
For how many integer values of 'x' is x^2-3x-19 divisible by 289?
Q 16,17,18 :-)
@heylady said:For how many integer values of 'x' is x^2-3x-19 divisible by 289?
Is it 0 ?
x*(x -3) = 289k + 19
x*(x -3) = 289k + 19
RHS mod 17 = 2
Now, varying the form of x from 17m to 17m + 16
Only 17m + 10 makes the LHS mod 17 = 2
=> (17m + 10)*(17m + 7) = 289k + 19
=> 289*m(m+1) + 51 = 289k
=> 17*m(m+1) + 3 = 17k
Now, the above equation has no solution as LHS mod 17 =/= RHS mod 17
=> 0 solutions ?
=> 0 solutions ?
@shattereddream said:How many digits cannot be the unitμ°½β¬β’s digit of the product of 3 three-digit numbers whose sum is 989? Options- 2,4,1,3
OA- 2
@albiesriram said:Q 16,17,18
1-> AE_ _ _ 6 cases
AG _ _ _ = 6 cases
AL_ _ _ 6 Cases
ANE _ _ 2 cases
ANGEL 1 case
ANGLE 1 case
Total 22 case
2-> Next No. is 2016 =2+0+1+6 =9
2016 is a multiple of 9
So ans is 6
AG _ _ _ = 6 cases
AL_ _ _ 6 Cases
ANE _ _ 2 cases
ANGEL 1 case
ANGLE 1 case
Total 22 case
2-> Next No. is 2016 =2+0+1+6 =9
2016 is a multiple of 9
So ans is 6
@marchex those numbers are 3 & 7
: ques. 18
36+x^2=y^2
=> y^2-x^2 = 36
(y+x)(y-x)=36
y=10, x=6 holds good
hence, 10 squares
36+x^2=y^2
=> y^2-x^2 = 36
(y+x)(y-x)=36
y=10, x=6 holds good
hence, 10 squares
@albiesriram said:Q 16,17,18