how about approximating 233/3221.Can someone explain it stepwise.
@realslimshady said:puys help me out with these1.(47^47)-(13^13) is divisible bya.7b.9c.10d.4e.none of these2.if (N5)^2 + (N^3) + (1N) is divisible by 7 and N is non zero single digit, then what can be it's valuea.2b.3c.5d.7e.9Please show your approach as well. thanks
1) 10
2) 7
@realslimshady said:puys help me out with these1.(47^47)-(13^13) is divisible bya.7b.9c.10d.4e.none of these2.if (N5)^2 + (N^3) + (1N) is divisible by 7 and N is non zero single digit, then what can be it's valuea.2b.3c.5d.7e.9Please show your approach as well. thanks
1)
47^47 - last digit 3
13^13 - last digit 3
so (47^47)-(13^13) is divisible by 10
2)
what is there to solve with 7 in the option
47^47 - last digit 3
13^13 - last digit 3
so (47^47)-(13^13) is divisible by 10
2)
what is there to solve with 7 in the option
@realslimshady said:puys help me out with these1.(47^47)-(13^13) is divisible bya.7b.9c.10d.4e.none of these2.if (N5)^2 + (N^3) + (1N) is divisible by 7 and N is non zero single digit, then what can be it's valuea.2b.3c.5d.7e.9Please show your approach as well. thanks
take in bold
@jaituteja said:how about approximating 233/3221.Can someone explain it stepwise.
you can do it percentage wise:
3221 ka 1% = 32.21 i am taking 32 here
32*7 = 224
233-224 =9
3221 ka .1 % = 3.221
3.22 *2 = 6.44
so in total it is:: 7% + 0.2% = 7.2% = .072
@krum said:1) 47^47 - last digit 313^13 - last digit 3so (47^47)-(13^13) isdivisible by 102)what is there to solve with 7 in the option
krum, in first ques u cant just tell checking the unit digit...it is divisible by 10..it could be divisible by 4 also..so we need to check its tens place also..which in this case is not coming..
@milestogo3 said:krum, in first ques u cant just tell checking the unit digit...it is divisible by 10..it could be divisible by 4 also..so we need to check its tens place also..which in this case is not coming..
by 4 it gives a remainder of 2 
@gs4890 said:by 4 it gives a remainder of 2
wohi bola, its not cming in this case. last two digits are 10. :D
@jaituteja said:how about approximating 233/3221.Can someone explain it stepwise.
3221 x 0.1 = 322
3221 x 0.03 = 96
thus, u'll get approx 7.2% :)
3221 x 0.03 = 96
thus, u'll get approx 7.2% :)
@milestogo3 said:krum, in first ques u cant just tell checking the unit digit...it is divisible by 10..it could be divisible by 4 also..so we need to check its tens place also..which in this case is not coming..
i assumed only 1 option to be correct
@gs4890 said:What are the no. of factors of 1000000 which are less then 1000 but do not divide 1000 ?@pankaj1988@anytomdickandhary@krum@mbajamesbond - aajao sir letz start nw
1000000 = (1000)^2 = {(2^3)*(5^3)}^2
this implies that all the factors will be of form (2^a)*(5^b)
if both a and b are less than equal to 3 then it will be a factor of 1000. {as 1000 = 8*125}
Hence we can conclude that all the factors of form (2^a)*(5^b) will not be a factor of 1000 iff atleast one of a or b is greater than 3
2^4, 2^5,2^6,
5*2^4, 5*2^5, 5*2^6,
25*2^4, 25*2^5
625
Hence total of 9 such factors
ATDH.
/good morning 
f(x) and g(x) are two quadratic functions such that f(3) = 0 , g(5) = 0 . If they have a common root and f(5) * g(7) = 12, what is the value of common root?
1) 4
2) 8
3) either 4 or 8
4) NOT
@rkshtsurana said:f(x) and g(x) are two quadratic functions such that f(3) = 0 , g(5) = 0 . If they have a common root and f(5) * g(7) = 12, what is the value of common root?1) 42) 83) either 4 or 84) NOT
4 to satisfy ho raha hain ; & m to lazy to check for 8 :P
OA batao bhai
OA batao bhai
what is the highest 3-digit number that divides the number 11111...1(27 times) perfectly without leaving any remainder?
please tell the procedure?
thanks in advance
@IIM-A2013 said:what is the highest 3-digit number that divides the number 11111...1(27 times) perfectly without leaving any remainder?
please tell the procedure?
thanks in advance
a 111 b 333 c 666 d 999
@IIM-A2013 said:what is the highest 3-digit number that divides the number 11111...1(27 times) perfectly without leaving any remainder?please tell the procedure?thanks in advance
ITS 999
for 999 , we make bunches of 3 digits...and add dem and check remainder
so 9 *111 = 999/999 reminder 0