@jain4444
327??
@jain4444 said:How many positive integers nā¤1000 cannot be written in the form a^2āb^2āc^2 where a,b and c are non-negative integers subject to a ā„ b+c?
If c = 0, then we can get all odd numbers
Since we can get all odd numbers from a^2 ā b^2, we can say that we can get all even numbers also by putting c = 1
So, all numbers from 1 to 1000 are possible
Consider all quadruples of real numbers (x1,x2,x3,x4),such that x1 ā¤ x2 ā¤ x3 ā¤ x4 and each of the four numbers plus the product of the other three equals 130. If S is the sum of all possible values of x4, what is S ?
@jain4444 said:13^71 mod 57E(57) = 3613^36k mod 57 = 1let 13^71 mod 57 = rr*13 mod 57 = 157k + 1 = 13r=> k = 5 and r = 2257*Q + 22 = 13^71unit digit of 13^71 is 757*Q + 22 = xxxx7so , unit digit of Q = 5
@jain4444 said:13^71 mod 57E(57) = 3613^36k mod 57 = 1let 13^71 mod 57 = rr*13 mod 57 = 157k + 1 = 13r=> k = 5 and r = 2257*Q + 22 = 13^71unit digit of 13^71 is 757*Q + 22 = xxxx7so , unit digit of Q = 5
Please could u explain the underlined step?
@KhannaiiM said:Please could u explain the underlined step?
We have got
13^36k mod 57 = 1
hence , 13^72 mod 57 =1
13^71 *13 mod 57 =1
let 13^71 mod 57 = r
r*13 mod 57 = 1
if 13r when divided by 57 leaves remainder as 1
so
13r -1 must be div by 57
therefore
13r-1=57k
13^36k mod 57 = 1
hence , 13^72 mod 57 =1
13^71 *13 mod 57 =1
let 13^71 mod 57 = r
r*13 mod 57 = 1
if 13r when divided by 57 leaves remainder as 1
so
13r -1 must be div by 57
therefore
13r-1=57k
@Vipul24 said:in last steps how u attain the value 16/169??ar PFX=16/169 ar FBC??
PFX~BFC
px/bc=4/13
@Shray14 said:@Dexian@mailtoankit@Subhashdec2Sorry for the late replying..its ans among the options is NOT.My approach..5A no varies from 0001 to 9999=10*10*10*9and till 5s=10*10*10*9*18till 5S-2234 it will be 10*10*10*9*18+2235..
bhai 1 - 9999 = 9999 numbers..why you took 9000?
A class of 78 students attended a test which consisted of two parts. 16 students failed in the
first part, while 20 failed in the second.The number of students clearing at least one part is??
1] 32
2] 36
3] 52
4] 68
first part, while 20 failed in the second.The number of students clearing at least one part is??
1] 32
2] 36
3] 52
4] 68
someone please solve this
In a chess tounament everybody plays everybody else.Everyone scored the same number of points, except four players whose total score were 17.5.Which of the following can be the number of players participating in the tourmanent. Assume that for each win a player scores 1 point , for draw 1/2 point and zero for losing.
a. 27
b. 29
c. 31
d. 42
@falcao said:A class of 78 students attended a test which consisted of two parts. 16 students failed in thefirst part, while 20 failed in the second.The number of students clearing at least one part is??1] 32 2] 36 3] 52 4] 68someone please solve this
Let a be students failing in only part 1
b be students failing only in part 2
c who failed both
d who failed none
a+b+c+d=78
a+c +b+c -c +d=78
16 + 20+d-c=78
d-c=42
c can be 16 at max
d=58
78-16=62
which is not in the options...:P
@falcao said:A class of 78 students attended a test which consisted of two parts. 16 students failed in thefirst part, while 20 failed in the second.The number of students clearing at least one part is??1] 32 2] 36 3] 52 4] 68someone please solve this
52 ?
failed in both = x
16 - x + x + 20 - x + n = 78
n - x = 42
using options,only one possibility when x = 10
so n = 52
pata nahi sahi hai ki nahi...
Let a = 1! + 2! + 3! + . . . . .. . .. 99! + 100!
Number of digits in the number ā¬Ėa ā¬ā¢ are
how to do this type of questions
Number of digits in the number ā¬Ėa ā¬ā¢ are
how to do this type of questions
@mailtoankit said:52 ?failed in both = x16 - x + x + 20 - x + n = 78n - x = 42using options,only one possibility when x = 10so n = 52pata nahi sahi hai ki nahi...
bhai x to 16 tak ja skta h
@Subhashdec2 said:bhai x to 16 tak ja skta h
haan....but options mein nahi tha...isiliye..x = 10 liya
@mailtoankit said:haan....but options mein nahi tha...isiliye..x = 10 liya
ye sahi h... to minimum kaise hua fir...:P
@Subhashdec2 said:Let a be students failing in only part 1b be students failing only in part 2c who failed bothd who failed nonea+b+c+d=78a+c +b+c -c +d=7816 + 20+d-c=78d-c=42c can be 16 at maxd=5878-16=62which is not in the options...
16 students failed in the first part => 78-16 cleared the first part = 62
20 failed in the second part => 78 - 20 cleared the second part = 58
-
58 + 62 = 120
Excess number of students = 120 - 78 = 42
-
So no. of students who cleared only part 1 = 62 - 42 = 20
No. of students who cleared only part 2 = 58 - 42 = 16
-
No. of students who cleared 'atleast' one part = 20 + 16 = 36
20 failed in the second part => 78 - 20 cleared the second part = 58
-
58 + 62 = 120
Excess number of students = 120 - 78 = 42
-
So no. of students who cleared only part 1 = 62 - 42 = 20
No. of students who cleared only part 2 = 58 - 42 = 16
-
No. of students who cleared 'atleast' one part = 20 + 16 = 36
-
Am I faulting somewhere?
'Atleast' does sounds fishy to me. It should have been 'only' in one part.
@mailtoankit @Subhashdec2 @Psychamour bhai log answer nahi pata.kuch ka 52 aa raha hai kuch ka 68 and here @Psychamour has calculated 36.by the way,it is question of the day on the facebook page of ims learning resources.i was getting 52 somehow but decided to ask here.don't know the answer and neither has the admin given it as of yet
What is the least number which when divided by 42,72, and 84 leaves the remainders 24, 56 and 68 respectively?