try dis now If 5x + 2y + z = 81, where x, y and z are distinct positive integers, then find the difference between the
maximum and the minimum possible value of x + y + z. :)
is the ans 55?
Max value = 75 and min value = 20
is the ans 55?
Max value = 75 and min value = 20
E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1
please help me with this...am not able to understand....
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
I saw the solution... but am not able to understand...please help
wht does E(17)=16 mean
I saw the solution... but am not able to understand...please help
wht does E(17)=16 mean
its called euler number
read Euler's Theorem and then these types of Qs will be a cake-walk for you also :)
ATB !!!
is the ans 55?
Max value = 75 and min value = 20
yes it is 55 .. you can refer to my post also on the previous page for further verification
dharmesh.patel Saysplease help me with this...am not able to understand....
E(17) = 16
(38^16!)^1777 mod 17=
remainder = 1
i hope u know euler's rule
and u understand that 38^16 mod 17 = 1
and 16!= 16*k so 38^16! mod 17 = 1
hope u got the point.
hi to all
which is d largest four digit no. which when divided by 6 leaves remainder 5 and when divided by 5 leaves remainder 3
:shocked:
6x+5=5y+3
equation satisfies for x=3
this smallest no is 23
and largest 4 digit no is 30(332) +23 = 9983 .. Answer
Originally Posted by MIOCHE View Post
each coefficient in the equation ax^2+bx+c is determined by throwing an ordinary dice(six faced).find the probability that the eqn will have real roots?
plz ans wid full appraoch
I am listing all the combinations
For b=2; (1,1)
For b=3;(1,1)(1,2)(2,1)
For b=4;(1,1)(1,2)(1,3)(1,4)(2,1)(2,2)(3,1)(4,1)
For b=5;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(3,1)(3,2)(4,1)(5,1)(6,1)
For b=6;(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(3,1)(3,2)(3,3)(4,1)(4,2)(5,1)(6,1)
Total Combinations:1+3+9+13+17=43
So this was a very tedious method..Anyone having a shorter approach?
You made it tedious otherwise it is a oral ques can be done in 1 min or less depends on ur speed .........
assume value of 'a' constant and multiply that value by 4 and for different values of b check it .........
for eg a = 1 then 'ac' can have values in multiple of 4 which must be less then or equal to b^2.....
take another case for a = 3 then ac will bounce in the multiple of 12 hence if b = 4 i.e. b^2 = 16 then we can have only one value of c as 1 but for b = 5 i.e. b^2 = 25 we can two values of c i.e. 1 n 2........
Let me show you how I calculated.........which can be done orally.......
for
a = 1 , b=1 = zero value
a = 1 , b=2 = 1 value
a = 1 , b=3 = 2 values
a = 1 , b=4 = 4 values
a = 1 , b=5 = 6 values
a = 1 , b=6 = 6 values
For a = 1 total 19 values.......
a = 2 , b=1 = zero value
a = 2 , b=2 = 0 value
a = 2 , b=3 = 1 values
a = 2 , b=4 = 2 values
a = 2 , b=5 = 3 values
a = 2 , b=6 = 4 values
for a = 2 total 10 values.......
a = 3 , b=1 = zero value
a = 3 , b=2 = 0 value
a = 3 , b=3 = 0 values
a = 3 , b=4 = 1 values
a = 3 , b=5 = 2 values
a = 3 , b=6 = 3 values
for a = 3 total 6 values........
a = 4 , b=1 = zero value
a = 4 , b=2 = 0 value
a = 4 , b=3 = 0 values
a = 4 , b=4 = 1 values
a = 4 , b=5 = 1 values
a = 4 , b=6 = 2 values
for a = 4 total 4 values
a = 5 , b=1 = zero value
a = 5 , b=2 = 0 value
a = 5 , b=3 = 0 values
a = 5 , b=4 = 0 values
a = 5 , b=5 = 1 values
a = 5 , b=6 = 1 values
for a = 5 total 2 values
and for a = 6 we'll have 2 value
So a total = 19 + 10 + 6 + 4 + 2 + 2 = 43.........
Probability that the eqn will have real roots
= P(a or b or c) + P((a,b) or (b,c) or (c, a)) + P(a,b,c)
= 1/6 + 1/6^2 + 1/6^3
= 1/6 + 1/36 + 1/216
= 43/216
any observations pls let me know..
viveknitw Saysfind -12121212...................300digits% 999
121 + 212 = 333
333* 50 = 16650
16650
16 + 650 = 666
answer is 666
another method,
= 12(10^298 + 10^296 + 10^294 + .....10^4 + 10^2 + 1) % 999
= 12(10 + 10^2 + 1 + 10 + 10^2 + 1 +......10 + 10^2 + 1) % 999
= 12(111*50) % 999
= 66600 % 999
= 666
any observations pls let me know
if a n degree equation has n+1 roots...then what can we say about the nature of the equation?.....for eg. if a quadratic equation happens to have three roots
barclays_boss Saysif a n degree equation has n+1 roots...then what can we say about the nature of the equation?.....for eg. if a quadratic equation happens to have three roots
Is this possible? I doubt .... please provide an example if you have one ...
The age of Confusius' teacher was a number divisible by
a. 7
b. 9
c. 13
d. 15
a. 7
b. 9
c. 13
d. 15
wat to find?
barclays_boss Saysif a n degree equation has n+1 roots...then what can we say about the nature of the equation?.....for eg. if a quadratic equation happens to have three roots
this thing cannot exist...
a. 7
b. 9
c. 13
d. 15
65 can be the age of the teacher...but wat is the question askin for??
Abhinav90 Saysthis thing cannot exist...
I have edited my post pls see again. Any way I have solved it.
another method,
= 12(10^298 + 10^296 + 10^294 + .....10^4 + 10^2 + 1) % 999
= 12(10 + 10^2 + 1 + 10 + 10^2 + 1 +......10 + 10^2 + 1) % 999
= 12(111*50) % 999
= 66600 % 999
= 666
any observations pls let me know
Another Method:
121212...300 digits % 999 = 33333...150 digits % 999 = 111..150 digits % 333
Now 333=37*9..Hence Chk Individually...
1111..150 times % 37 = 0 i.e. 37a(Completely Divides) ------(1)
1111...150 times%9 = 9b + 6------(2)
From (1) and (2),9b+6 = 37a....--->a=6
Hence,37 * 6 * 3 = 666...
But,I personally feel,method shown by naga is less tedious...
The age of Confusius' teacher was a number divisible by
a. 7
b. 9
c. 13
d. 15
my take option c) 13..
jain_ashu SaysI have edited my post pls see again. Any way I have solved it.
that was not intended for u...It was for barclay user...
The age of Confusius' teacher was a number divisible by
a. 7
b. 9
c. 13
d. 15
teacher age is 65
65 is multiple of 13
so c)13
