In a sports contest all the medals were awarded over six days.On the first day 1 medal and 1/7th of remaining medals were awarded.On the second day 2 medals and 1/7th of remaining medals were awarded and so on till the fifth day.On sixth day, all the remaining medals were awarded.
What could be the minimum number if medals that were awarded during the six days?
1- 42
2- 36
3- 30
4-none of these
Happy Solving 😃
In a sports contest all the medals were awarded over six days.On the first day 1 medal and 1/7th of remaining medals were awarded.On the second day 2 medals and 1/7th of remaining medals were awarded and so on till the fifth day.On sixth day, all the remaining medals were awarded.
What could be the minimum number if medals that were awarded during the six days?
1- 42
2- 36
3- 30
4-none of these
Happy Solving :)
Hit And Trial Method
We can have total number of medals n = 15,22,29,36 and so on because on the first day after subtracting '1', the number is divisible by 7. (14+1, 21+1....)
I.) n = 15;
Day 1 : 1 + 14/7 = 3; Left -> 15-3 = 12
Day 2 : 2 + 10
II.) n = 22;
Day 1 : 1+21/7 = 4; Left -> 22 - 4 = 18.
Day 2 : 2+16
III.) n = 29;
Day 1 : 1 + 28/7 = 5; Left -> 29-5 = 24.
Day 2 : 2 + 22
IV.) n = 36
Day 1 : 1 + 35/7 = 6;
Day 2 : 2 + 28/7 = 6;
Day 3 : 3 + 21/7 = 6;
Day 4 : 4 + 14/7 = 6;
Day 5 : 5 + 7/7 = 6;
Day 6 : 6
Thus, Minimum Medals = 36
In a sports contest all the medals were awarded over six days.On the first day 1 medal and 1/7th of remaining medals were awarded.On the second day 2 medals and 1/7th of remaining medals were awarded and so on till the fifth day.On sixth day, all the remaining medals were awarded.
What could be the minimum number if medals that were awarded during the six days?
1- 42
2- 36
3- 30
4-none of these
Happy Solving :)
2)36
P.S.-Fractions me medal nahi de sakte hain yr.
Answer is 21.3
Students who has at least 29 questions wrong = 60-2n= 60-2*29= 2
Students who has at least 28 questions wrong = 60-2n= 60-2*28= 4
.
.
.
Students who has at least 1 questions wrong = 60-2n= 60-2*1= 58
therefore there are 2 people of each of 1,2 ,3.........29 marks
till now we know marks of 58 people and remaining 42 people have 30 marks each...
average= (2*1+2*2+..........+2*29+ 42*30)/100= 21.3
ans for the other question is 3 cm2
or we take the total marks as 3000 and subtract 2(1+..29) from it and divide it by 100.
hey guys i want to know your views on this method , iknow it is a bit strange but one of my friend told me it could be seen as a remainder theorem application , i dont know whether it is correct or not.
Q) find the remainder (83^261)/17
soln) (83^261)/17 = ((16x5 + 3)^261)/(16+1)
putting 16 as -1 in numerator ,as from deno 16+1=0
then numerator becomes (-2)^261 and the units digit of which is 2
2 is the final remainder answer
..........
hey guys i want to know your views on this method , iknow it is a bit strange but one of my friend told me it could be seen as a remainder theorem application , i dont know whether it is correct or not.
Q) find the remainder (83^261)/17
soln) (83^261)/17 = ((16x5 + 3)^261)/(16+1)
putting 16 as -1 in numerator ,as from deno 16+1=0
then numerator becomes (-2)^261 and the units digit of which is 2
2 is the final remainder answer
Answer : I think this ques. can be solved in a better way.
17 is a prime number. so phi(17) = 16.
Use of fermat's theorem Nr^phi(Dr)/Dr = 1 provided Nr & Dr are co-primes.
therefore (83^261)/17 = (83^(16*16+5))/17 = (83^5)/17 = ((-2)^5)/17 = (-32)/17 = 2
can anyone tell me how this last step was simplified?
y = ( root(2)+1)^6
x = (1 - root2)^6
x + y = 2*(1 + 15*2 + 15*4 + 8 ) = 198
can anyone tell me how this last step was simplified?
y = ( root(2)+1)^6
x = (1 - root2)^6
x + y = 2*(1 + 15*2 + 15*4 + 8 ) = 198
You can approach the question like this : --
Let a = 1 + root(2)
b = 1 - root(2).
a + b = 2, a-b = 2root(2), ab = -1
then
x = b^6, y = a^6
then x + y = a^6 + b^6 = (a^2)^3 + (b^2)^3
= (a^2 + b^2)(a^4 + b^4 - a^2.b^2)
=( (a+b)^2 - 2ab)((a^2-b^2)^2 + a^2.b^2)
=( (a+b)^2 - 2ab)((a+b)^2.(a-b)^2 + a^2.b^2)
= (4+2)(4.8 + 1)
= 6X33
= 198.
I Hope I am clear in explaining.
Thanks.
Answer : I think this ques. can be solved in a better way.
17 is a prime number. so phi(17) = 16.
Use of fermat's theorem Nr^phi(Dr)/Dr = 1 provided Nr & Dr are co-primes.
therefore (83^261)/17 = (83^(16*16+5))/17 = (83^5)/17 = ((-2)^5)/17 = (-32)/17 = 2
Where can I find more info. on this theorem?
dcount123 SaysWhere can I find more info. on this theorem?
HERE YOU GO:
Remainders Reloaded: Euler, Fermat and Wilson's Theorems for CAT 2011 | PaGaLGuY.com - India's biggest website for MBA in India, International MBA, CAT, XAT, SNAP, MAT
The hundreds place digit of 1993 913 1083 is :
OPTIONS 1)6
2)4
3)0 4)None of these
The hundreds place digit of 199^3 91^3 108^3 is :
OPTIONS
1)6
2)4
3)0
4)None of these
please ignore previous post.
anurag"big cat" Saysplease ignore previous post.
Hundreds digit means we need to calculate the remainder with 1000.
=> 199^3 mod 1000 = 39601*199 mod 1000 = 601*199 mod 1000 = 599.
91^3 mod 1000 = 8281*91 mod 1000 = 821*91 mod 100 = 571.
108^3 mod 1000 = 11664*108 mod 1000 = 664*108 mod 1000 = 712.
So, it should be 599 - 571 - 712 = -684.
Hundreds digit = 6, hence option 1.
answer should be 4
the numbers are -4 ,4,7,11,12
wat about 11? How will u achieve that with addition of 2 digits.
N is a set of all natural numbers less than 500 which can be written as sum of two or more consecutive natural numbers .Find the maximum numbers of elements possible in N ?
It would be great if sombody shares the solution as well .....
N is a set of all natural numbers less than 500 which can be written as sum of two or more consecutive natural numbers .Find the maximum numbers of elements possible in N ?
It would be great if sombody shares the solution as well .....
All powers of 2 cannot be expressed as a sum of two or more consecutive numbers.
so -> 1,2,4,8,16,32,64,128,216 cannot be expressed.
Hence 500 - 9 = 491 numbers can be.
N is a set of all natural numbers less than 500 which can be written as sum of two or more consecutive natural numbers .Find the maximum numbers of elements possible in N ?
It would be great if sombody shares the solution as well .....
Powers of 2 (including 1)cannot be written as sum of two or more consecutive natural no.s
Hence, 1,2,4,8,16,32,64,128,256
Total no.s : 499-9=490
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
The answer is one.. Can someone explain how??
a. 1
b. 16
c. 8
d. 13
The answer is one.. Can someone explain how??
find the remainder when (38^16!)^1777 is divided by 17
a. 1
b. 16
c. 8
d. 13
The answer is one.. Can someone explain how??
{(34 + 4)^16!}^1777 / 17
{(4)^16!}^1777 / 17
((16)^(16!/2)}^1777 / 17
(17 - 1)^(16!/2)^1777 / 17
(- 1)^(16!/2) ^1777
16! /2 = positive and even value
and we know any power of even value wud only result in even no.
so,remainder should be 1.:)