@sonamaries7 said: someone plz help solving the qs:what is the remainder when 67raised to the power of 53 is divided by 101? (67^53 divided by 101)Plz provide detailed solns ...Thanks
Hi,
Is the Answer 2?
@[419941:Enceladus] I cannot undersatnd how 6X mod 49 gives 41 could you please elaborate
what is the answer of (67^53)/101??? is it 86?
@[547670:koyal1990] What is your approach . My answer is -86 or say 15.
Can you please explain your approach.
Thanks
@[532679:devendersnegi]- 101 and 67 being co primes, i used euler's method. euler's no of 101 is 100. so we can write the number as [67^100]^1/2 (which is 50) *67^3. by indices rules, it boils down to 67^53. now the 1st part remainder is 1, next part was a lil calculation heavy. i tried to minimize the cal by taking 67^2 which is 4489 and dividing it by 101. this gives a remainder of 45. then 45*67 divided by 101 gives 86. i am not very sure of the method though. how did u solve it n what is the ans?
@[515126:kejriwalsonal1]:thanx:)
hey all.. im a little new to eulers method and chinese remainder method.. So I would be obliged if anyone can explain what the following statements mean.. along with the approach to get the final answer !
(36^41) mod 7 = 1
(36^41) mod 11 = 3
(36^41) mod 11 = 3
also.. fyi.. i found this in this thread..
http://www.pagalguy.com/forums/quantitative-ability-and-di/number-system-questions-and-discussions-t-43728/p-1642745?page=3
thank you ! :)
Plz help in solving dis ques...............
How many numbers less than 500 will have digit sum equal to 5???
@Rshmi05 said:Plz help in solving dis ques...............
How many numbers less than 500 will have digit sum equal to 5???
Take No. of nos. having digit sum 5 from 1) 20 to 30______ 2)40 to 50_______ 3)60 to 80
You'll get Your ans.
I have counted and it came around 20.
But options are
39,72,55 ,81
@Rshmi05 said:Plz help in solving dis ques...............
How many numbers less than 500 will have digit sum equal to 5???
method 1 :
single digit - 5
two digits :
a + b = 5 or 14
a + b = 5 --> 50,14,41,23,32
a + b = 14 --> 77,68,86,59,95
three digits :
a + b + c = 5 or 14 or 23
count and find the numbers
method 2 :
digit sum is nothing but divisibility by 9 .. so all number of form 9k +5 will have digit sum 5
so 500/9 = 55 .
so answer is 55
single digit - 5
two digits :
a + b = 5 or 14
a + b = 5 --> 50,14,41,23,32
a + b = 14 --> 77,68,86,59,95
three digits :
a + b + c = 5 or 14 or 23
count and find the numbers
method 2 :
digit sum is nothing but divisibility by 9 .. so all number of form 9k +5 will have digit sum 5
so 500/9 = 55 .
so answer is 55
@[169132:naga25french] .................ok..........got it...thanks a lot..............
@[169132:naga25french]- why is digit sum divisibility by 9? cud u please explain the second method plz? TIA... 
P is a prime number greater than 37.then the largest number that will always divide (P-1)X(P+1) is
a) 32 b) 16 c) 8 d) 4 e) 24
@lopagargg said: P is a prime number greater than 37.then the largest number that will always divide (P-1)X(P+1) isa) 32 b) 16 c) 8 d) 4 e) 24
it will be 24
Take the least prime number greater than 37, which is 41
P = 41
(p-1)*(p+1) = 40*42 = 4*10*2*21 = 2*2*2*5*2*3*7 = 2^4*3*5*7
32 = 2*16 = 2^5
24 = 2*12 = 2*2*2*3 = 2^3*3
m and n are natural number with m x n =160000.if neither m nor n is a multiple of 10,then what is the value of m+n
a) 784 b) 824 c) 881 d) 924 e) 961
Solve: ||||x – 1| – 2| – 3| – 4|
a. 4 b. 10 c. 14 d. 15
(Please tell me how to approach such multiple Modulus sums)
Ans is C) 881
160000 = 2^4*2^4*5^4
for no to be not a multiple of 10, 2 and 5 should not be together.
so numbers are 2^8 and 5^4
=> 256 + 625 = 881