find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

Hey i am getting 1??

paridhigarg Says

Couldn't solve it..what is the answer..?

Answer is 1.
38 is a co prime to 17 so 38 raised to the power 16 will leave 1 as the remainder when divided by 17 as E(17) is 16.


Refer to the Euler theorem.

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

Couldn't solve it..what is the answer..?

a stdent took 5 test and each sccre is out of 100. if his score were in ratio of 6:7:8:10:11.in how many test did he score above 60%.
1. he score more than 80% in exactly 2 of test
2.all score were whole no withot any frction

is it 16?
let me know

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@mods close this thread,two threads running a request to puys to post in new thread.

ProcMock 11
Q-The sum of thirty-two consecutive natural numbers is a perfect square. What is the least possible sum of the smallest and the largest of the thirty-two numbers?

a81
b36
c49
d64

Please post your queries in Quant thread Part 8.......
@Mod Please close this thread

ProcMock 11
Q-The sum of thirty-two consecutive natural numbers is a perfect square. What is the least possible sum of the smallest and the largest of the thirty-two numbers?

a81
b36
c49
d64

Q) After the addition of 35 liters of water to a can of diluted milk, the concentration of milk in the can becomes 30%. Now, further 40 liters of water is added to the can and the concentration of milk in the can gets reduced by 10 percentage points. How many more liters of water must be added to the can now such that the concentration of milk in the can becomes 8%?
Q) When Sunil will be as old as his father is at present, Sunil will be five times as old as his son is at present. Also, by then Sunil's son will be six years older than the age of Sunil at present. The sum of the ages of Sunil's father and Sunil is 85 years at present. What is the sum of the ages of Sunil's son and Sunil at present?

Q)Let f(x) be a polynomial of degree 51 such that when f(x) is divided by (x - 1), (x - 2), (x - 3),...and (x - 51), it leaves 1, 2, 3,... and 51 respectively, as the remainders. Find the value of f(52) + f(0).

How many integral values of x are possible for which (2^70 + 2^1039 + 2^x) equals the square of a whole number?

A 0 B 1 C 2 D 3

only 1 value
ie (2^35+2^1003)^2 = (2^70+2^1039+2^2006)

so x = 1003

-------------------------------------------------

How many integral values of 'x' are possible for which (2^70 + 2^1039 + 2^x) equals the square of a whole number?

A 0 B 1 C 2 D 3

Distance between A and B is 80 kms. A bus left A at constant speed towards B.30 mins later car left from A to B. Car overtook the bus in 30 mins and continued towards B.With out stopping at B, the car turned back and again encountered the bus 80 mins after he had left A. Speed of bus?

Ans:40 kmph

Plz post the method

My method(its not working though ) :banghead:
Speed of Bus=B
Speed of Car=C

Since bus left 1/2 hr early distance covered by bus is B/2 kms

Now car over takes the bus in 1/2 hr

So (B/2)/(C-B) = 1/2
=>C=2B

When the car overtakes the bus bus would have covered distance of B kms
So remaining distance is 80-B

when car reaches point B bus would have covered half of 80-B
So remaining is (80-B)/2

Given total time car takes to meet bus for second time = 80 mins
So 80/C + /(C+B)= 80 mins
=> 80/2B + /(2B+B)= 4/3 hr

Solving this I get sum stupid values for B :banghead:

Think bhai, can yu point out my mistake

as car takes 30 minutes to cover the distance travelled by bus in 60 minutes
so, speed of car=twice the speed of bus
now in total both travel 160km in 80 minutes
so bus travels 160*1/3 km in 80 minutes
so, speed=160*1/3*60/80=40km/hr

as car takes 30 minutes to cover the distance travelled by bus in 60 minutes
so, speed of car=twice the speed of bus
now in total both travel 160km in 80 minutes
so bus travels 160*1/3 km in 80 minutes
so, speed=160*1/3*60/80=40km/hr

puys plz help me out wid dis prob...find da number of whole no. solutions of da equation 2x+3y+5z=100....

Q-1 How many 2 digit positive integers are there which are one and half times larger than the product of their digits??

Q-2 How many positive integers x between 1 and 1000,both inclusive, is 4x^6+x^3+5 is divisible by 7 ??


Thanx

In bold part, you have made a small mistake.. They meet for the first time after travelling certain distance..
So, we need to remove that from 80 first...

Here is my appraoch:

My answer is 80/3 kmph.

Say, they meet for first time after 'x' km.
Now, car and bus together travel 160 km.

Excluding first x of each, in remaining distance, car travels double the distance of bus.

Now, remaining distance = 160-2x
Bus will travels 1/3 of it and car will trave; 2/3 of it.

So, bus travels (160-2x)/3 = (160/3 - 2x/3) in 80 mins.

Now, bus takes 60 mins for x. So, it will take 40 mins from 2x/3.
So, bus travels 160/3 in 40+80 i.e 120 mins.
So, in one hour bus will travel 80/3 km.

So, speed of bus = 80/3 kmph.

Vc=2Vb

Vc*(80/60)-80+Vb*(110/60)=80.
ans is not cmng.pls tell where d approach s wrong

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13

ans is 1...

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Thread to be closed.

Cheers!

1) One cubic metre piece of copper is melted and recast into a square cross-section bar, 36m long. An exact cube is cut off from this bar. If cubic metre of copper cost Rs.108, then the cost of this cube is?
2) If 'h' be the height of a pyramid standing on a base which is an equilateral triangle of side 'a' units, then the slant height is? (the formula) Please post the methods too.

tarun g Says

Solve in base 7 the pair of equations 2x-4y=33 and 3x+y=31 where x,y and the coefficients are in base 7.

Combining the two equations to get the value of x :-

2x - 4 (31 - 3x) = 33;

Now convert base 7 to decimal.

Eg.

31 = (3 * 7^1 + 1 * 7^0) = 22
lly
33 = 24

All the one digit numbers below 7 remain same

The equation becomes:-


2x - 4 (22 - 3x) = 24;

this yields x= 8.

Now in base 7, we can write 8 (in decimal) as 11.


Hence x = 11..
Similarly find y.


Just to elucidate, successive numbers in the base 7 number system will be : -

0 1 2 3 4 5 6 10 11 12 13 14 15 16 20.....


Hope it helps.
Please point out if there is a faster or better way to do it..

Plz help me out with these problems on Number system

1.Find the remainder of 2^2+22^2+222^2+2222^2+......(222.....49 twos)^2 is divided by 9

2.n=202*20002*200000002*20000....2(15 zeroes)*20000...2(31 zeroes).find the sum digits in this multiplication
a.112
b.160
c.144
d.cant determine

plz provide ur answers with explanation.

Solve in base 7 the pair of equations 2x-4y=33 and 3x+y=31 where x,y and the coefficients are in base 7.

When 2 of a,b,c is 5
possible cases 3C2x2 coz the third no can be 0 nd 1 only

When 1 of a,b,c is 5
possible cases 3C1 x 5 x 5 coz d other no.'s can be anything but 5

When none of the no's is 5
5 x 5 x 5 = 125

So, total = 125 + 6 + 75 = 206...

OA??

Seems to be the correct answer. Thanks

Surely is the correct approach and hence yields the right answer.
Thanks !

Distance between A and B is 80 kms. A bus left A at constant speed towards B.30 mins later car left from A to B. Car overtook the bus in 30 mins and continued towards B.With out stopping at B, the car turned back and again encountered the bus 80 mins after he had left A. Speed of bus?

Ans:40 kmph

Plz post the method
My method(its not working though ) :banghead:
Speed of Bus=B
Speed of Car=C

Since bus left 1/2 hr early distance covered by bus is B/2 kms

Now car over takes the bus in 1/2 hr

So (B/2)/(C-B) = 1/2
=>C=2B

When the car overtakes the bus bus would have covered distance of B kms
So remaining distance is 80-B

when car reaches point B bus would have covered half of 80-B
So remaining is (80-B)/2

Given total time car takes to meet bus for second time = 80 mins
So 80/C + /(C+B)= 80 mins
=> 80/2B + /(2B+B)= 4/3 hr

Solving this I get sum stupid values for B :banghead:

Think bhai, can yu point out my mistake

In bold part, you have made a small mistake.. They meet for the first time after travelling certain distance..
So, we need to remove that from 80 first...

Here is my appraoch:

My answer is 80/3 kmph.

Say, they meet for first time after 'x' km.
Now, car and bus together travel 160 km.

Excluding first x of each, in remaining distance, car travels double the distance of bus.

Now, remaining distance = 160-2x
Bus will travels 1/3 of it and car will trave; 2/3 of it.

So, bus travels (160-2x)/3 = (160/3 - 2x/3) in 80 mins.

Now, bus takes 60 mins for x. So, it will take 40 mins from 2x/3.
So, bus travels 160/3 in 40+80 i.e 120 mins.
So, in one hour bus will travel 80/3 km.

So, speed of bus = 80/3 kmph.

How many values of (a,b,c) are possible such that (a+b+c -6) is a multiple of 4 where all a ,b ,c are <=6 and greater than 0?

Distance between A and B is 80 kms. A bus left A at constant speed towards B.30 mins later car left from A to B. Car overtook the bus in 30 mins and continued towards B.With out stopping at B, the car turned back and again encountered the bus 80 mins after he had left A. Speed of bus?

Ans:40 kmph

Plz post the method

My method(its not working though ) :banghead:
Speed of Bus=B
Speed of Car=C

Since bus left 1/2 hr early distance covered by bus is B/2 kms

Now car over takes the bus in 1/2 hr

So (B/2)/(C-B) = 1/2
=>C=2B

When the car overtakes the bus bus would have covered distance of B kms
So remaining distance is 80-B

when car reaches point B bus would have covered half of 80-B
So remaining is (80-B)/2

Given total time car takes to meet bus for second time = 80 mins
So 80/C + /(C+B)= 80 mins
=> 80/2B + /(2B+B)= 4/3 hr

Solving this I get sum stupid values for B :banghead:

Think bhai, can yu point out my mistake

In a number system the product of 44 and 11 is 3414. The number 3111 of this system, when converted to the decimal number system, becomes

The answer is 406.

Did hit and trial to find out the base..
484 % 5 = 4
96 % 5 = 1
19 % 5 = 4
3 % 5 = 3

Hence the base is 5.

3111 = 3X5^3 + 1X5^2 + 1X5 + 1 = 375 + 25 + 5 + 1 = 406

In a class of 12 students , 7 boys and 5 girls.The class has 4 sessions each day,one each of arithmetic ,algebra,geometry and probability.These classes are to be held one after the other in 4 distinct time slots and can be in any sequence .Further there are 2 teachers available and they can teach any topic.
In how many distinct ways can a session be scheduled for two consecutive days if both the teachers have to have equal no. of sessions..?

I tried doing it lyk dis:
session arrangement for 1st day -4!
session araangement for second day-4!
teachers choosing subject - 2^4=16
so 4!*4!*16=9216...which is wrong!!!!!!!!!!!!!!!!!!!!

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laugh-a-riot Says

Yes Answer is 36..but can u pls explain how do we arrive at that??..Thnx in advance..

You have to make Cases:

Case 1: x>0,y>0 and x>y
Eqn reduces to x=3

Case 2: x>0, y>0 and x<0,y<0 and x>y
Eqn reduces to y=-3

Case 4: x<0, y<0 and x36 units

From remaining 80 numbers i.e. 77 + 3 numbers, 11 + 1 numbers are divisible by 7.

@think can u explain the bold part plz

We will have 80 remaining numbers i.e 3 + 77 remaining numbers.

Now, out of 77 numbers, 11 are surely divisible by 7.
Now, we just have to check if first 3 numbers have any one divisible by 7.
203 will there as it is not divisible by 2, 3 or 5 but it is divisible by 7.
So, 1+11 = 12 such numbers..

I should have written 3+77 as it looks simpler..

varun.tyagi Says

Is it 36 units

Yes Answer is 36..but can u pls explain how do we arrive at that??..Thnx in advance..

Compiled question and solution for reference

find the area bounded in the x-y plane by the curve x+y+x-y=6. Please can someone explain the approach?
a.24
b.36
c.49.
d.48

my take 36

find the area bounded in the x-y plane by the curve x+y+x-y=6. Please can someone explain the approach?
a.24
b.36
c.49.
d.48

Is it 36 units

It is 68.

We have 300 numbers (from 201 to 500, ignore 200 as it is divisible by 2)
So, 150 are divisible by 2.
From remaining 150 odd numbers; 50 (i.e 1/3 of 150) are divisible by 3.
From remaining 100 numbers, 20 are divisible by 5.
From remaining 80 numbers i.e. 77 + 3 numbers, 11 + 1 numbers are divisible by 7.

So, totally, 150+50+20+12 = 232 numbers out of 300 are divisible by atleast one of 2, 3, 5 and 7.

Remaining are 68 which is the needed answer.

@think can u explain the bold part plz

find the area bounded in the x-y plane by the curve x+y+x-y=6. Please can someone explain the approach?
a.24
b.36
c.49.
d.48