Number System - Questions & Discussions

MBA CAT 2022
1329
Hello everyone,


I have grt amount of difficulty while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??

Lets take fr Eg:: 3218/5439

How will be your approach to this???

For such kind of examples, you can work with approximations. You'll need to multiply a bit, but I guess multiplication is way easier than handling such long divisions.

Taking your example of 3218/5439, we first assume 3218/5000. This is pretty simple and the answer is 0.6436 (you don't need to actually divide this. Just take the double of 3218 in this case).

Now since we've approximated, we have to compensate as well. Since 3218/5439 would obviously be less than 3218/5000 (which we found earlier), we need to subtract 439/5439 from 0.6436.

Now before you start thinking that the task is actually getting tougher, just read on 😃 We're just dealing with approximations here and are fine with a few points of difference here or there.

We know that 1% of 5439 = 54.39. So we can safely assume that 8% would give us somewhere around 439. We're not interested what would be the exact value of 8% of 5439.

So just take 8% of 0.6436 = 0.051488

And subtract this from 0.6436 = 0.5921.... This is the final answer :)

If you check with a calculator, the answer is 0.5916.... pretty close :)

This method may seem cumbersome but believe me, once you know what to do and have sufficient practice, you can do it quickly enough. I would just reiterate the steps again:

1. 3218/5439 = ?
2. 3218/5000 = 0.6436
3. 439/5439 = 8%
4. 8% of 0.6436 = 0.051488
5. 0.6436 - 0.051488 = 0.5921

Hope you understood the method
1330

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

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

option 2..

1331
find the remainder when (38^16!)^1777 is divided by 17

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

option 2..

i am getting 1 as ans...this can be written as 38^16k mod 17 and since E(17)= 16 also since 38 and 17 are co prime, rem of 38^16k/17 shud be 1..
1332

Ramu has to guess his friend's 8 digit phone number. He remembers the following.
(1) the first four digits are either 2756 or 2766.
(2) the number is odd.
(3) the digit appears exactly once.
What is the maximum no of distinct trails he may have to make before he gets the correct number?
1 . 3401
2 . 2792
3 . 3524
4 . 2435
5 . None of these

1333
CATch_22 Says
i am getting 1 as ans...this can be written as 38^16k mod 17 and since E(17)= 16 also since 38 and 17 are co prime, rem of 38^16k/17 shud be 1..


it shud be 1...Euler theorem can be applied directly...
1334

find the remainder when (38^16!)^1777 is divided by 17
38 ^( 16!x1777) mod 17
= 4 ^(16!x1777) mod 17
=1 mod 17
since 4 mod 17=4
16 mod 17=16
64 mod 17 =13
52 mod 17 =1
thus cycle is 4
answer is 1

1335

guys can some one teach the eulers method for finding the remainder.. plz

thanx in advance

1336

i am new here.
so, bear with any ignorance on my part.

is the answer 28?

seems like i have gatecrashed into the wrong question, i was referring to another one.
this (the getting-used-to) will take more than a while. 😞

1337

i was referring to the other question.

1338
Hello everyone,


I have grt amount of difficulty while i am solving DI section especially bcos of long divison. CAn u pls help me with different methods of divison you are using??

Lets take fr Eg:: 3218/5439

How will be your approach to this???


a very simple technique which i use is
first i see the number
here it is slight greater thn 3000/greater than 5400
now we use appx..
first step:-32/54=.592
now if the options are not close u will have no difficulty in eliminating it..
otherwise
since the denominator increase is 39 after 54 (54--39) while the numerator is increasing by 18(32--1..so denominator increase is definitely more

so the ans will decrease very slightly..


now again if the qstn comes like this
3789/5439

take numerator as round 38/denominator as 54..u will definitely get close to the number
1339
find the remainder when (38^16!)^1777 is divided by 17

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

option 2..


i'm getting ans 1 too...E(17)=16
SO REMAINDER IS 1
1340
guys can some one teach the eulers method for finding the remainder.. plz

thanx in advance


Euler's method:

(X)^A
------ (REM) = 1
Z

provided
1) X and Z are coprimes to each other, that is there is no common factor other than 1.
2) A is the Euler's number of Z.

To find Euler's Number.

Let Z = b^m*c^n*d^p.... where (b,c,d...are the prime factors)

Then Euler number for Z is given by = Z (1-(1/b))*(1-(1/c))*(1-(1/d))....

In other words, Euler's Number of a number is nothing but the number of coprimes to the number counted from 1 till the number-1. Confused, see below example.:D (if you understand once, remainder questions will be 30 secs question for you:D)

Lets take an example.

Euler's Number of 6:

6 = 2*3

therefore Euler's number is = 6 * (1-1/2) * (1-1/3) = 6*(1/2) * (2/3) = 2
Using the second definition, Number of coprimes to 6 from 1-5 are 1,5 = 2 (Fermet Little number)

For prime numbers Euler's number = Number -1 (Fermet Little number)

Therefore Euler's number for 17 = 16.
If you count from 1-16 , all are coprime to 17, hence the Euler's number of 17 is 16.

One major application of Euler's number.
Suppose Question comes that - How many numbers are there which are less than 450 NOT containing the multiples of 2,3 and 5

The Usual method is to find the number of multiples of 2 till 450(A), 3 till 450(B),5 till 450(C),6 till 450(D),10 till 450(E),15 till 450(F),30 till 450(G)

and then use the formula = 450 - (A+B+C-D-E-F+G).

Using Euler's number = 400 (1-(1/2))*(1-(1/3))*(1-(1/5)) = 450*1/2*2/3*4/5 = 120. (But this method can be used only when the number is a multiple of the numbers mentioned, that is 450 is a multiple of 2,3,5)

Here I am using the second definition of Euler's number.

Hope it helps:D
1341
Ramu has to guess his friend's 8 digit phone number. He remembers the following.
(1) the first four digits are either 2756 or 2766.
(2) the number is odd.
(3) the digit appears exactly once.
What is the maximum no of distinct trails he may have to make before he gets the correct number?
1 . 3401
2 . 2792
3 . 3524
4 . 2435
5 . None of these


I am getting option 5.

My approach.

if the number we consider is 2756
as the number is odd, the last digit I can fill up by 1/3/9 (5 and 7) is already used.
5th,6th and 7th can be filled with 0/1/3/4/8/9.
Therefore 8th place can be filled by 3 ways,
7th place can be filled by (2+3) = 5 ways (as after filling 8th place, I am left with only 2 odd numbers)
6th place = 4 ways and 5th place = 3 ways.

Therefore 3*4*5*3 = 180 ways

Similarly for 2766, I am getting 4*5*6*4 = 480 ways. (I am considering that rest numbers are not repeating)

Therfore the maximum number of attempts required is 480+180 = 660. In the 660th attempt, the person in question will guess the correct phone number.

Let me know if I am correct.
1342

Puys,

The following was one of my most favourite question.

What is the remainder of (6^83 + 8^83)/49

Options
1) 0
2) 35
3) 14
4) 48
5) 42

1343
Puys,

The following was one of my most favourite question.

What is the remainder of (6^83 + 8^83)/49

Options
1) 0
2) 35
3) 14
4) 48
5) 42



solved many times in same thread .. remainder is 35
remainder = -6 - 8 = -14 = 35
1344
Puys,

The following was one of my most favourite question.

What is the remainder of (6^83 + 8^83)/49

Options
1) 0
2) 35
3) 14
4) 48
5) 42



Lets solve it for one more time.

E(49)=42.

Let 6^83=x
8^83=y

Now 6x=49n+1 so n=5, =>x=41
Similarly 8y=49b+1 so b=7 => y=43

Putting the value of x and y.

(41+43) % 49 Ans is 35.
1345
Puys,

The following was one of my most favourite question.

What is the remainder of (6^83 + 8^83)/49

Options
1) 0
2) 35
3) 14
4) 48
5) 42


(6^83 + 8^83)/49 using reverse eular theorem
(6^-1 +8^-1)/49=(41 +43)mod 49=35 mod 49
1346

a, b, c, d and e be non-negative real numbers such that a + b + c + d + e = 10. Let X be the maximum of the numbers a + b, b + c, c + d and d + e. The least possible value of X lies in the interval






1347
a, b, c, d and e be non-negative real numbers such that a + b + c + d + e = 10. Let X be the maximum of the numbers a + b, b + c, c + d and d + e. The least possible value of X lies in the interval








The least value of X will be for

a = c = e = 10/3 and b = d = 0

So Least Value lies in
1348
The least value of X will be for

a = c = e = 10/3 and b = d = 0

So Least Value lies in


I am just trying to understand working principle.

You are trying to determine which number we can maximise out of a+b, b+c,c+d,d+e and keeping the rest at 0.

as a,c,e are there in those 5 sums, we are trying to maximise a,c,e keeping the constraints of a+b+c+d+e = 10.

Therefore a= c=e =10/3 and rest b=d = 0

is this the way you did it?