Quant by Arun Sharma

MBA CAT 2024
2278
How is it possible to get 0 there...It must be 1 only...Correct me if 'm wrong..
I thought 1 term is (6*1*1) and second term is (4*5*1)....

the correct approach is
6*1*1-4*5*1=6-0
hence the answer is 6
u hav 2 consider d unit digit of the product which is 20
2279
the correct approach is
6*1*1-4*5*1=6-0
hence the answer is 6
u hav 2 consider d unit digit of the product which is 20

Hi

Infact i go with the same view.

It has to be
6*1*1-4*5*1=6-0 only :)

Happy Learning !
2280

th last two digits in the multiplication of 35*34*33*32*31*30*29*28*27*26?
00
40
30
10
none of these

how do i use the eulers method to solve a question?
thanx in advance

2281
th last two digits in the multiplication of 35*34*33*32*31*30*29*28*27*26?
00
40
30
10
none of these

how do i use the eulers method to solve a question?
thanx in advance


There is a 0 in 30 and a 5 and an even number.So,last 2 digits have to be 00

option (1)

Some literature on Euler Theorem here:

Euler's theorem - Wikipedia, the free encyclopedia

Basically if we have to find the remainder of

a^n mod b

where a and b are coprime,and b's Euler number is say x.

Now,if n is a multiple of x then the remainder is 1.

Difficult?Lets see an example

Find the remainder of 29^72 mod 11

Now,Euler number of any integer n=A^x*B^y*C^z where A,B,C are the prime factors of n is given by:

n(1-1/A)(1-1/B)(1-1/C)

For prime numbers,Euler number is equal to (n-1)

So,Euler number of 11 is 10.

So,as 29 and 11 are co prime.If the numerator is of the format 29^10x,then the remainder will be 1.

So, 29^70 mod 11 is 1

So,it effectively becomes 29^2 mod 11

841 mod 11

the answer is 5.

Now,there are some variations to these kind of sums,you can find them on the net.

EDIT: Coprime numbers are those which do not share any common factors except 1.
2282
the unit digit in the expression 36^234* 33^512* 39^180 - 54^29* 25^123* 31^512.
options:-
8
0
6
5
4

I keep getting the answer as 4. correct answer(bk) 6.
plz explain.
thanx in advance :)


units digit of 36^234=6
units digit of 33^512=1
units digit of 39^180=1
6*1*1=6
units digit of 54^29=4
units digit of 25^123=5
units digit of 31^512=1
units digit of(4*5*1)=20=>0
hence 6-0=0
2283

A quantity P varies directly as the sum of Q and R..If Q increases by 2 and R increases by 2,then how much does P increase???

2284

the remainder when 1234567891011......484950 is divided by 16 is
3
4
5
6

find the remainder when the no represented by 22334 raised to power (1^2 + 2^2 +3^2+.+66^2) is divided by 5
2
4
0
none of these

Arun, bikas and chetakar have a total f 80 coins among them. arun triples the number of coins with the others by giving them some coins from his whole colection. next, bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at the beginning?
11
10
9
12

here i got the answer as 20. is that correct??

thanx in advance 😃

2285
the remainder when 1234567891011......484950 is divided by 16 is
3
4
5
6


divisibility test of 16 is finding whether the last 4 digits are divisible by 16 or not.

4950 mod 16

remainder comes out to be 6.

find the remainder when the no represented by 22334 raised to power (1^2 + 2^2 +3^2+.+66^2) is divided by 5
2
4
0
none of these


22334^(1^2+2^2+...66^2) mod 5

(22330+4)^(1^2+2^2...) mod 5

4^(1^2+2^2+...66^2) mod 5

now,1^2+2^2+...66^2 is 11*67*133 which is odd

And,4^odd number mod 5 is always 4

So,answer should be 4.

Arun, bikas and chetakar have a total f 80 coins among them. arun triples the number of coins with the others by giving them some coins from his whole colection. next, bikas repeats the same process. after this bikas now has 20 coins. find the number of coins he had at the beginning?
11
10
9
12

here i got the answer as 20. is that correct??

thanx in advance :)


let they have a,b,c coins at the start

a+b+c=80

A has (a-2b-2c)
B has 3b
C has 3c

after second transaction,

A has 3(a-2b-2c)
B has 3b-2a+4b+4c-6c
C has 9c

7b-2a-2c=20

2a+2b+2c=160

9b=180

b=20

So,you are right here 😃
2286
😃 I cant believe i missed the 1st one. thanxx


~~~~now,1^2+2^2+...66^2 is 11*67*133 which is odd~~~~ i didnt get that. can u xplain again??
thanx a ton
2287
:) I cant believe i missed the 1st one. thanxx


~~~~now,1^2+2^2+...66^2 is 11*67*133 which is odd~~~~ i didnt get that. can u xplain again??
thanx a ton


That is the formula to calculate the sum of squares of first n natural numbers which is given by

n(n+1)(2n+1)/6

as 11,67,133 are all odd,the resultant has to be odd.

Hope that helps 😃
2288

find the remainder when the no represented by 22334 raised to power (1^2 + 2^2 +3^2+.+66^2) is divided by 5
2
4
0
none of these



From fermat little theorem
U know 22334^4/5= 1
now expanding 1^2 + 2^2 +3^2+.+66^2 u get 98021
98021 can be put in the form 5k+1
so 22334^(4k+1)/5 = 4^3/5 so answer is 4..

To find reminder learn Fermat little theorem and Euler theorem, two most effective tool to find reminder.
2289

what should be added to two numbers which are in the ratio 4:5 so that the ratio becomes 5:6????"m getting 1 but OA says it cant be determined...Can someone tel which is the correct answer and approach too...

2290
Ganu02 Says
what should be added to two numbers which are in the ratio 4:5 so that the ratio becomes 5:6????"m getting 1 but OA says it cant be determined...Can someone tel which is the correct answer and approach too...


a/b=4/5

(a+x)/(b+x)=5/6

6a+6x=5b+5x

5b-6a=x

20b-24a=4x

a=4x

So,if x=1;a=4,b=5
if x=2;a=8,b=10
and so on.

So,it cannot be determined.
2291
How is it possible to get 0 there...It must be 1 only...Correct me if 'm wrong..
I thought 1 term is (6*1*1) and second term is (4*5*1)....

yup..u r rite..i mde a mistak whle writing..:-(
2292
Ganu02 Says
what should be added to two numbers which are in the ratio 4:5 so that the ratio becomes 5:6????"m getting 1 but OA says it cant be determined...Can someone tel which is the correct answer and approach too...

here..the ratio of 2 no's is given.the no's can be anything..
for 4:5 =>no's can be (4,5),(8,10),(12,15)..and so on..as no increases the additive no will also increase..and hence the ans..:)
2293
:) I cant believe i missed the 1st one. thanxx


~~~~now,1^2+2^2+...66^2 is 11*67*133 which is odd~~~~ i didnt get that. can u xplain again??
thanx a ton
ans shuld be none of this..m getting 1..
=>22334^(1^2+2^2+3^2+4^2+0+1^2...up to 13 terms+1^2)mod 5=22334^((13*13)+1)mod 5=22334 mod 5=4
2294
ans shuld be none of this..m getting 1..
=>(1^2+2^2+3^2+4^2+0+1^2...up to 13 terms+1^2)mod 5=((13*13)+1)mod 5=1


dude when u expand 1^2+2^2+...66^2

u get n(n+1)(2n+1)/6
= 66(67)(133)/6
=11(67)(133)=98021

From format little theorem u know 22334^4/5= 1
we need to find the reminder of 22334^98021/5 can be put in the form =22334^(4k+1)/5
=(22334^4k x 22334^1)/5
here 22334^4k/5 = 1(from format theorem we just arrived at)
so 1 x 4 = 4 is the reminder
hope that helps
2295
dude when u expand 1^2+2^2+...66^2

u get n(n+1)(2n+1)/6
= 66(67)(133)/6
=11(67)(133)=98021

From format little theorem u know 22334^4/5= 1
we need to find the reminder of 22334^98021/5 can be put in the form =22334^(4k+1)/5
=(22334^4k x 22334^1)/5
here 22334^4k/5 = 1(from format theorem we just arrived at)
so 1 x 4 = 4 is the reminder
hope that helps

thnx yaar..
i dint see that no 22334..
2296

Hi,

I am struggling solving inequality questions corresponding to modulus function.
I am mentioning one problem from LOD1.

1. (mod(x-1) -3)(mod(x+2) -5)
I am able to break into different parts and tried to evaluate answers corresponding to different scenarios.

I want to know how to evaluate answers from different scenarios.Sometime its confusing if we have so many different scenarios.

Please suggest me approach.

Thanks in advance

2297

dear it is summtion N^2=n(n+1)(2n+1)/2 so it will be 66*67*133/2=33*67*133 =odd
i think it is clear