Number System - Questions & Discussions

MBA CAT 2022
3578
@Estallar12 said:
LCM (20, 24, 40) = 480. (Total units of work).Total Working Hours = 12*8 = 96.=> 480/96 = 5 units per working hour.So, 5/20, 5/24 and 5/40 for Man, Woman and Boy. For the New Task -=> x*(1/4) + 6*(5/24) + 2*(1/8) = 480/60=> x/4 = 8 - (1/4 + 5/4) = 8 - 3/2 = 13/2=> x = 26 Men.
bro 480/60 kaise aaya???
3579
What is the remainder when 1*1+11*11+111*111+1111*1111+ .....+ (2001 times 1) * (2001 times 1) is divided by 100?
3580


@sumit99 said: bro 480/60 kaise aaya???

For the task,
total unit.=480
and
total work time =60 hrs ( 12 days*5hr/day)
3581


@sumit99 said:If 3 cats catch 3 rats in 3 minutes, how many cats will catch 100 rats in 100 minutes?

100 cats?
I know it can't be so easy.
Someone Please share the approach if solved.
3582
@cadmium said:
@sumit99 said:If 3 cats catch 3 rats in 3 minutes, how many cats will catch 100 rats in 100 minutes?100 cats?I know it can't be so easy.Someone Please share the approach if solved.
3 cats only.

3 cats take 3 minutes to catch 3 rats. That means each cat takes 3 minutes to catch one rat.

Now we have 100 rats to be caught. So one cat would catch 100/3 rats in 100 minutes. So for 100 rats, we would need 100/(100/3) = 3 cats. (Assuming the 100th rat is being eaten by all the three rats simultaneously. yucks! )
3583

1. what is the remainder when 2^1100 is divided by 101?

2. What is the remainder when 971(30^99+61^100)*(114 8)^56 is divided by 31?

Can anyone help please?

3584


@Quantohelp said:So one cat would catch 100/3 rats in 100 minutes.

How come know this ?
3585


@Quantohelp said:So one cat would catch 100/3 rats in 100 minutes.

How we came to know this ?
3586

can someone help me with solving the below question:


Find the remainder in the following cases:
1. 54^124 divided by 17
2. 67^99/7
3587
@mba.keeda said:
can someone help me with solving the below question:Find the remainder in the following cases:1. 54^124 divided by 172. 67^99/7
Use euler:

E(17) = 16

124 = 16*7+ 12

So 54^124/17 is same as 54^12/17 = 3^12/17 = 27^4/17 = 10^4/17 = 100^2/17 = 15^2/17 => 225/17 => 4..ANS :)


For 67^99/7 euler co-eff is 6 [E(7) = 6]

given equation is: 67^3/7 = 4^3/7=> 1..ANS 😃
3588
@bpolagani Remainder when 2^1100 is divided by 101
Since the divisor 101 is prime make use of the Fermat's rule.
Fermat's theorem says that if p is prime and p does not divide a, then a^(p-1) when divided by p the remainder is 1.
Hence, 2^100 divided by 101, the remainder is 1 => 2^1100 when divide by 101 the remainder is 1.

Remainder when 971(30^99+61^100)*(114 8)^56 is divided by 31
Consider 30^99 + 61^100
30 divided by 31, the remainder is -1 and hence 30^99 divided by 21 the remainder is -1
61 divided by 31, the remainder is -1 and hence 61^100 divided by 31, the remainder is 1.
Thus the remainder of 30^99 + 61^100 divided by 31 is -1+1 = 0
Hence the remainder of the given product must also be 0

3589

is there any separate link of number system , where i will get all the question in a bunch?

3590

Can somebody please help with this question??


A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits.
Out of all such possible two-digit numbers, a number is randomly picked. What is the probability that this number is divisible by 4?

1. 3/8
2. 5/12
3. 2/7
4. 7/12

Thanks in advance

3591
@shreyaverma86 what's the OA??
2/7 ??
3592

@saurav205 Answer is 3/8 ... Here is the solution which somebody helped out with 😃 10a+b mod a+b = k ---- i
10b+a mod a+b = k ---- ii
ii - i
9(a-b) mod a+b = 0
the numbers satisfying the condition are 12,21,18,81,36,63,45,54,15,51,24,42,27,72,48,84 out of which only 12,36,24,72,48 and 84 are divisible by 4.. therefore req probability is 6/16=3/8
3593
@shreyaverma86 said:
Can somebody please help with this question??A two-digit number having distinct digits when divided by the sum of the digits gives the same remainder as when a two-digit number that is formed by reversing the digits of original number is divided by the sum of the digits. Out of all such possible two-digit numbers, a number is randomly picked. What is the probability that this number is divisible by 4?1. 3/82. 5/123. 2/74. 7/12Thanks in advance
Interesting question:
Lets say the number is of the form AB
then 10A + B = X1 * (A+B) + R
and 10B + A = X2 * (A+B) + R
or 9(A-B) = (X1-X2) * (A+B)

A-B cannot be >= 9

Considering A-B = 1, 9 has to factored into (X1-X2) and (A+B)

So (X1-X2) can be 1, then numbers are 54 and 45
if (X1-X2) is 3, numbers are 12 and 21
if (X1-X2) is 9, no two digit numbers are possible.

Again lets take a case of (A-B) > 1 lets say (A-B) = 2

then (X1-X2) and (A+B) can be factors of 18, which is (X1-X2) = 1 (not possible as A+B cannot be 18 ), 3, 6, 18 (not possible as A+B = 1)

If (X1-X2) = 6, then we get A and B as fractions so only possibility is A+B should be = 6
The numbers are : 42 and 24

Lets consider A-B = 3, then only (X1-X2) = 3 is possible
Numbers are 63 and 36

if A-B = 4, then 36 and (X1-X2) = only 6 is a possibility.
Numbers are 51 and 15

if A-B = 5 then (X1-X2) = 5 and
the numbers are 72 and 27

If we see a pattern, when there is a odd number, only one possibility is there, hence if A-B = 7, numbers are 81 and 18

if A-B > 6 or any even numbers, there are no such numbers possible.

hence the number of numbers divisible = 12, 24, 36, 72, 48, 84

Hence 6/16 or 3/8

P.S. Any shorter method would be appreciated.

Edit : Missed out one condition when A-B = 4 and X1-X2 = 3, Numbers are 48 and 84

3594
@shreyaverma86 said:
@saurav205Answer is 3/8 ... Here is the solution which somebody helped out with 10a+b mod a+b = k ---- i10b+a mod a+b = k ---- iiii - i9(a-b) mod a+b = 0the numbers satisfying the condition are 12,21,18,81,36,63,45,54,15,51,24,42,27,72,48,84 out of which only 12,36,24,72,48 and 84 are divisible by 4.. therefore req probability is 6/16=3/8
I dont understand why they have
@kinji.at.pg1 said:
How can 48 and 84 satisfy? 48 is divisible by 12 but 84 is not.
12*7 = 84..
3595
@saurav205 said:
I dont understand why they have 12*7 = 84..
he he he Offcourse....Severe lack of practice....
3596

pls anyone explain me remainder theorm ..how to calculate remainders 😞 😞 :'(

3597
@Ankurb said:
Let's start things afreshQIf n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) areI. 18 II. 28 c. 38a. Only Ib. Only I & IIc. Only IId. Only I, II & IIIe. None of the above
c....????