Number System - Questions & Discussions

MBA CAT 2022
1553

Could anyone help me in following question
Q. Find the remainder of
2133
divided by 133.
This is the time practice test question with 128 answer.
Please explain the solution in detail.

Thanks in advance

1554

This is 2 raise to power 133 divided by 133

1555
Could anyone help me in following question
Q. Find the remainder of
2133
divided by 133.
This is the time practice test question with 128 answer.
Please explain the solution in detail.

Thanks in advance


2^133 / 133

133 = 19*7

2^133 % 19 = 2^18k * 2^7 % 19 = 14 ( fermat theorem and remainder thm application )

2^133 % 7 = 2^6k * 2 % 7 = 2 ( fermat theorem and remainder thm application )



so by chinese remainder thm ,

19a + 14 = 7b + 2


so final remainder is 128 ..
1556
2^133 / 133

133 = 19*7

2^133 % 19 = 2^18k * 2^7 % 19 = 14 ( fermat theorem and remainder thm application )

2^133 % 7 = 2^6k * 2 % 7 = 2 ( fermat theorem and remainder thm application )



so by chinese remainder thm ,

19a + 14 = 7b + 2

so final remainder is 128 ..


Hi,

Can you explain please how you got 128 after 19a+14=7b+2..?
1557
Hi,

Can you explain please how you got 128 after 19a+14=7b+2..?


19a + 14 = 33,52,71,90,109,128

7b + 2 = 9,16,23,30,37,44,51,58,65,72,79,86,93,100,107,114,121,128

common to both is 128 .. hence the answer
1558

What is the unit digit of 7^5^6^13.

Its 7 or 9

1559
What is the unit digit of 7^5^6^13.

Its 7 or 9


5^ anything will end in 25 and so is in the 4k+1 format.

So, 7^(4k+1)

will give the last digit as 7.
1560
Hey puys ,

Please help on these problems.
Source from Biju material.

Question :1
find b in terms of a.Given that (4096)^0.5=(ab^3)^1%3.
a)sqr rt a b)a^2 c) a^8 d) a^4

Question :2
Find the sum of the series N=1/3+1/15+1/35+.......100 terms
a)99/199 b)1/199 c)100/201 d)none of these

Question :3
A fibonacci series is a number series in which the nth term is given by
Tn=Tn-1+Tn+1 .Also T1=1 T2=1

the 66 th term of the series is
a)odd b)even c)cannot be determined d) none of these

Question:4

N=(323232...50 digits )in base 9.Find the remainder when the number is divided by 8?

a)1 b)-1 c)0 d)none of these

Please post the detail solution.
Thanks in Adance

Regards
Rajiv


Answer to Q no. (2) is 100/201

This is how to go about it:
Sum of first number=1/3
Sum of first two numbers=1/3 + 1/15 =6/15 =2/5
Sum of first 3 numbers=1/3 + 1/15 + 1/35 =3/7
Observation:
Sum of first n numbers=n/(2n+1)

So, sum o first 100 numbers would be 100/201

Hope this is fine!!
1561

Please solve this:
10000! = (100!)K P, where P and K are integers. What can be the maximum value of K?
give the solution also

options are:
a. 104 b. 103 c. 102 d. 105


Thanks

1562

ans to Q4

In base 10 , remainder to 9 will be the digit sum of the divident. Simillarly in base 9 , remainder to 8 will be the digit sum..So 323232...fifty digits, digit sum will be 5*25 = 125= 8, therefore ans 0

Hope this helps

1563

ans to Q4

In base 10 , remainder to 9 will be the digit sum of the divident. Simillarly in base 9 , remainder to 8 will be the digit sum..So 323232...fifty digits, digit sum will be 5*25 = 125= 8, therefore ans 0

Hope this helps

1564
brij: what kinji@PG says is perfectly right.....only that u cannot use euler here as 1990(denominator) and 2( the number whose power is being raised) are not coprimes....so break 1990=199*10( product of coprimes)

now 2^1990/(199*10)
consider: 2^1990/199 = remainder 29
consider 2^1990/10 = remainder 4

here say 199=a that gives remainder 29 and 10=b that gives remainder 4
so chinese says: remainder of 2^1990/ 199*10 = a(4)x+b(29)y where ax+by=1 i.e 199x+10y=1, we get x=-1,y=20

substituting , we get remainder is
199(4)(-1)+10(29)(20) = 5004
but since 5004 in greater than 1990, we divide it by 1990 to get the remainder we get 1024.


hi... can you please tell me how this happened ?
2^1990/199 = remainder 29
Also ... where can I find chinese rem. theorem?
I've an idea abt CRT bt I'm confused?
PLs help...
1565

Hi

Can anyone please help me with finding the last two digits type of problems. As I understand it requires us to divide the number by 100 and finding the remainder. But sometimes things can get really long is there any shortcut.

Lets use two examples from Arun Sharma

Find the last two digits of the given numbers.

a) 101X102X103X197X198X199

b) 65X29X37X63X71X87X62

Someone please help !

1566
Hi

Can anyone please help me with finding the last two digits type of problems. As I understand it requires us to divide the number by 100 and finding the remainder. But sometimes things can get really long is there any shortcut.

Lets use two examples from Arun Sharma

Find the last two digits of the given numbers.

a) 101X102X103X197X198X199

b) 65X29X37X63X71X87X62

Someone please help !


a) divide the numbers separately :

you get remainders as 1,2,3, -3,-2 ,-1

multiply them u get 36

b)
first take out 5 both from 65 and 100

so now we have 13*29*37*63*71*87*62 MOD 20

after you solve it...finally multiply 5 again to get the last two digits

Hope im clear
1567
Please solve this:
10000! = (100!)K P, where P and K are integers. What can be the maximum value of K?
give the solution also

options are:
a. 104 b. 103 c. 102 d. 105


Thanks


The largest prime no in 100! is 97. Thus it is enough to find the exponent of 97 in 10000! ie 104.

So K=104.
10000!=(100!^104)*P
1568
hi... can you please tell me how this happened ?
2^1990/199 = remainder 29

PL:



Hi ! manmishr...

Use Euler's theorum here :

And know this that for a prime number 'P' the euler's number is '(P-1)'

So 199 being a prime number,its euler number is =198

and since 2 and 199 are co-prime,we can use the euler's theorem here

therefore ;

2^1980 * 2^10 MOD 199= 1*1024 MOD 199 = 29
1569
Hi,

Can you explain please how you got 128 after 19a+14=7b+2..?


Dear,

we can solve with fermt theorem only

2^108/133=1 so 1*2^25/133=256^3*2/133=(-10)^3*2/133=(-10)*200/133=(-10)67/133=123*67/133=8241/133=remainder 128

If not clear please mention.
1570
Answer to Q no. (2) is 100/201

This is how to go about it:
Sum of first number=1/3
Sum of first two numbers=1/3 + 1/15 =6/15 =2/5
Sum of first 3 numbers=1/3 + 1/15 + 1/35 =3/7
Observation:
Sum of first n numbers=n/(2n+1)

So, sum o first 100 numbers would be 100/201

Hope this is fine!!



Dear ,

my answer also same

1/2(1-1/3)+1/2(1/3-1/5)+................+(1/199-1/201)=1/2(200/201)=100/201
1571
hi... can you please tell me how this happened ?
2^1990/199 = remainder 29
Also ... where can I find chinese rem. theorem?
I've an idea abt CRT bt I'm confused?
PLs help...


Dear,
with euler theorem u can find above ,see euler no will be 198 for 199 so 2^198/199 =1 so 2^1980/199=1
2^1990/199=(2^1980/199)*(2^10/199)=1*29=29

i think it is clear now.
1572
Hi

Can anyone please help me with finding the last two digits type of problems. As I understand it requires us to divide the number by 100 and finding the remainder. But sometimes things can get really long is there any shortcut.

Lets use two examples from Arun Sharma

Find the last two digits of the given numbers.

a) 101X102X103X197X198X199

b) 65X29X37X63X71X87X62

Someone please help !


a) last two digits is the remainder when divided by 100

So 1X2X3X97X98X99 is equivalent to 1x2x3x-3x-2x-1 = -36 therefore remainder is 64...hence the answer will be 64...

b)Use the same negative remainder concept to reduce the bigger number to smaller numbers and then multiply and get the answer. will not take more than 2 minutes...ans i got is 90...

Hope this helps