Number System

[swati_swati]

Originally Posted by GauravShah

2000/13 gives remainder as 11.

Now 11^1000 = 121^500

121/13 gives remainder as 4

i.e to find the remainder of 4^500/13 = 2^1000/13

Remainder cycle for 2/13 is

2^1/13 = 2
2^2/13 = 4
2^3/13 = 8
2^4/13 = 3
2^5/13 = 6
2^6/13 = 12
2^7/13 = 11
2^8/13 = 9
2^9/13 = 5
2^10/13 = 10
2^11/13 = 7
2^12/13 = 1
2^13/13 = 2

i.e a cycle of 12.
1000/12, remainder is 4.

i.e. same as 2^4/13, heance remainder for 2000^1000/13 is 3

Gaurav.

2000^1000/13=(2002-2)^1000/13=(-2)^1000/13
since even power,henceitwill be2^1000/13
and rest will bsame as above.

[swati_swati]

Originally Posted by tackledude

Guys can u suggest me a good book for QA. Iam planning to join a good institute in Chennai, any suggestions!

Regards,
PREM

hi!prem
if u want to start with abhijit guha is good one to start.right from easy to tough ones for starters.after u complete it ,try arun sharma from TMH publication.moreover career launcher material is best for quant.
ALL THE BEST!

[tackledude]

Originally Posted by swati_swati

hi!prem
if u want to start with abhijit guha is good one to start.right from easy to tough ones for starters.after u complete it ,try arun sharma from TMH publication.moreover career launcher material is best for quant.
ALL THE BEST!

THX u so much........R u from Chennai?

[Homer]

Originally Posted by tackledude

What is the remainder of 2000^1000 divided by 13?

Reminds me of a question from CAT...anyway here is another way of doing it.

2000^1000/13 = (2002-2)^1000/13 =.... (-2)^1000/13 = 2^1000/13
= 2^(6x166+4)/13
= 64^166 . 16 /13 = (65-1)^166. 16 /13 = [{....(-1)^166}.16] / 13
= ....16/13 = 3 as remainder.

Hope i managed to explain well....

[swati_swati]

Originally Posted by tackledude

THX u so much........R u from Chennai?

hi
sorry but i'm not from chennai.i'm studying in pantnagar(uttaranchal).b.s.c3rdyear.but does it makes any difference.(just joking!!).

[tackledude]

Q:1 The rate for admission to an exhibition was Rs.5/- and was later reduced by 20%. As a result, the sales proceeds increased by 44%. The percentage increase in attendance was.



Q:2 If the price of one kg of cornflakes is increased by 25%, the increase is Rs.10/-. Find the new price of the cornflakes per kg?



Q:3 When the price of sugar was increased by 32%, a family reduced its consumption in such a way that the expenditure on sugar was only 10% more than before. If 30 kg were consumed before. Find the new consumption.

[GauravShah]

Originally Posted by tackledude

Q:1 The rate for admission to an exhibition was Rs.5/- and was later reduced by 20%. As a result, the sales proceeds increased by 44%. The percentage increase in attendance was.

Lets Sales be S, Rate be R and number be N.

S = R x N

Initially, R1 = 5. So S1 = 5N1

After reduction, R2 = 4 and S2 = 1.44S1

i.e 4*N2 = 1.44*5*N1
i.e N2 = 7.2*N1/4
i.e N2 = 1.8N1 or increase of 80%

Originally Posted by tackledude

Q:2 If the price of one kg of cornflakes is increased by 25%, the increase is Rs.10/-. Find the new price of the cornflakes per kg?

Very simple

25% -> Rs.10. so 100% -> Rs.40

So initial price was Rs. 40. New price Rs.50

Originally Posted by tackledude

Q:3 When the price of sugar was increased by 32%, a family reduced its consumption in such a way that the expenditure on sugar was only 10% more than before. If 30 kg were consumed before. Find the new consumption.

Again.
Price P, Consumption C and Expenditure E

We have E = P x C

First case C1 = 30. So E1 = P1 * 30

Now P2 = 1.32*P1 and E2= 1.1*E1

Now P2*C2 = E2

1.32*P1*C2 = 1.1*P1*30

So C2 = 33/1.32 = 25 Kgs

HTH

Gaurav.

[rohit_vij]

Originally Posted by deepamkataria

Hi

Instead of applying higher level mathematics. This one should be simple and quick

1. Divide 30 by 17. Remainder is -4 (30-34).
2. Again square -4. Remainder now is -1 (16-17).
3. Now power left is 38 (as you have already square). So answer is 1. Had power reamaining be odd (instead of 38 :even the result would be -1 .i.e. 16 (17-1).


The genral formula is

X = M (Mod N) Where M is a reamainder when X is divided by N

X^y = M^y(Mod N). So proceed step by step.

hi all
there is another way of solving such questions
it goes by (a^p - a) is divisible by p if p is a prime no:
now provided if a and p are co_prime then a^(p-1) when divided by p wld leave remainder 1. this wld aid in shortening lot many such questions where primes are involved.

[deepamkataria]

I think the answer should be 4 and not 5 or 6. If the answer is not correct.Please explain the solution in detail.

Thanks

Regards
Deepam

Originally Posted by deepamkataria

The solution will be on the same lines of the MOD Theorm.

Find seprately the remainder when 5555^2222 is divided by 7 which is (-1)^1111 = -1 or the remainder is 6. And, the remainder when 2222^5555 is divided by 7 is (-1)^1851 * (-1)^2 = -1*1 = -1 or the remainder is again 6. So total remainder is 12 which when divided by 7 gives the final remainder as 5.

So the final answer is 5. Do check it and let me know if this is correct.

[sorabh]

1) 121212.......upto 300 digits is divided by 99, what is the remainder ?
a)15
b)18
c)12
d)22

2) What is the remainder when 1567*1432*1652 is divided by 25 ?
a)22
b)8
c)3
c)none of these

3) How many rectangles can one make from a regular chess board ?
a)1296
b)1332
c)1242
d)1221

Thanks in advance.