Folks, keep tuned to this thread for a Challenge problem tonight 😃
@bodhi_vriksha said:It's 369. Basically 729x=001mod1000 => x =369mod1000 (by backtracing)
sorry...... got 369......... lekin answered 569............ offc me aaj thoda time crunch hai boss..........
@bodhi_vriksha said:Another quick one:What is the remainder when 2^2001 is divided by 2001?
3*23*29 = 2001
2^2001%3 = 2
2^2001%23 = 2^21%23 = 12
2^2001%29 = 2^13%29 = 14
=> 3p+2 = 23q+12 = 29r+14
so 449....
@bodhi_vriksha said:Another quick one:What is the remainder when 2^2001 is divided by 2001?
449 ?
2^2001 mod 2001 = 2^2001 mod 3*23*29
2^2001 mod 3 ----> 2 mod 3 = 2----> 3k + 2
2^2001 mod 23 ---> 12 mod 23 = 23a + 12
2^2001 mod 29 ----> 14 mod 29 = 29b + 14
3k + 2 = 23a + 12 = 29b + 14 = 449 ?
Good job there everyone!
@bodhi_vriksha said:Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....Find the last three digits of 3^1994.
last 3 digits = remainder with 1000
1000 = 125*8
3^1994%8 = (9)^997%8 = 1
3^1994%125 = 3^94%125 = 3^4*(243)^18%125 = 81*(-7)^18%125 = 81*24^3%125 = 81*74%125 = 119
=> 8p+1 = 125q+119
so 369...
@mailtoankit said:369 ?3^2*(3^4)^498 = 9*41 = 369...
Can you please explain me this in detail plz... got a low wattage bulb in my brain :(
@Vinaysastra said:Can you please explain me this in detail plz... got a low wattage buld in my brain
haha..same here!!
basically to find out the last three digits..you have to divide the given number by 1000
now 3^1994 = 3^2*3^1992 = 9*(3^4)^498 = 9*(81)^498....now In (81)^498 last digit will always be 1....to find the second last digit...multiply the digit at tens place in bracket i.e. 8 , to the last digit of the exponent i.e. again 8 = 8*8 = 64(this method is used only in the case where last digit is 1)...so this will give us the last two digits of original no. as 41....
now this 9*(81)^498 will become = 9*41 = 369....which are the final last three digits
PS - you can check more about finding the last one,two,three digits at totalgadha.com..
@mailtoankit said:haha..same here!!basically to find out the last three digits..you have to divide the given number by 1000now 3^1994 = 3^2*3^1992 = 9*(3^4)^498 = 9*(81)^498....now In (81)^498 last digit will always be 1....to find the second last digit...multiply the digit at tens place in bracket i.e. 8 , to the last digit of the exponent i.e. again 8 = 8*8 = 64(this method is used only in the case where last digit is 1)...so this will give us the two digits of original no. as 41....now this 9*(81)^498 will become = 9*41 = 369....which will the final last three digitsPS - you can check more about finding the last one,two,three digits at totalgadha.com..
Thanks A Lot :)
A magazine company prints 5000 copies at 5,00,000 every month.In July,they distribute 500 free copies.Besides they were able to sell 2/3rds of the remaining magazines at 20% discount.Besides the remaining magazines were sold at the printed price of Rs 200.Find the percentage profit for the magazine company(assume a uniform 20% of the sale price as the vendor's discount and assume no other source of income).
A.56
B.24.8
C.28.5
D.22.6
Correct Answer is 24.8. I wanted to know the procedure.
Thanks in advance.
A.56
B.24.8
C.28.5
D.22.6
Correct Answer is 24.8. I wanted to know the procedure.
Thanks in advance.
@mohnish_khiani said:A magazine company prints 5000 copies at 5,00,000 every month.In July,they distribute 500 free copies.Besides they were able to sell 2/3rds of the remaining magazines at 20% discount.Besides the remaining magazines were sold at the printed price of Rs 200.Find the percentage profit for the magazine company(assume a uniform 20% of the sale price as the vendor's discount and assume no other source of income).A.56B.24.8C.28.5D.22.6Correct Answer is 24.8. I wanted to know the procedure.Thanks in advance.
look the CP of a single copy is 500000/5000=100
and the SP of each copy is 200.
The vendor gets 20% discount on 200 i.e 40.. the vendor sells 4500 copies
of these 4500 copies 2/3(4500)=3000 copies are sold at 160 and the vendor gets a discount of 32 ie 20% of 160.. and the vendor takes his share of 32/copy sold.. hence the company gets 28/copy as profit.. hence a total of 28*3000=84,000
By selling 500 free copies the company incurs a loss of 100*500=50,000
The net profit is 34,000
Now the remaining 1500 copies are sold at 200 and hence the company gets 60/copy
Total profit is 60*1500=90,000.
the final net profit is 1,24,000
The cost price is 5,00,000
Profit%= (124000/500000)*100=24.8
and the SP of each copy is 200.
The vendor gets 20% discount on 200 i.e 40.. the vendor sells 4500 copies
of these 4500 copies 2/3(4500)=3000 copies are sold at 160 and the vendor gets a discount of 32 ie 20% of 160.. and the vendor takes his share of 32/copy sold.. hence the company gets 28/copy as profit.. hence a total of 28*3000=84,000
By selling 500 free copies the company incurs a loss of 100*500=50,000
The net profit is 34,000
Now the remaining 1500 copies are sold at 200 and hence the company gets 60/copy
Total profit is 60*1500=90,000.
the final net profit is 1,24,000
The cost price is 5,00,000
Profit%= (124000/500000)*100=24.8
@mohnish_khiani said:A magazine company prints 5000 copies at 5,00,000 every month.In July,they distribute 500 free copies.Besides they were able to sell 2/3rds of the remaining magazines at 20% discount.Besides the remaining magazines were sold at the printed price of Rs 200.Find the percentage profit for the magazine company(assume a uniform 20% of the sale price as the vendor's discount and assume no other source of income).A.56B.24.8C.28.5D.22.6Correct Answer is 24.8. I wanted to know the procedure.Thanks in advance.
at one place please !!
Solution :
cost price per copy = 100
they distribute freely = 500*100 = 50000 Rs = this is a loss to the company
out of 4500, 3000 are sold at 160 so profit =60Rs. out of this vendor gets 20% = 32
hence company gets = 28
total profit = 28*3000 = 84000
rest 1500 at 200 so profit = 100.
Vendor gets a discount = 40
so total profit = 1500*60 = 90000
total profit = 84000+90000-50000 = 124000
Profit % = 124/500 = 24.8
@bodhi_vriksha said:Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....Find the last three digits of 3^1994.
369
@chillfactor said:
Find the number of non-negative integral solutions for the equation 3a + 4b + 12c = 432
(a/4 + b/3) + c = 36
the only way we can have a integral solution to above eqn is that
a=4k and b = 3p
hence we seek a solution for (4k)/4 + (3p)/3 + c = 36
=>k + p + c = 36
no. of non -ve solution for above eqn is (36+3-1)C(2) = 38C2 =342 solutions
ATDH.
@mailtoankit said:449 ?2^2001 mod 2001 = 2^2001 mod 3*23*292^2001 mod 3 ----> 2 mod 3 = 2----> 3k + 22^2001 mod 23 ---> 12 mod 23 = 23a + 122^2001 mod 29 ----> 14 mod 29 = 29b + 143k + 2 = 23a + 12 = 29b + 14 = 449 ?
can you explain the last step(in bold) ?
problem with basics..
@rnishant231 said:can you explain the last step(in bold) ?problem with basics..
3k+2 = 23a+12
so:
k =( 23a+ 10) /3
now both k and a can only be +ive intg so do hit and trial
Taking mod 3
k = (2a + 1)/3
so for what value of a will k be interger???
a=1
so we get a=1 ,k=1
so no : 23+12 = 52 and so on
Q. The coefficient of x^15 in the product (1-x)(1-2x)(1-2^2x)...........(1-2^15x) is equal to
a) 2^105-2^121
b) 2^121-2^105
c) 2^120-2^104
d) none of these
Q. The coefficient of x^49 in the product of(x-1) (x-3).........(x-99) is
a) -99^2
b) 1
c) -2500
d) none of these
A person has taken a loan of 10000 rs for 3 yrs at 10% p.a but at the end of 1st yr he paid back rs 5000 , at the end of 2nd yr he paid back rs 3000 . What amount should he pay back in 3rd yr to completely pay back the debt.
@getupsid said:Q. The coefficient of x^15 in the product (1-x)(1-2x)(1-2^2x)...........(1-2^15x) is equal to a) 2^105-2^121b) 2^121-2^105c) 2^120-2^104d) none of theseQ. The coefficient of x^49 in the product of(x-1) (x-3).........(x-99) is a) -99^2b) 1c) -2500d) none of these
Q1) a
Q2) c
both are based on the same logic, choose x in all but one of the terms and simplify the result obtained on multiplying out all of the terms