the ans is the remainder on dividing the expression by100.
use the remainder thm.
Puys some help on this... It's easy buy unable to do somehow..........
Find the last two digits if 15*37*63*51*97*17
a)35 b)45 c)55 d)85 e)75
Pls explain
Use remainder on division by 100...
15*37*63*51*97*17=15*37*63*51*(-3)*17
Put 63=7*9
-(15*7*37*9*51*3*17)= -(105*111*17*9*51)= -(5*11*17*9*51)
= -(55*9*17*51)= -(495*17*51)=-(-5*17*51)=85*51
= -15*51=15*49=735....
Hence a)....
Puys some help on this... It's easy buy unable to do somehow..........
Find the last two digits if 15*37*63*51*97*17
a)35 b)45 c)55 d)85 e)75
Pls explain
(15x37x63x51x97x17) /100
Simplify by 5 (3x37x63x51x97x17)/20 (dont forget to multiply by 5 at the end )
(3x-3x3x-9x-3x-3)/20
(27x9x9)/20
(81x27)/20
(1x7)= 7x5=35
(15x37x63x51x97x17) /100
Simplify by 5 (3x37x63x51x97x17)/20 (dont forget to multiply by 5 at the end )
(3x-3x3x-9x-3x-3)/20
(27x9x9)/20
(81x27)/20
(1x7)= 7x5=35
definitely a simpler method.....
(15x37x63x51x97x17) /100
Simplify by 5 (3x37x63x51x97x17)/20 (dont forget to multiply by 5 at the end )
(3x-3x3x-9x-3x-3)/20
(27x9x9)/20
(81x27)/20
(1x7)= 7x5=35
Wonderful!!!!!!!!! thanks a ton!!!!!!!!! can u jst tell me y did u take 5 der?? jst cos al d numbers are close to some multiple of 20????? and how did the first term become 3?????? from 15??????
chandrakant.k SaysWonderful!!!!!!!!! thanks a ton!!!!!!!!! can u jst tell me y did u take 5 der?? jst cos al d numbers are close to some multiple of 20????? and how did the first term become 3?????? from 15??????
15 and 100 have both been simply divided by 5.
I chose 20 cuz its quite easy to find the remainders with that number..You can simplify it further as per your convenience..just remember to multiply the final answer with the number by which u divide(i.e 5 in this case)
for 1 st
2*pi*r=44
r=7
pi*r^2=>154
=>154/4
=>38.5
volume is lbh
=>10x8x0.5x2 + 5 x 8 x0.5x2 + 9 x 5 x 0.5 x 2
=165
can u put more light on the values 8,9, and 5 that are mentioned in the equation.
Please clarify this doubt:-
A team of miner planned to mine 1800 tons of ore during a certain number of days.Due to technical difficulties in one-third of the planned no. of days, the team was able to achieve an o/p of 20 tonnes less than planned o/p.To make up for this,the team overachieved for the rest of the dys by 20 tonnes.The end result was that the tem completed the task one day ahead of time.How many tons of ore did the team initially plan to ore per day?
a)50
b)100
c)150
d)200
e)250
Hi,
Can you plz help me in solving the below problem:
Q) given a series 1,2,5,10,17,26,37,......... what is the sum of the 1st fifty terms of the series?
Hi,
Can anyone plz help me in solving the below problem:
Q) given a series 1,2,5,10,17,26,37,......... what is the sum of the 1st fifty terms of the series?
Hi,
Can anyone plz help me in solving the below problem:
Q) given a series 1,2,5,10,17,26,37,......... what is the sum of the 1st fifty terms of the series?
let Sn = 1+2+5+10+...+tn
Sn = 1+2+ 5 +...+tn-1+tn
---------------------------------
Sn-Sn = 1+(1+3+5+...tn-1)-tn
or tn = 1+ (1/2)*(n-1)
or tn = 1 + (n-1)(n-1) = 1+(n-1)^2 = n^2 - 2n +2
Sn = sum(tn) = n(n+1)(2n+1)/6 - n(n+1) +2n
therefore, S50 = (50*51*101/6) - (50*51) + (2*50) = 40475
What's the approach for the following type of questions:
1.Find the remainder (32^32^32)/9
2.(50^51^52)/11
Please help.
What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!) ?
Options :
a.1
b.7!
c.8!
d. 9!
What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!) ?
Options :
a.1
b.7!
c.8!
d. 9!
2(8!)-21(6!)
= 2*8(7!)-3(7)(6!)
= 16(7!) - 3(7!)
= 13(7!)
14(7!)+14(13!)
= 14(7!)+14(13!)
= (1+13)(7!) + 14(13!)
= 7! + 13(7!) + 14(13!)
=> Remainder will be 7!
What's the approach for the following type of questions:
1.Find the remainder (32^32^32)/9
2.(50^51^52)/11
Please help.
Check up on Euler No.
Basically, for a number n, if E = k
Then for any no. p coprime to n, we can say that p^k/n leaves remainder 1
1.
E = 9(1-1/3) = 6
We need to represent 32^32^32 in form of 32^(6k+y) coz we know that 32^6k would leave remainder 1 on division with 9.
32^32Mod6 = 32 * 32^31Mod6
=> 16 * 32^31Mod3
=> 1 * -1Mod3
=>2Mod3
Multiplying back by 2, we get 32^32 = 6k+4
So 32^32^32 = 32^(6k+4)
32^(6k+4)Mod9 = 32^4Mod9 = 5^4Mod9 = 625Mod9 = 4
2.
E = 11(1-1/11) = 10
50^51^52 = 50^(10k+1)
50^(10k+1)Mod11 = 50Mod11 = 6
let Sn = 1+2+5+10+...+tn
Sn = 1+2+ 5 +...+tn-1+tn
---------------------------------
Sn-Sn = 1+(1+3+5+...tn-1)-tn
or tn = 1+ (1/2)*(n-1)
or tn = 1 + (n-1)(n-1) = 1+(n-1)^2 = n^2 - 2n +2
Sn = sum(tn) = n(n+1)(2n+1)/6 - n(n+1) +2n
therefore, S50 = (50*51*101/6) - (50*51) + (2*50) = 40475
Hi, Thanks a lot fo rthe quick reply..... But i couldnot understandf the 2nd line of the solution... tn = 1+ (1/2)*(n-1) can u plz elaborate
let Sn = 1+2+5+10+...+tn
Sn = 1+2+ 5 +...+tn-1+tn
---------------------------------
Sn-Sn = 1+(1+3+5+...tn-1)-tn
or tn = 1+ (1/2)*(n-1)
or tn = 1 + (n-1)(n-1) = 1+(n-1)^2 = n^2 - 2n +2
Sn = sum(tn) = n(n+1)(2n+1)/6 - n(n+1) +2n
therefore, S50 = (50*51*101/6) - (50*51) + (2*50) = 40475
prabinpatodia SaysHi, Thanks a lot fo rthe quick reply..... But i couldnot understandf the 2nd line of the solution... tn = 1+ (1/2)*(n-1) can u plz elaborate
Sn-Sn = 1+(1+3+5+...tn-1)-tn
or tn = 1+(1+3+5+...tn-1)
or tn = 1+ (1/2)*(n-1)
the bold part is nothing but the sum of the AP series mentioned in the previous line in Italics...hope this clears
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Please help me out man!
In how many ways can you distribute 32 apples among 4 different students?
I know you'll grin seeing this question! But I really forgot the method. Please help me. 
In how many ways can you distribute 32 apples among 4 different students?
I know you'll grin seeing this question! But I really forgot the method. Please help me.
Hey !
Is the answer to your question 32^4 ?? Since each apple will have an option of going to either of the 4 guys ... I'm not very sure ..please confirm ...