2> sum of digits of N=8
3>a*b*c=1*5*2=10
@RRD2 said:Sove for X and Y-X^1/2 + Y = 2X + Y^1/2=3
x=(2-y)^2
(2-y)^2+y^1/2=3
put y^1/2=t
y=t^2
now solve to get t and the substitution
(2-y)^2+y^1/2=3
put y^1/2=t
y=t^2
now solve to get t and the substitution
@amresh_maverick said:1> unit digit of 1^2! + 2^3! + 3^4! +......+50^51! ?2> N =2^2999 * 5^3002 , some of digits of N ?3> 523abc is divisible by 7,8 9. Find the value of a*b*c ?PS: 2 min mein solve karo, main 5 min mein dinner karke aata hun
OA :
1> unit digit =3
2> sum of digits =8
3> Cannot be determined
1> unit digit =3
2> sum of digits =8
3> Cannot be determined
@amresh_maverick said:OA :1> unit digit =32> sum of digits =83> Cannot be determined
Sir jee pahle waale question ka solution please.
or tell me where i went wrong.
here is what I did :
1+4 + 1+ after this all numbers that are there will have there power in the form of 4k..so used cyclicity.
1^2! = 1
2^3! =4
3^4! = 1
4^4k = 6
5^4k = 5
6^4k =6
7^4k = 1
8^4k = 6
9^4k = 1
added all these 31.
10^4k =0
then the cycle repeats ... for 11-19 , 21-29,31-39,41-49
so have sets 31*5 =5 as the unit digit.
Point out the mistake...
@saurav205 said:Can you share the approach for the 1st question.
1 + 4 + 1 + 6 + 5 + 6 + 1 + 6 + 1 = 31
so 31*5 = 5 unit digit ....
@saurav205 said:Sir jee pahle waale question ka solution please.or tell me where i went wrong.here is what I did :1+4 + 1+ after this all numbers that are there will have there power in the form of 4k..so used cyclicity.1^2! = 12^3! =43^4! = 14^4k = 65^4k = 56^4k =67^4k = 18^4k = 69^4k = 1added all these 31.10^4k =0then the cycle repeats ... for 11-19 , 21-29,31-39,41-49so have sets 31*5 =5 as the unit digit.Point out the mistake...
Remaining 4 sets mein 2^4k ka value aa jayega. = 6 instead of 4.
So 4 times replace 4 by 6, add 8, last digit becomes 3.
regards
scrabbler
So 4 times replace 4 by 6, add 8, last digit becomes 3.
regards
scrabbler
@mailtoankit said:1 + 4 + 1 + 6 + 5 + 6 + 1 + 6 + 1 = 31so 31*5 = 5 unit digit ....
wohi toh..5 aa raha hai...
you had mentioned 2 as the answer..
thats why I asked.
But seems like 5 is also wrong...the OA is 3..dont know how..
@scrabbler said:Remaining 4 sets mein 2^4k ka value aa jayega. = 6 instead of 4.So 4 times replace 4 by 6, add 8, last digit becomes 3.regardsscrabbler
Oh yes...
Thanks.
@saurav205 said:Sir jee pahle waale question ka solution please.or tell me where i went wrong.here is what I did :1+4 + 1+ after this all numbers that are there will have there power in the form of 4k..so used cyclicity.1^2! = 12^3! =43^4! = 14^4k = 65^4k = 56^4k =67^4k = 18^4k = 69^4k = 1added all these 31.10^4k =0then the cycle repeats ... for 11-19 , 21-29,31-39,41-49so have sets 31*5 =5 as the unit digit.Point out the mistake...
The mistake is the bi-cycle will repeat frm 11 to 50
U took 2^3! ends in 4 - correct
But for 11 to 20
12^13! ends in 6 (u took it as ends in 4)
hope u get the point
U took 2^3! ends in 4 - correct
But for 11 to 20
12^13! ends in 6 (u took it as ends in 4)
hope u get the point
1> Last two digits of 78^379
2> 123456789123456789..... upto 180 digits when divided by 11 will rem ?
A right angled triangle with hypotenuse 10 inches & other two sides of variable length is rotated about its longest side thus giving rise to a solid. Find the maximum possible area of such a solid?
@MANJULNEOGI said:A right angled triangle with hypotenuse 10 inches & other two sides of variable length is rotated about its longest side thus giving rise to a solid. Find the maximum possible area of such a solid?
I think it will be cone with r=5 and height =5
Also, volume hoga naa ? correct me
Also, volume hoga naa ? correct me
@amresh_maverick said:I think it will be cone with r=5 and height =5Also, volume hoga naa ? correct me
r aur H mein relation kya hai bhai?
@amresh_maverick said:1> Last two digits of 78^3792> 123456789123456789..... upto 180 digits when divided by 11 will rem ?
OA :
1> 92
2> 0
1> 92
2> 0
@amresh_maverick said:1> Last two digits of 78^379
78^379 = 8^379 = 2^1137 = 2^7 * (2^10)^113 = 8*24^113 = 8*24 = 92?
second waala explain karna ....