@Jackson1 said:there is a formula for that- (n-1)!*(sum of all numbers)*(111...n times)...
Great buddy...I was unaware of any such formula.Thanks a lot.
@saurabhlumarrai said:Its correct. approach please...
each digit will come in all the places for 4! times.........
like 1XXXX - 4! numbers
also X1XXX - 4! numbers
so 1 is valued at 10000 for 4! times and so on for other places as well...
same for other digits as well..
so (1+3+5+7+9) *(10000+1000+100+10+1)*4!= a)6666600
like 1XXXX - 4! numbers
also X1XXX - 4! numbers
so 1 is valued at 10000 for 4! times and so on for other places as well...
same for other digits as well..
so (1+3+5+7+9) *(10000+1000+100+10+1)*4!= a)6666600
@Dexian said:each digit will come in all the places for 4! times.........
like 1XXXX - 4! numbers
also X1XXX - 4! numbers
so 1 is valued at 10000 for 4! times and so on for other places as well...
same for other digits as well..
so (1+3+5+7+9) *(10000+1000+100+10+1)*4!= a)6666600
awesome approach dude...
@saurabhlumarrai said:awesome approach dude...
are kahe ka awsome bhai...
i used this formula: (n-1)!*(sum of all numbers)*(111...n times)...
and xplained this as approach...
no big deal...
i used this formula: (n-1)!*(sum of all numbers)*(111...n times)...
and xplained this as approach...
no big deal...
@saurabhlumarrai said:A Five digit number is formed by using 1,3,5,7 and 9 without repeating any of them.What is the sum of all such possible numbers?a)6666600b)6666660c)6666666d)None of these
6666600 ?
(5 - 1)!*(1 + 3 + 5 + 7 + 9)*(11111) = 6666600
@Vipul24 said:N=12345678910111213........40. What will be the reminder when N is divided by 9?
1 ?
1 + 2 + ................9 = 9*10/2 = 45
10 + 11 + 12 +.....19 = 10*11/2 = 55
20 + 21 + ........... 29 = 11*12/2 - 1 = 65
30 + 31 +.............40 = 12*13/2 - 1 - 2 + 4 = 75
(45 + 55 + 65 + 75) mod 9 = 244 mod 9 = 1
@jain4444 said:If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
y*log4 = log5---> y = log5/log4
(2x)*log3 = log4----> x = log2/log3
z*log5 = log8 ----> z = 3log2/log5
9^(log5/log4*log2/2log3*3log2/log5)*3^2
9^(3log2/2log3 + 1)
9^(log8 + log9/2log3)
9^(log72/log9)
9^(log(base9)72)
= 72 ?
@jain4444 said:If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
getting 72...used log..
@Vipul24 said:N=12345678910111213........40. What will be the reminder when N is divided by 9?
40.41/2 mod 9
20.41 mod 9
20.41 mod 9
2.5 mod 9
1 mod 9
1 mod 9
remainder 1
@saurabhlumarrai said:A Five digit number is formed by using 1,3,5,7 and 9 without repeating any of them.What is the sum of all such possible numbers?a)6666600b)6666660c)6666666d)None of these
(1+3+5+7+9)*4! *11111 = 6666600
even if u dont know the formula..find the sum of all five digit numbers and divide by 2..bcz digits given are all odd
even if u dont know the formula..find the sum of all five digit numbers and divide by 2..bcz digits given are all odd
@Logrhythm said:@jain4444 said:If, 4^y=5.3^(2x)=4.5^z=8.Find value 9^(yxz) *3^2=...?
getting 72...used log..
boss,no need of using log-
Simple method:
3^(2x)=4
=>9^x=4
Applying power of y on both sides,we get:
9^xy=4^y
but 4^y=5(given)
so,9^xy=5
Now,Applying power of z on both sides,we get:
9^xyz=5^z
but agin 5^z=8
so 9^xyz=8
and 9^(yxz) *3^2=8*9=72
easy and effective way
@Vipul24 said:Last 2 digits for 37^64??
(37*37)^32 = .. {(1369)(1369)}^16= (61)^16= 61 last two digits..
@Vipul24 said:Last 2 digits for 37^64??
37^64/100
split 100 in 25*4
euler of 4 = 2
64/2=0 so rem of 37^64/4=1
euler of 25 = 4*25/5=20
64/20=4
37^4/25=11
so only 61 satisfies
@pakkapagal said:The ratio of the monthly income of A and B is 3:4.The ratio of the monthly expenditure of A and B is 4:5 .Which of the following represents a possible value of the ratio of their savings?a)9:10 b)3:4 c)13:20 d)4:53x-4y/4x-5y stuck here
see.. let's take total salary is 700.
A would earn 300, B 400.
let's say they spend 450 together.. A spends, 200, and B 250
so savings ratio will be .. 100/150= 2/3 ... somewhere around .6666..
by options.. 9/10=.90
3/4=.75
13/20= .65
4/5=.80
best option would be 13:20.
A would earn 300, B 400.
let's say they spend 450 together.. A spends, 200, and B 250
so savings ratio will be .. 100/150= 2/3 ... somewhere around .6666..
by options.. 9/10=.90
3/4=.75
13/20= .65
4/5=.80
best option would be 13:20.
@Vipul24 said:N=12345678910111213........40. What will be the reminder when N is divided by 9?
sum till 1 to 9 = 45
so 45*4+10+20+30+4=244
244/9 rem = 1
@Vipul24 said:N=12345678910111213........40. What will be the reminder when N is divided by 9?
sum till 1 to 9 = 45
so 45*4+10+20+30+4=244
244/9 rem = 1
@pakkapagal said:The ratio of the monthly income of A and B is 3:4.The ratio of the monthly expenditure of A and B is 4:5 .Which of the following represents a possible value of the ratio of their savings?
a)9:10 b)3:4 c)13:20 d)4:5
3x-4y/4x-5y stuck here
3x-4y/4x-5y
divide the equation by x
let y/x= a
3-4a/4-5a
a is possible only for 13:20
How many ordered pairs (P, Q) are there such that the unit €™s digits of P^P and Q^Q are the same? P and Q are natural numbers less than 10 and are not necessarily distinct.
@amresh_maverick said:How many ordered pairs (P, Q) are there such that the unit €™s digits of P^P and Q^Q are the same? P and Q are natural numbers less than 10 and are not necessarily distinct.
22?
@amresh_maverick said:How many ordered pairs (P, Q) are there such that the unit €™s digits of P^P and Q^Q are the same? P and Q are natural numbers less than 10 and are not necessarily distinct.
1^1=1
2^2=4
3^3=7
4^4=6
5^5=5
6^6=6
7^7=3
8^8=6
9^9=9
(6,4),(4,6)(8,6)(6,8)(4,8)(8,4) (4,4)(6,6)(8,8)???
9 such pairs??
EDIT : (1,1) (2,2)(3,3)(5,5)(7,7)(9,9) also.
total : 15.
correct me if i am wrong.
2^2=4
3^3=7
4^4=6
5^5=5
6^6=6
7^7=3
8^8=6
9^9=9
(6,4),(4,6)(8,6)(6,8)(4,8)(8,4) (4,4)(6,6)(8,8)???
9 such pairs??
EDIT : (1,1) (2,2)(3,3)(5,5)(7,7)(9,9) also.
total : 15.
correct me if i am wrong.