vibhor bhai sir na bola karo yaar 
Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
@Aman.Malhotra @ScareCrow28
A,B,C,D,E,F,G are 7 DISTINCT natural nos less than 13
A*B*C=D*E*F=G*B*D=G*A*E
1.Amongst the 5 unused nos,how many are prime ?
a)2
b)3
c)4
d)1
e)5
2.What is the value of G ?
a)4
b)6
c)8
d)12
e)2
b)6
c)8
d)12
e)2
3.What is the value of A+E?
a)5
b)7
c)9
d)8
e)CBD
b)7
c)9
d)8
e)CBD
@jain4444 said:@Aman.Malhotra@ScareCrow28vibhor bhai sir na bola karo yaar Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
67,69,71 ?
3abcd3
So it's 3 consecutive 2 digit numbers with units digit 7,9,1
70^3 = 343000
So the numbers should be 67,69,71
3abcd3
So it's 3 consecutive 2 digit numbers with units digit 7,9,1
70^3 = 343000
So the numbers should be 67,69,71
@jain4444 said:@Aman.Malhotra@ScareCrow28vibhor bhai sir na bola karo yaar Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
67*69*71=328233
@jain4444 said:Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
the is of the form AAA where A is less than equal to 4
@jain4444 said:@Aman.Malhotra@ScareCrow28vibhor bhai sir na bola karo yaar Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
Since last digit is 3...so only possibility of the last digit of numbers is: 1, 7 and 9
Product = 6 digit number
Since, 70^3 = 343000
Hence numbers are close to 70
Only possibility being: 67, 69 and 71 ..
@jain4444 said:@Aman.Malhotra@ScareCrow28vibhor bhai sir na bola karo yaar Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
60^3
Last digit is 3 so we should not consider nos ending in 5
Hence starting with
67*69*71= 328233
@jain4444 said:Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
25a+5b+c=64c+8b+a
8a=21c+b
a=3
c=1
b=3
331
@Budokai001 said:A,B,C,D,E,F,G are 7 DISTINCT natural nos less than 13A*B*C=D*E*F=G*B*D=G*A*E1.Amongst the 5 unused nos,how many are prime ?a)2b)3c)4d)1e)52.What is the value of G ?a)4b)6c)8d)12e)23.What is the value of A+E?a)5b)7c)9d)8e)CBD
2*3*8 = 12*1*4 = 8*6*1 = 4*2*6
used => 1 2 3 4 6 8 12
unused => 5 7 9 10 11
A = 2 , G = 6 , E = 4 , B = 8 , C = 3 , D = 1 , F = 12
1) 3
2) 6
3) 4 + 2 = 6
@jain4444 said:@Aman.Malhotra@ScareCrow28vibhor bhai sir na bola karo yaar Find the 6-digit number beginning and ending in the digit 3 that is the product of three consecutive odd integers.
328233 is the product of 69 67 and 71
all correct
Find the 6-digit number beginning and ending in the digit 2 that is the product of
three consecutive even integers.
three consecutive even integers.
@jain4444 said:all correctFind the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.
Working on same lines..
Numbers should be ending in 4, 6 and 8 only
60^ = 216000
So, numbers are 64, 66 and 68
Product = 287232
@jain4444 sir bro (as per above comment
)
)A = 8 , G = 6 , E = 1 , B = 2 , C = 3 , D = 1 , F = 12
is also possible
here A+E=9 so last one will be CBD
f(n)=f(n-2)+f(n+2)
If it is given that f(x)=f(x+p)
What is min value of p ?
a)12
b)6
c)8
d)4
e)5
@jain4444 said:let 1gm = 1 rsselling quantity => x*120/100 = 800 => x = 666.66 so , profit = 666.66*A% = 1000 => A = 50%
first profit 200/800*100 equal to 25%
second profit 20%
using successive percentage 20 + 25 + 20*25/100 equal to 50
@Aman.Malhotra
second profit 20%
using successive percentage 20 + 25 + 20*25/100 equal to 50
@Aman.Malhotra
@YouMadFellow said:2^11 = 2048 (occurs only once in S)
=> K = (1 + 4011)*(2006/2) = 4024036 ?
@vijay_chandola said:odd terms in the series= 2007, 2009, 2011....... 4011
1+ 3+ 5+7+9+11+13+15+.......... 4009+4011= 2006*{2*1+(2006-1)*2}/2= 4024036
You guys missed one 1003 which should be added as we have 2006 and 4012 both.
So, answer will be 4024036 + 1003 = 4025039
@bullseyes said:it is positive integral but u can show for both
1/z - 1/y = 1/x
zy - xy + zx = 0
(y + x)(z - x) = -x^2 = f*(-x^2/f), where f is factor of -x^2
So, y = f - x and z = x - (x^2/f)
When we are looking for integral solutions
y = f - x, and f is a factor of -x^2
We know that x^2 has, say 'k' factors, so, y can take '2k' values (as f can be a negative factor also). But for f = x, y = 0, which is not allowed, so we need to remove this case
So, in case of integral solutions, this equation will have 2k - 1 solutions
In case of positive integral solutions,
y = f - x > 0
f > x
So, we need to consider only those factors of x^2 which are greater than x
Also, x - (x^2)/f > 0
=> f > x (same as above, so once y > 0, then z > 0 will be automatically taken care of)
So, no of positive integral solutions will be given by no of factors of x^2 greater than x
@jain4444 said:all correctFind the 6-digit number beginning and ending in the digit 2 that is the product ofthree consecutive even integers.
64*66*68 equal to 287232
@jain4444 said:Find a,b and c if the number "abc" in base 5 is equal to the number "cba" in base 8.
24a=63c+3b
a,b,c
so,
b=3,c=1,a=3
@Budokai001 said:f(n)=f(n-2)+f(n+2)If it is given that f(x)=f(x+p)What is min value of p ?a)12b)6c)8d)4e)5
b) 6 ??
f(n) is an odd function
Put, n= 4
f(4) = f(2) + f(6)
=> f(6) = f(4) - f(2)
Also, f(2) = f(0) + f(4)
=> f(4) - f(2) = -f(0) = f(0)
And f(0)
Hence, f(6) = f(0)
Hence pperiod = p =6
How many ways can the digits 1,2,3,4,5,6,7,8,9 be arranged so that no even digit is in its original position?