Explanation of the previous question:
nth term of the series 2,3,6,11,......=n^2-n+2=n(n-1)+2
==> nth term of the given series={n(n-1)+2}/n!=1/(n-2)!+2/n!
=> summation=(e-1)+2*(e-1)=3(e-1)
Find the number of quadratic polynomials ax2 + bx + c such that:
a) a, b, c are distinct.
b) a, b, c E {1, 2, 3, ...2008}
c) x + 1 divides ax2 + bx + c
a) 2013018 b) 2013021 c) 2014024 d) 2018040
a) a, b, c are distinct.
b) a, b, c E {1, 2, 3, ...2008}
c) x + 1 divides ax2 + bx + c
a) 2013018 b) 2013021 c) 2014024 d) 2018040
solve
edit.
options:
12, 11, 10, 9.
@nick_baba said:Find the number of quadratic polynomials ax2 + bx + c such that:a) a, b, c are distinct.b) a, b, c E {1, 2, 3, ...2008}c) x + 1 divides ax2 + bx + ca) 2013018 b) 2013021 c) 2014024 d) 2018040
c???
a+c=1
a+c+x=2006
total sol = 2008c2
ans = 2008c2 - 1004
@albiesriram said:solveedit.options:12, 11, 10, 9.
10?
Avg age of boys = 18
No of boys = 10
hence, no of girls = 20
Avg age of girls = 21
To make the avg age of boys+teacher = 21
Sum total of boys' age+teacher's age = 21*11 =231
Hence, age of teacher = 231-180 = 51 (between 49 and 54)
@nick_baba said:Find the number of quadratic polynomials ax2 + bx + c such that:a) a, b, c are distinct.b) a, b, c E {1, 2, 3, ...2008}c) x + 1 divides ax2 + bx + ca) 2013018 b) 2013021 c) 2014024 d) 2018040
2014024
b = a + c
b = 3 => 2 * 1
b = 4 => 2 * 1
b = 5 => 2 * 2
b = 6 => 2 * 2
.
.
b = 2007 => 2 * 1003
b = 2008 => 2 * 1003
Total = 2 * 1003 * 1004 / 2 *2 = 2014024
b = a + c
b = 3 => 2 * 1
b = 4 => 2 * 1
b = 5 => 2 * 2
b = 6 => 2 * 2
.
.
b = 2007 => 2 * 1003
b = 2008 => 2 * 1003
Total = 2 * 1003 * 1004 / 2 *2 = 2014024
@grkkrg said:2014024b = a + cb = 3 => 2 * 1b = 4 => 2 * 1b = 5 => 2 * 2b = 6 => 2 * 2..b = 2007 => 2 * 1003b = 2008 => 2 * 1003Total = 2 * 1003 * 1004 / 2 *2 = 2014024
@soumitrabengeri said:10?Avg age of boys = 18No of boys = 10hence, no of girls = 20Avg age of girls = 21To make the avg age of boys+teacher = 21Sum total of boys' age+teacher's age = 21*11 =231Hence, age of teacher = 231-180 = 51 (between 49 and 54)
can u elaborate more??? how did u arrive to this soln??? hit n trial ??
Some geometry..
@Brooklyn said:can u elaborate more??? how did u arrive to this soln??? hit n trial ??
Yes..went through the options and started substituting random values..crude method..but always seems to work for me
Please let me know if there is a better solution for this
@grkkrg said:2014024b = a + cb = 3 => 2 * 1b = 4 => 2 * 1b = 5 => 2 * 2b = 6 => 2 * 2..b = 2007 => 2 * 1003b = 2008 => 2 * 1003Total = 2 * 1003 * 1004 / 2 *2 = 2014024
dost why have u multiplied 2 in the last of Total = 2 * 1003 * 1004 / 2 *2 = 2014024
@Brooklyn said:c???a+c=1a+c+x=2006total sol = 2008c2ans = 2008c2 - 1004
bhai thoda ise logically bhi explain kar skte ho??
some progressions...
edit :- typo error sorry..
sorry!!
sorry!!find 1^2+2^2 +3^2+5^2+8^2+...upto 100 terms..
options..1) 13581
2) 14796
3) 16234
4) 10100
4) none of these
@nick_baba said:dost why have u multiplied 2 in the last of Total = 2 * 1003 * 1004 / 2 *2 = 2014024
the summation is repeated twice :)
@nick_baba said:bhai thoda ise logically bhi explain kar skte ho??
dekho:
we knw x=-1 is a soln of the q
this gives us: a - b + c=0
ie a+c=b
now b can be any value from 1 to 2008
and a+c=b
so we get :
a+c =
adding a variable x that takes care of extra value, we get:
a+c+x=2008
now a,c >0
so
a+c+x=2006
total int soln=2008c2
now nos such as 2008 can also be written as 1004 + 1004
but we have constratint a,b,c different
so take these out , ie all nos which are multiple of 2 => 2008/2 = 1004
Hope this helps 😃