Last two digits kabhi aaya hai kya CAT/XAT etc. me?? Padhna hai ya it will be time waste to learn?? Help me.
@TootaHuaDil said:Find the number of possible values of n = 125 * m if n is less than or equal to 1000 and m is not divisible by 5
7
@TootaHuaDil said:Find the number of possible values of n = 125 * m if n is less than or equal to 1000 and m is not divisible by 5
7?
m = 1,2,3,4,6,7,8
@bs0409 said:A square is inscribed in a circle and the circle is inscribed in a regular octagon. Find the ratio of the area of the square to that of the octagon.
rt2+1 : 4
@mohitjain said:
is it (rt2+1):4
@mailtoankit said:(rt2 + 1)/4 ?side of big square = xso side of octagon = a + rt2*a + a = x-----> a = x/(2 + rt2)area of octagon = x^2 - 4*1/2 *a^2 = x ^2 - 2*x^2 (1/(2 + rt2))^2area of smaller square = x^2/2ratio = x^/2 / x^2(1 - 2 (1/(2 + rt2))^2solving this eqn...will get(rt2 + 1)/4
@ChirpiBird said:rt2+1 : 4
ANS- 1/(4(rt2-1))
@TootaHuaDil said:Last two digits kabhi aaya hai kya CAT/XAT etc. me?? Padhna hai ya it will be time waste to learn?? Help me.
last two digits is nothing but the remainder when you divide the number by 100.
suppose 31^786/100.. find rem.
this is same as find the last two digits of 31^786. :)
similarly, last three digits k liye 1000 se divide karo.
basically, dont consider it as a separate topic, it is a part of the remainder topic only.
suppose 31^786/100.. find rem.
this is same as find the last two digits of 31^786. :)
similarly, last three digits k liye 1000 se divide karo.
basically, dont consider it as a separate topic, it is a part of the remainder topic only.
@TootaHuaDil said:The quotient obtained by dividing n by 121 and the number 11 are co-prime to each other. Find the number of possible values of n if n is less than or equal to 10000.
75 values?
:neutral:
@ChirpiBird said:plz can u tell me how you solved?
let say x is the quotient when we divide n by 121.
=> n = x*121
now, n
=> n can be 1*121, 2*121, 3*121...... up to 82*121
now, x is co-prime to 11 => x cannot take 11, 22, 33, 44, 55, 66, 77 = 7 values.
hence, 82-7 = 75 values
@vijay_chandola said:let say x is the quotient when we divide n by 121.=> n = x*121now, n => n can be 1*121, 2*121, 3*121...... up to 82*121now, x is co-prime to 11 => x cannot take 11, 22, 33, 44, 55, 66, 77 = 7 values.hence, 82-7 = 75 values@TootaHuaDil
isme 11 ke sath quotient ko co-prime hona tha ya 11 se divide kar ke mile quotient k sath hona tha?? (I don't know) :O
Find the last two non-zero digits of 36!-24!...Ye factorial ke last two digits kaise nikalte hain??
@TootaHuaDil said:The quotient obtained by dividing n by 121 and the number 11 are co-prime to each other. Find the number of possible values of n if n is less than or equal to 10000.
isn't it just the possible number of multiples of 121 till 82 and also which are co prime to 11??
the only primes are 11 22 33 44 55 66 77
so 82-7 = 75
@TootaHuaDil said:isme 11 ke sath quotient ko co-prime hona tha ya 11 se divide kar ke mile quotient k sath hona tha?? (I don't know)
by 121 and 'number' diya h.
Should be co-prime of 11.
Otherwise question doesn't make sense.
2*121 = 22*11
3*121 = 33*11
4*121 = 44*11
kabi co-prime aayega hi nahi
:neutral:
@TootaHuaDil said:The quotient obtained by dividing n by 121 and the number 11 are co-prime to each other. Find the number of possible values of n if n is less than or equal to 10000.
There are lot of numbers. Difficult to count.
Basically n shld be of the form 121p+11r+s where r,s
@TootaHuaDil said:Find the last two non-zero digits of 36!-24!...Ye factorial ke last two digits kaise nikalte hain??
@vijay_chandola @chandrakant.k @Logrhythm @bs0409 koi ye bhi bana do