Find the probability that the number of form 7^n+7^m is divisible by 5.
where m & n are chosen from 1 to 100.
@The_Loser
Is is 1/4??
7^x will always end in 4 different Integers namely 7, 9, 3 n 1..
Thus, n and m can be chosen in 4*4 = 16 ways..[Am reducing the cases]
Now, Favorable cases will be 7 n 3...9 n 1..
Thus, these can be chosen in 2*2 ways [for n and m]
Thus, P(E)= 4/16 = 1/4..
PS: My concrn is related to the limit of n and m..i.e. from 1 to 100..bt i thnk it shld hold for any range..
@rkshtsurana said:f(x) and g(x) are two quadratic functions such that f(3) = 0 , g(5) = 0 . If they have a common root and f(5) * g(7) = 12, what is the value of common root?1) 42) 83) either 4 or 84) NOT
Ans. 3) Both 4 and 8.
@The_Loser said:Find the probability that the number of form 7^n+7^m is divisible by 5.where m & n are chosen from 1 to 100.
1/4 ???
@The_Loser said:Find the probability that the number of form 7^n+7^m is divisible by 5.where m & n are chosen from 1 to 100.
7 ends in 7,9,3,1
there would be total 16 pairs
7,3
3,7
9,1
1,9
will be divisble by 5
4/16=1/4??
How many numbers from 1 to 1000 are divisible by 60 but not by 24?
60=2^2*3*5
24=2^3*3
so 60*1,60*3.... these are the required numbers
60*15=900 last no
900=60+(n-1)120
n=8
@DreamOvrReality said:What is the remainder when 1! + 2 × 2! + 3 × 3! + 4 × 4! + … + 12 × 12! is divided by 13?1) 12) 123) 114) 0I got it as 1) 1, the OA is 2) 12....I want to know why?My approach:n*n! = (n+1-1)*n! = (n+1)! - n!So 13!-12! remains.12!/13 = -1hence the remainder is +1.
add and subtract (1! + 2! + 3! + 4! + … + 12!) to original expression. We get (2! + 3! + 4! + … + 13!) - (1! + 2! + 3! + 4! + … + 12!) = 13! - 1! => Remainder will be -1 or 12
@DreamOvrReality said:What is the highest possible value of 'n' for which 3^(1024) – 1 is divisible by 2^n?1) 132) 103) 114) 12
by euler's theorem 3^(2^(n-1)) = 1 mod(2^n). Clearly n=11
on checking we find that n=12 also satisfies this condition but n=13 does not. So n=12 is the answer here and not n=11
@Rakaesh said:If sin(x-2y)=cos(4y-x),find cot 2y.I am having problem in solving it.Can anyone give me some idea?
sin(x-2y)=cos(4y-x) means that x-2y and 4y-x are complementary angles.
So x-2y+4y-x=90 => y=45 => cot(2y)=0
(1) All the divisors of 360, including 1 and the number itself, are summed up. The sum is 1170. What is the sum of the reciprocals of all the divisors of 360?
(2) How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?
(3) A number when devide by N leaves remainder 4.When one third of yhus number is devide by N,it leaves 29.what is the least value of such number greater than 1000?
(4) If K = 231 — 319, then how many positive divisors of K^2 are less then K but do not divide K?
(5) N is a (n + 1) digit positive integer which is in the form anan - 1an - 2 . . . a2a1a0, where ai (i = 0, 1, 2, . . ., n) are digits and an0, thus N = an 10n + an-1 10n - 1+ . . . + a1 10 + a0, where 0 ai 9 and an0. We define F (N) = (an + 1) (an - 1 + 1) . . . (a1 + 1) (a0 + 1) For Example F(3407) = (3 + 1) (4 + 1) (0 + 1) (7 + 1) = 160. Identify the number of two digit numbers such that F(N) = N + 1.
a. 9
b. 1
c. 6
a. 9
b. 1
c. 6
(6) 123456789101112131415161718192021222324252627.......upto4041424344 div by 45? what will be the quotient and remainder?
@TootaHuaDil said:(2) How many integers between 1 and 1000, both inclusive, can be expressed as the difference of the squares of two non negative integers?
750.
@TootaHuaDil said:(3) A number when devide by N leaves remainder 4.When one third of yhus number is devide by N,it leaves 29.what is the least value of such number greater than 1000?
1002.??