How many 4 digit positive integral numbers are there in base 7 if you are counting the numbers in the same base system?
Bill & Clinton take the square of a certain no. in decimal system and express it in base 5 and 6 respectively. Then Bush comes and he takes the two representations and assuming that the expressions are in base 10 adds the nos. Which of the following cannot be the value of the unit's digit of the sum obtained?
a) 0 b) 2 c) 8 d) 6 e) 3
a) 0 b) 2 c) 8 d) 6 e) 3
If binary nos from 100 to 1000000 are written find the no of 1s in it.
@ravi.theja said:Solve for x:(a/ax-1)+(b/bx-1)=a+b
Multiply the original equation by (ax-1)(bx-1) first to get the whole thing over a common denominator.
- a·b·x^2 ·(a + b) + x·(a^2 + 4·a·b + b^2) - 2·(a + b) = 0
The standard quadratic formula is very messy to apply, but trying factorization instead, the above equation does actually factorize cleanly into:
(-(a + b)·x + 2)·(a·b·x - (a+b)) = 0
hence x = (a+b)/(a·b) or x = 2/(a+b)
how many numbers(n) are there between 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural numbers??
what annual payment will discharge a debt of Rs6450 due in 4 years at 5% simple interest??
a)1400rs b)1500 c)1550 d)1600
@TootaHuaDil said:how many numbers(n) are there between 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural numbers??
7 such numbers??
12,42,72,102,132,162 and 192...
@TootaHuaDil said:how many numbers(n) are there between 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural numbers??
All numbers of the form 2p, 3p and (5p-1)/2 are excluded where p is a prime.
for 2p conditions there are 25 primes
for 3p conditions there are 18 primes
for (5p-1)/2 conditions there are 22 primes
so there are total 65 numbers that we should exclude, so 200-65 = 135
@naveenkrs said:All numbers of the form 2p, 3p and (5p-1)/2 are excluded where p is a prime.for 2p conditions there are 25 primesfor 3p conditions there are 18 primesfor (5p-1)/2 conditions there are 22 primesso there are total 65 numbers that we should exclude, so 200-65 =135
i may have not understood your question but do you mean that every number of the form 2k, 3k and 5k-1/2 where k is a non-prime satisfies this condition??
@Logrhythm said:i may have not understood your question but do you mean that every number of the form 2k, 3k and 5k-1/2 where k is a non-prime satisfies this condition??
yeah, I have solved it this way. Tell me if I am wrong here.
@TootaHuaDil said:oa is 2
oh sry...i did not read the part "composite natural numbers"
12, 42, 72, 102 and 132 are all invalid..
only 162 and 192 are valid...hence only 2...
@naveenkrs said:yeah, I have solved it this way. Tell me if I am wrong here.
yes that is a wrong method...
n/2 and n/3 means n is div by 6...
(2n+1)/5 means n is of the form 5k+4/2
so we need to find numbers of the form -> 6x = (5y+4)/2
rest u can do....
@blowcat13 said:what annual payment will discharge a debt of Rs6450 due in 4 years at 5% simple interest??a)1400rs b)1500 c)1550 d)1600
1500?
principal amount = x
x + (x + x*5*1/100) + (x + x*5*2/100) + (x + x*5*3/100) = 6450
x = 1500
@TootaHuaDil said:how many numbers(n) are there between 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural numbers??
n/2 and n/3 are integer, so n = 6k
(2*6k + 1)/5 = (12k + 1)/5 is integer
12k + 1 = 5r
5r = 10k + (2k + 1)
k = 5p + 2
n = 30p + 12.
(2*6k + 1)/5 = (12k + 1)/5 is integer
12k + 1 = 5r
5r = 10k + (2k + 1)
k = 5p + 2
n = 30p + 12.
12,42,72,102,132,162,192... are the 7 numbers.....but only in the case of 162 and 192 we get i all three primes 😃 so ans is 2...no.'s are 162 and 192
@Logrhythm now is it ok :)
@Logrhythm now is it ok :)