Is integer N a prime number?I. N = p! + 1 for some prime number pII. q! + 1 a) if the question can be answered using statement I alone but not using II aloneb) if the question can be answered using statement II alone but not using I alonec) if the question can be answered using either statement aloned) if the question can be answered using statements I and II together but not using eitheralonee) if the question cannot be answered even by using both statements I and II together
Is integer N a prime number?I. N = p! + 1 for some prime number pII. q! + 1 a) if the question can be answered using statement I alone but not using II aloneb) if the question can be answered using statement II alone but not using I alonec) if the question can be answered using either statement aloned) if the question can be answered using statements I and II together but not using eitheralonee) if the question cannot be answered even by using both statements I and II together
statement 1 5!+1=121 not a prime no. and 3!+1=7 prime no. so nothing can be said,
statement 2 well n>q!+1 so it can be q!+2mod2=0, q!+3mod3=0.....q!+(q-1)modq-1=0 so it cannot be prime no. so from statement b alone we can answer however nothing can be said by 1st statement. so optB
Consider the following quadratic equation: The roots of this equation lie in the interval ( €“4, 5). Find the sum of [p] where p takes all its possible values.[x] indicates the Greatest Integer Function less than or equal to x.OPTIONS1) 0 2) 4 3) 5 4) None of these
According to the survey, the population of a city increase by 10% every year for two years and then decreases by 10% every year for two years. If population just before 4 years was 100000, what was it after four years? 108900980109610092400980100
1). P(17) = a.(17)^2 + b.(17) + c = 10 P(24) = a.(24)^2 + b.(24) + c = 17
Subtracting 1st equation from the second one, we get 41a + b = 1 => the two ordered pair of solutions for (a,b) are (-1, 42) and (1, -40)
Now, P(n) = a.n^2 + b.n + c = n + 3 => a.n^2 + (b-1)n + c - 3 = 0
We require n1 + n2, which is the sum of roots = - (b-1)/a We get the value as 41 as the unique answer.
2). x2 − 2px − 1 + p2 = 0
∴ x2 − 2px + p2 − 1 = 0
∴ (x − p)2 − 1 = 0
∴ (x − p + 1)(x − p − 1) = 0 So, the roots of the equation are (p – 1) and (p + 1), both of which lie in the interval (–4, 5).
Case 1:
p– 1 > –4, and p + 1 =>p > –3, and p =>–3 p When –3 p p] = –3; When –2 ≤ p p] = –2; When –1 ≤ p p] = –1; When 0 ≤ p p] = 0; When 1 ≤ p p] = 1; When 2 ≤ p p] = 2; When 3 ≤ p p] = 3;
Thus, the sum of [p] is 0, as p assumes values in –3 p
Using statement I alone, N may or may not be prime, e.g. 3! + 1 = 7 (prime), 5! + 1 = 121 (not prime) Using statement II alone, N can be equal to q! + 2 or q! + 3 or €Ś q! + q which are divisible by 2, 3, €Ś q respectively. So, N cannot be prime. Hence, [2].
What is the 2037th positive integer that can be expressed as the sum of two or more consecutive positive integers? (The first three are 3 = 1+2, 5 = 2+3, and 6 = 1+2+3.)
A Question to be answered::What is the 2037th positive integer that can be expressed as the sum of two or more consecutive positive integers? (The first three are 3 = 1+2, 5 = 2+3, and 6 = 1+2+3.)No OA...Please discuss in details. (Edited) Is the answer 2049 ???
all numbers except for powers of 2
upto 2037th term, 2^0-2^10 - 11 numbers taking next 11 numbers we get 2037+11=2048 but 2048=2^11 =>2049
A Question to be answered::What is the 2037th positive integer that can be expressed as the sum of two or more consecutive positive integers? (The first three are 3 = 1+2, 5 = 2+3, and 6 = 1+2+3.)No OA...Please discuss in details. (Edited) Is the answer 2049 ???
Consider the increasing sequence - 1, 3, 4, 9, 10, 12, 13 €Ś and so on. The sequence consists of all those positive integers which are powers of 3 or sum of distinct powers of 3.
A Question to be answered::What is the 2037th positive integer that can be expressed as the sum of two or more consecutive positive integers? (The first three are 3 = 1+2, 5 = 2+3, and 6 = 1+2+3.)No OA...Please discuss in details. (Edited) Is the answer 2049 ???
basically the numbers that dont have any odd factors cant be expressed as a sum of 2 or more consecutive no.s so you have exclude only powers of 2.
and so you will need to subtract upto 2^11 so actually 12 numbers so 2037th terms must be 2037+12=2049
a) if the question can be answered using statement I alone but not using II alone b) if the question can be answered using statement II alone but not using I alone c) if the question can be answered using either statement alone d) if the question can be answered using statements I and II together but not using either alone e) if the question cannot be answered even by using both statements I and II together
@krum: :thumbsup: for the second question's solution :D
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