We know that it should be an AP, 1st element is 1, and last element 1000. a = 1, a + (n-1) d = 1000 ; (n-1) d = 999. So 999 is the product of (n-1) and d. So, we need to see the factors of 999 to see what all values/how many of those values (n-1) and d can assume. Prime factorize 999. The prime factors are 3, 3, 3, and 37. (999 = 3 * 3 * 3 * 37). (Other than 1 and 999)So, we can have(n - 1) = 3, d = (3 * 3 * 37)(n - 1) = 3 * 3, d = (3 * 37)(n - 1) = 3 * 3 * 3, d = (37)(n - 1) = 37, d = (3 * 3 * 3)(n - 1) = 37 * 3, d = (3 * 3)(n - 1) = 37 * 3 * 3, d = (3)
Other than this, we can have 1 and 999 also as factors of 999.If we have n-1 as 1, n is 2. But that means only 2 terms in AP. (Not possible according to question)If we have n-1 = 999, n = 1000. d = 1. This is possible.So, we have totally, 7 possible combinations.
The factors of 999 are 1,3,9,27,37,111,333,999. We can assign each of these values to d or (n-1), and find the other one correspondingly and proceed. Again in this case, (n-1) cannot be 1. All other cases are possible. Both the approaches are essentially the same. Just that you need not spend time multiplying the factors in the 1st approach.
f(7)^2 - f(6)^2 = 517
(f(7) + f(6)) (f(7) - f(6)) = 517 = 11 x 47
As 11 and 47 cannot be further factorized and f(7) or f(6) should be natural numbers, we can say
f(7) + f(6) = 47 (Largest prime factor should come from their sum)
f(7) - f(6) = 11 (smallest prime factor should come from their difference)
solving, f(7) = 29 and f(6) = 18
f(10) = f(8) + f(9)
= f(7) + f(6) + f(7) + f(8)
= f(7) + f(6) + f(7) + f(7) + f(6)
3f(7) + 2f(6) = 3 x 29 + 2 x 18 = 123.
Let the two digit number be represented as ab.
a^2 – b^2 = (a – b)(a + b)
Now, |a^2 – b^2| will be a prime number only when |a – b| = 1 and (a + b) comes out to be a prime number
Two digit numbers in which |a – b| = 1 are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87,89, 98
From these numbers, the numbers in the form of 'ab' in which (a + b) comes out to be a prime number are 12, 21, 23, 32, 34, 43, 56, 65, 67, 76, 89 and 98.
There are 12 two digit numbers such that absolute difference of the square of the two digits results in a prime number.
Q100) A and B play a game of dice between them. The dice consist of colors on their faces (instead of numbers). When the dice are thrown, A wins if both show the same color; otherwise B wins. One dice has 4 red face and 2 blue faces. How many red and blue faces should the other dice have for both the to players have same chances of winning?
a) 3 red and 3 blue faces
b) 2 red and remaining blue
c) 6 red and 0 blue
d) 4 red and remaining blue
Q97) The percentage profit earned by selling an article at Rs. 1,920 is equal to the percentage loss incurred by selling the same article at Rs. 1,280. At what price (in Rs.) should the article be sold to make a profit of 25%?
Q96) From a rectangular sheet of dimensions 30 cm × 20 cm, four squares of equal size are cut from the four corners. Then the resulting four sides are bent upwards to give it the shape of an open box. If the volume of the box is 1056 cm^3, what is the length of the side of the squares cut from the corners?
Q91) A four-letter code has to be formed using the alphabets from the set (a, b, c, d) such that the codes formed have odd number of a’s. How many different codes can be formed satisfying the mentioned criteria?
Q100) The following table gives the concentration of 8 different samples of sugar solutions. Exactly 3 samples are mixed to form a 15% sugar solution. How many such combinations of samples are possible ?
Q100) There is a 3 digit no. ABC. The no. Is a perfect square and the no. of factors of ABC is also a perfect square.
If A + B + C is also a perfect square then what is the no of factors of the six digit number ABCABC.what is the no of factors of the 6 digit no ABCABC if the cube of the product of the digits of the no ABC is not divisible by 5?