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 A be the point at which the goat is tied.Now the radius of the circle is 21So Total area of the circle will be pie r^2 = pie * 21 *21But it covered 3/4 th of the circle + Some extra partSo it will be 3/4 (pie * 21^2) + Red part + Yellow part= 3/4( pie * 21^2) + 1/4(pie * 14^2) + 1/4(pie * 7^2)= 1232OA = A
[credits : @hemant_malhotra]
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.
The perfect squares that have the last two digits in the form of bb will end in either 00 or 44. So, we need to consider 38^2, 62^2 and 88^2. Only 88^2 will satisfy the condition and will be equal to 7744 (remember! if you don’t already). So, a + b = 11.
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) Moiz has a machine into which he can put any number of one rupee coins. If he inserts n rupees, the machine returns 2n rupees. Each time he uses the machine, however, he must insert more money than he did on the previous use. If he starts with exactly Rs 1 and use the machine once, he will have Rs 2. On his next use of the machine, he is forced to insert Rs 2 yielding Rs 4, and on his third use of the machine, he can insert either Rs 3 or Rs 4 yielding a total of Rs 7 or Rs 8. The largest 2 digit integer Z such that it is impossible to obtain exactly Z rupees with the machine, starting with Rs 1 is
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?