QBM002
M151. What is the total number of different divisors including 1 and the number that can divide the number 3600?
M152. A set X consists of 100 natural nos, each of which is a perfect cube. The maximum no of elements of X that one can always find such that each of them leaves the same remainder when divided by 17 is
M153. Let x and y be positive whole numbers, and let p be any odd prime.
It is well known that x3 + y3 is never equal to an odd prime.
But given that n is a positive integer which contains an odd factor greater than one, prove that xn + yn = p has no solutions.
M154. The largest number amongst the following that will perfectly divide 101100 - 1 is
(1) 100 (2) 10,000 (3) 100100 (4) 100,000
M155. (x, y) is a pair of non-negative integers such that (x+ y - 5)^2 = 9xy.
What are the possible values of (x,y)
QM156.
QM157. How many four digit numbers exist which can be formed by using the digits 2, 3, 5 and 7 once only such that they are divisible by 25?
M158. A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
M159. Given that x and y are integer, how many different solutions does the equation |x| + 2|y| = 100 have?
M160. A number consisting entirely of the digit one is called a repunit; for example, 11111.
Find the smallest repunit that is divisible by 63.
END
@Abhishek
Q.151
3600 =24 x 32 x 52
total factors = 5x3x3 = 45.
Hey..... yea..sorry... i thought 45 does not include 1 and the number 3600..jus referred a book n confirmed..
45 is the answer..
Thanks
Abi
n/a
Hey..... yea..sorry... i thought 45 does not include 1 and the number 3600..jus referred a book n confirmed..
45 is the answer..
Thanks
Abi
n/a
Ans: 111111
clue... the number should be divisible by 7 and 3
Abi
n/a
Here unit place should be = 5 i.e 1 way
Ten place can be 2 or 7 i.e 2 way
Balance two number cane be arranged thus 1x2x2x1 =4 Ans
Here |x| + 2|y| = 100 or |y| = (100 -|x|)/2.
Thus Here |y)| .=0 thue the values for
x=0 y=-100 and +100 = 2sol
x=2 and x=-2 y=-46and+46 thus 4 ways
... x=100 or x=-100 then y = 0 2 ways
Solution = 2+4+4+...+2 where 2n-2 =100 or n =51 term is 100 thus 4+..4+(50times) = 4x50=200 Ans.
M158. A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
n = q.d + r
Then r< d and (d - r) is ethier 1 or 2.
i cant understand........., how it solve?????????
M155. (x, y) is a pair of non-negative integers such that (x+ y - 5)2 = 9xy.
What are the possible values of (x,y)?
Sol.
only two possible pairs of (x, y) are (0, 5) and (5, 0)
total no.of factors = 45 + 2 = 47
Abi
n/a
n/a