Re: New Set: Number Properties!!!
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01 Dec 2020, 04:50
1.
St1 -- x could be 1 or -1
St2 -- again x could be 1 or -1
Answer - E
2. As n has only 2 positive factors, n must be a prime number as it has 1 and itself as its factors.
St1 -- n/2 is one of the factors of n, n can be only be 2. SUFFICIENT
St2 -- As LCM of n and n+10 is an even number. Apart from 2, all other prime numbers are even. So, if n is not 2, we will have n (Odd) and n+10 (another odd number), their LCM cannot be even. Hence n = 2.
Answer - D
3. x and y are consecutive perfect squares
St1 - Both x & y have 3 positive factors -> x & y are squares of prime numbers. If x = 2^2, and y =3^2, then remainder is 1, but if x=3^2 and y = 5^2, then remainder is 7. NOT SUFFICIENT
St2 - Again x & y are square of prime numbers, which is same as St1.
Answer - E
4. As each digit is a positive multiple of 4, it can be 4 or 8
St1 - We can have unit digit as 4, then tens and hundreds digit will be also 4 & 4 (number = 444), but, we can also have unit digit as 8, and tens and hundreds digit can be 4 & 8, 8 & 4 and 8 & 8. NOT SUFFICIENT
St2 - Sum of digits of multiple of 3 is also divisible by 3. Hence sum of digits of the number should not be divisible by 3. So, 444 and 888 are out. We can have any combination of 8,8,4 or 4,4,8 . NOT SUFFICIENT
Combining two statements, we are still left with 848 and 488. Hence E
5.
St1 -- As a<b<c, median of a!, b! & c! will be b!. As b! is odd, we only have 0! and 1! as odd number (i.e. 1). So b can be 0 or 1, but if b will be 0, then a will be -ve integer, and we do not have factorial of -ve number defines, thus a = 0, and b= 1, but c can be any integer. NOT SUFFICIENT
St2 -- c! is a prime number, thus c=2, as 2! = 2, and that is the only prime possible for factorials. but, a & b can be any integer.
Combining, we have a=0, b=1 and c=2. Hence C.
6.
St1 -- we can have {2,3,-2} or {-2,-3,-2}. NOT SUFFICIENT
St2 -- We can have {-3,-2,-1} or {1,2,3}. NOT SUFFICIENT
Combining, as we cannot have all elements +ve due to St1, and St2 restricts all elements to be of same sign. Thus, all elements will be -ve. SUFFICIENT
7.
St1 -- x = 12k + 6 = 6(2k+1), it will be a square number if 2k+1 is an odd power of 6, as 2k+1 will always be a odd number. x is not a square of an integer. SUFFICIENT
St2 -- x = 14k + 2 = 2(7k+1). For k=1, x is a square, but for k=2, it is not. NOT SUFFICIENT
Hence, A
8.
St1 -- That is already given in the question, no additional info provided. NOT SUFFICIENT
St2 -- A terminating decimal is one, which only has 2 and 5 or their multiple in denominator. So, all of them can be 1/3, thus median will be 1/3 > 1/5. Also, all elements can be 1/7, thus median will be 1/7<1/5 NOT SUFFICIENT
Combination, has same result as St2, cause St1 provided no additional info. Hence E
9.
St1 - As ab=2, a & b can be both +ve or both -ve. If both are -ve, we will have [a] or [b] to be <0, thus [a] + [b] cannot be 1. If both are +ve, then for [a] + [b] = 1, we need to have one of them as 0, and other 1, but if one of the number is less than 1 to have [] as 0, we need to have other number >2 to have ab=2, thus [a]+[b] != 1. SUFFICIENT
St2 -- 0<a<b<2, we can have multiple values for a &b, which can provide [a]+[b] as 0 or 1 or 2. NOT SUFFICIENT.
Hence A
10.
St1 -- if 9 is not a factor of N, x can be 1 only. If x=1, then y=5 to have 12 positive factors. SUFFICIENT
St2 -- if 125 is a factor of N, then y>=3, we can have y=3, x=2, or y=5, x=1. NOT SUFFICIENT
Hence A
11.
St1 -- |x - 2| < 2 - y. As |x-2| >=0, thus 2-y>0 -> y<2 -> y=1 (+ve integer). If y=1, then |x-2| < 1 -> 1< x 3 -> x=2 9 (prime number). SUFFICIENT
St2 -- x+y-3 = |1-y|. For y<=1 -> x+y-3 = 1-y -> x+2y=4. As x & y are +ve integers, only possible solution is x=2 & y=1. For y>1 -> x+y-3 = y-1 -> x=2. SUFFICIENT.
Hence C