Number Properties - Test 1

n³ - n = n (n² - 1) = n (n + 1) (n - 1)
So n³ - n can be expressed as a product of three consecutive numbers, then at least one of the three consecutive numbers should be a multiple of 3. Hence n³ - n can be divisible by 3.

(6x + 5) (6y + 7) = 36xy + 42x + 30y + 35
So (6x + 5) (6y + 7)/6 = 6xy + 7x + 5y + 35/6, the remainder should be the same as 35/6, which is 5.

49 = 77, 77 = 711
So 77 is the smallest common multiple of 1, 7, 11 and 77, then we just need find out the smallest common multiple of 49 and 77, which should be 7711 = 539.

21 is the greatest common factor of x and y. So 3 and 7 are two common 2 prime factors of x and y. Hence, x still has one prime factor a and y still has two prime factors b and c. The prime factors of x*y should be a, b, c, 3 and 7, the total number is 5.

120 = 524 = 45*6. So 120 can be expressed as the product of its consecutive factors.

assume the number of chairs is x.
So 80 < x < 120 and x is a common multiple of 9 and 12. The least common multiple of 9 and 12 is 36, so the common multiple of 9 and 12 between 80 and 120 can only be 36*3 = 108. 

1. only, 8^ (- x) = 2^ (- 3x) < 2^ (- 12) so - 3x < - 12, x > 4. Because x should be less than 7 and x is a prime number, x must be 5, sufficient
2. only, x > 4 and x < 7, x is prime number, x must be 5, sufficient.

assume y = 235*7 = 210, y has 4 prime factors and y is the least possible value which has 4 different prime factors. So if x has more than 4 different prime factors, x must be more than 210. 

1. only, x < 210, x cannot have more than 4 prime factors, sufficient. 2) only, x > 210, x may have more than 4 different prime factors, but x may have only 4 prime different factors, insufficient.

8x 9y 70 = 8x 9y00 + 70
8x 9y00 can be divisible by 4, but 70 cannot be divisible by 4. So 8x 9y 70 cannot be divisible by 4.

24 = 234, if x²  is divisible by 24, x² must have factors of 2 and 3.
1)  only, x does not have factor of 3, so x² does not have factor of 3, x² cannot be divisible by 24, sufficient.
2)  only, x is odd, so x does not have factor of 2, so x²  does not have factor of 2, x² cannot be divisible by 24, sufficient.

9! - 2* (7!)²
= (7!) (89 - 27!) = - 18*(7!) - 18 cannot be divisible by 5. 7! = 1234567 can only be divisible by 5^1. Since n is a positive integer, so the greatest possible value of n should be 1.

1)  only, the remainder is 42, so x + 10 must be greater than 42. x + 10 > 42, x > 32, x must be greater than 20, sufficient.
2)  only, the same reason as 1), bx > 36, we don't know the value of b, so we cannot know if x would be greater than 20 or not, insufficient.

2^1 = 2, 2 ^ 2 = 4, 2 ^ 3 = 8, 2 ^ 4 = 16, 2 ^ 5 = 32, 2 ^ 6 = 64… 
We can find out that the units digit of 2^n is the same as the units digit of 2^ (n + 4)
So the units digit of 2^60 would be the same as the units digit of 2^4, which is 6. Hence, the remainder of 2^60 divided by 10 will be 6.

1. only, a (50) = a (49) + 5 = a (48) + 5 + 5 = a (48) + 25 =…= a (1) + 495 = 3 + 495, so a (50) - 1 = 495 + 2, which cannot be divisible by 5, sufficient.
2. only, a (50) = a(49) + 5, if a (49) is not divisible by 5, then a(50) must not be divisible by 5, but a (50) - 1 might be divisible by 5. Because we don't know the remainder of a(50) divided by 5, we don't know if a (50) - 1 can be divisible by 5 or not, insufficient.

x = 5n, n is a positive integer.
1)  only, assume x = 4m + 3, m is a positive integer, so 5n = 4m + 3. If m = 3, n = 3, x = 15, the remainder will be 1. But if m = 8, n = 7, x = 35, the remainder will be 0. We don't know the value of x, so we don't know the remainder of x divided by 7, insufficient.
2)  only, assume x = 14p + 1, p is a positive integer, so x/7 = (14p + 1)/7 = 2p + 1/7, the remainder is 1, sufficient.

n can always be divisible by itself, so n is not an odd number.
n is not divisible by any odd number, so n's factors can only be 2 or 2m.
Hence, n can only be 2, 2^2 = 4, 2^3 = 8, 2^4 = 1^6 and 2^5 = 32. There are 5 possible values.

15 = 35, 21 = 37, 36 = 312
So the smallest common multiple of 15, 21 and 36 would be 35712 = 1260. D is right.

x = 9a + 3 = 15b + 3 = 21c + 3, a, b and c are all integers. So 9a = 15b = 21c, the least common multiple of 9, 15 and 21 will be 335*7 = 315. So 9a = 15b = 21c = 315, a = 35, b = 21, c = 15. Hence, x = 315 + 3 = 318.

x - 2 should not be negative, so x > = 2
1)  only, x can only be 2, then the remainder of 13 divided by 2 will be 1, which is not x - 2 = 2 - 2 = 0, so x cannot be less than 3, insufficient.
2)  only, 13 = nx + (x - 2), n is an integer. So (n + 1) x = 15 = 3*5, x can only be 3 or 5. x > 3, so x = 5, sufficient.

assume, there are x apples: 1)  only, x = 7 n + 3, we don't know the value of n, so we don't know the value of x, insufficient. 2)  only, x = 4m + 2, we don't know the value of m, so we don't know the value of x, insufficient. 1)  and 2) together, x = 7n + 3 = 4m + 2, so 7n + 1 = 4m. m can be 2 and n can be 1, then x = 10. But m can also be 9 and n can be 5, then x=35. So, there are at least two possible x value, insufficient.

2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64… We can find out that the units digit of 2ⁿ is the same as the units digit of 2^{(n + 4)}. So, the units digit of 2²⁰ would be the same as the units digit of 2⁴, which is 6. If the units digit of 2^{(20) - x} is 0 or 5, then it can be divisible by 5. So, x should be 1 or 6.

3¹ = 3, 3² = 9, 3³ = = 27, 3⁴ = 81, 3⁵ = 243, 3⁶ = 729…
We can find out that the units digit of 3^{(4n + 3)} should be the same as the units digit of 33, which should be 7. Hence, the units digit of 3^{(4n + 3)} - 7 should be 0, which means 3^{(4n + 3)} - 7 can be divided by 5. So the remainder of 3^{(4n + 3)} would be the same as the remainder of 7 divided by 5, which is 2.