Number System - Aptitude MCQ Questions and Solutions with Explanations

a² + b² + c² = 1
So, the maximum value of a² b² c² = $$\frac{1}{3}$$ × $$\frac{1}{3}$$ × $$\frac{1}{3}$$ = $$\frac{1}{27}$$
(∵ when sum of three positive quantities is fixed, the product will be maximum when the quantities are equal)
Hence, maximum value of abc = $$\frac{1}{{\sqrt {27} }} = \frac{1}{{3\sqrt 3 }}$$

21 = 3 × 7 is not a prime number because 21 is a composite number.

Let 3x + 7y = 11k
Then, y = $$\frac{(11k - 3x)}{7}$$
Then,
$$\eqalign{ & = 5x - 3y \cr & = 5x - \frac{{3(11k - 3x)}}{7} \cr & = \frac{{35x - 33k + 9x}}{7} \cr & = \frac{{44x - 33k}}{7} \cr & = \frac{{11(4x - 3k)}}{7} \cr} $$
Which is divisible by 11

Clearly, each of the 38 terms
$$\left[ {\frac{1}{4}} \right], \left[ {\frac{1}{4} + \frac{1}{{50}}} \right], \left[ {\frac{1}{4} + \frac{2}{{50}}} \right], $$      $$ ..... $$ $$ , \left[ {\frac{1}{4} + \frac{{37}}{{50}}} \right]$$   has a value lying between 0 and 1,
While each one of the 12 terms
$$\left( {\frac{1}{4} + \frac{{38}}{{50}}} \right),$$   $$\left( {\frac{1}{4} + \frac{{39}}{{50}}} \right),$$   $$.....,$$ $$\left( {\frac{1}{4} + \frac{{49}}{{50}}} \right)$$   has a value lying between 1 and 2.

Hence, the given expression
= (0 × 38) + (1 × 12)
= 12

(xⁿ + aⁿ) is divisible by (x + a) when n is odd.
∴ (12ⁿ + 1) is divisible by (12 + 1) i.e., 13 when n is odd.

(2¹² - 1) = (4096 - 1) = 4095, when is clearly divisible by 3, 5, 7 and 13 i.e., four numbers in all.

The smallest five digit number is 10000
The least number divisible by 88
Then the remainder is 56
= 10000 + (88 - 56)
= 10000 + 32
= 10032

Let the two consecutive integers be a and (a + 1).
Then,
(a + 1)² - a²
= a² + 1 + 2a - a²
= (2a + 1)
= (a + a + 1) = sum of given integers

The smallest 6-digit number is 100000
On dividing 100000 by 111, we get 100 as remainder.
So, the number to be added = (111 - 100) = 11
Hence, the required number = 100011

Clearly, the list of prime numbers from 2 to 99 has only 1 multiple of 2 and only 1 multiple of 5
So, number of zeros in the product = 1