Questions on Properties of Numbers

n × (n² – 1) = (n – 1) × n × (n + 1), which is a product of three consecutive numbers.

Since n is odd, the numbers (n – 1) and (n + 1) are both even. As they are two consecutive even numbers one of these numbers will be a multiple of 2 and the other will be a multiple of 4.

Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three.

Hence, the product of three numbers will be a multiple of 8 × 3 = 24.

Therefore, the highest number that always divides n × (n² – 1) is 24.

Case 1: If n is odd, n × (n² – 1) is divisible by 24 as proved in the earlier question.

Case 2: If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive positive integers is always a multiple of 3 and n is even, the product n × (n² – 1) is divisible by 6. Since n is even 5n is even.

If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n × (5n + 2) is a multiple of 8.

Hence, the product n × (n² – 1) × (5n + 2) is a multiple of 24.

In the 1^st 99 positive integers, digit 4 comes 20 times.

Similarly, from 100 to 199, digit 4 comes 20 times.

Now from 200 to 299, digit 4 comes again 20 times out of which we need to subtract 5 numbers (254, 264, 274, 284 and 294).

Therefore, total number of times that we write the digit 4 = 20 + 20 + 20 - 5 = 55.

From page number 1 to page number 9, we will use 1 digit per page $\Rightarrow $ digits used = 9.

From page number 10 to page number 99, we will use 2 digits per page $\Rightarrow $ digits used = 2 × 90 = 180.

From page number 100 to page number 252, we will use 3 digits per page $\Rightarrow $ digits used = 3 × 153 = 459.

Therefore, total number of digits used = 9 + 180 + 459 = 648

1 + 2 + 3 + … + n = $\frac{n(n+1)}{2}$= M² (say) $\Rightarrow $ n(n + 1) = 2 M²

Now n and n + 1 will have no factor in common. Since RHS is twice the square of a positive integer, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square.

As twice of a perfect square will be even, the other square will be odd. We start investigating the odd squares and their neighbours. The fourth such numbers we get is 288.

n⁴ + 6n - 6n³ - n² < 0 $\Rightarrow $ n (n² - 1)(n - 6) < 0. n cannot be equal to 1 or 0 or 6 because LHS becomes 0. Now n² - 1 will always be positive, therefore, n (n - 6) should be negative $\Rightarrow $   n = 2, 3, 4 and 5.

In order to write the above expression in the form (a + b)² = a² + 2ab + b², we note that 2⁸ = (2⁴)² and 2¹¹ = 2 × 2⁴ × 2⁶. Therefore, we need the square of 2⁶ $\Rightarrow $ 2ⁿ = (2⁶)² = 2¹² $\Rightarrow $ n = 12.

Nth term = (n² - 1) = (n + 1)(n - 1) $\Rightarrow $   series = 1 × 3 × 2 × 4 × 3 × 5 … × (n - 2) × (n) × (n - 1) × (n + 1) = 2 n (n + 1) × k² because all the other terms are squared.

The first value of n for which 2 n (n + 1) is a perfect square is n = 8.

p = $\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}-\sqrt{7}}$ = $\frac{{{(\sqrt{8}+\sqrt{7})}^{2}}}{(\sqrt{8}-\sqrt{7)}(\sqrt{8}+\sqrt{7})}$= ${{(\sqrt{8}+\sqrt{7})}^{2}}$ = $15+2\sqrt{56}$

Similarly, q = ${{(\sqrt{8}-\sqrt{7})}^{2}}$ = $15-2\sqrt{56}$

p² + pq + q² = ${{(15+2\sqrt{56})}^{2}}$+ ${{(15)}^{2}}-{{(2\sqrt{56})}^{2}}$+ ${{(15-2\sqrt{56})}^{2}}$ = 675 + 224 = 899

x² – y² = (x + y)(x – y) = 101. But 101 is a prime number and cannot be written as product of two numbers unless one of the numbers is 1 and the other is 101 itself. Hence, x + y = 101 and x – y = 1. à x = 51, y = 50.

\(\Rightarrow\) x² + y² = 51² + 50² = 5101.

When p = 3, the expression gives a prime number (17).

When p is not equal to 3, p² will be of the form 3k + 1 as every square number is of the form 3n or 3n + 1. Therefore, p² + 3p - 1 = 3k + 1 + 3p - 1 $\Rightarrow $ a multiple of 3. Therefore, for only p = 3, do we get a prime number from the expression.

The product (n - 1) (n - 2) (n - 3)..3×2×1 will not be divisible by n only when this product does not contain factors of n, i.e. n is a prime number. The prime numbers in the range given are 13, 17, 19, 23, 29, 31, and 37. 7 numbers in all.

4³ = 64. Digital sum of 64 is = 1. 4²⁴ = 4³ × 4³ × 4³ …× 4³ (8 times). Digital sums on both sides will be the same. $\Rightarrow $ digital sum of 4²⁴ = digital sum of 1 × 1 × 1 × 1… (8 times) = 1

It can be seen that

$[ \sqrt { 1 } ] = 1 , [ \sqrt { 2 } ] = [ 1.41 ] = 1 , [ \sqrt { 3 } ] = [ 1.73 ] = 1 , [ \sqrt { 4 } ] = 2$ and so on.

Therefore, from $[ \sqrt { 1 } ]$ to $[ \sqrt { 3 } ]$ , the value will be 1 , from $[ \sqrt { 4 } ]$ to $[ \sqrt { 8 } ]$ the value will be 2 , from $[ \sqrt { 9 } ]$ to $[ \sqrt { 15 } ]$ the value will be 3 and so on..

Therefore, the total value $= 3 \times 1 + 5 \times 2 + 7 \times 3 + \ldots + 13 \times 6 + 2 \times 7$

$= 217$ .

It is clear that $[ \sqrt { 25 } ] = 5 , [ \sqrt { 26 } ] = 5 , [ \sqrt { 27 } ] = 5$ and so on. The highest value of $x$ that we can take is 35 , since $[ \sqrt { 35 } ] = 5$ but $[ \sqrt { 36 } ] = 6$

Similarly, the highest value of y we can take is 48 , since

$[ \sqrt { 48 } ] = 6$ but $[ \sqrt { 49 } ] = 7$

Therefore, the greatest value of $x + y = 35 + 48 = 83$