When the second digit is 1, third digit can be 0, i.e., there is one such number.
When the second digit is 2, third digit can be 0 or 1, i.e., there are 2 such numbers.
When the second digit is 3, third digit can be 0, 1 or 2, i.e., there are 3 such numbers, and so on.
When the first digit is 7, second digit can be 1, 2, 3, 4, 5 or 6
So, there are (1 + 2 + 3 + 4 + 5 + 6) = 21 such numbers between 700 and 799
When the first digit is 8, second digit can be 1, 2, 3, 4, 5, 6 or 7
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7) = 28 such numbers between 800 and 899
When the digit is 9, second digit can be 1, 2, 3, 4, 5, 6, 7 or 8
So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 such numbers between 900 and 999
Hence, the required number
= (21 + 28 + 36)
= 85

On dividing 8112 by 99, we get 93 as remainder.
So, the required number to be subtracted is 93.

Clearly, 639 is not divisible by 7
Consider 2079
Sum of its digits
= (2 + 0 + 7 + 9)
= 18
So, it is divisible by both 3 and 9
Also, (79 - 2) = 77, which is divisible by 7
So, 2079 is divisible by 7
Also, (9 + 0) - (7 + 2) = 0
So, 2079 is divisible by 11
Hence, 2079 is divisible by each one of 3, 7, 9 and 11

Let n = 1 + x = 1 + m (m + 1) (m + 2) (m + 3), where m is a positive integer.
Then, clearly two of m, (m + 1), (m + 2), (m + 3) are even and so their product is even.
Thus, x is even and hence n = 1 + x is odd
Also, n = 1 + m (m + 3) (m + 1) (m + 2) = 1 + (m2 + 3m) (m2 + 3m + 2)
⇒ n = 1 + y (y + 2), where m2 + 3m = y
⇒ n = 1 + y2 +2y = (1 + y)2 , which is a perfect square.
Hence, I and III are true.

Required numbers are multiples of (10 × 13), i.e., 130
These numbers are 130, 260, 390, 520, 650, 780 and 910
They are 7 in number.

(1012 + 25)2 - (1012 - 25)2
= 4 × 1012 × 25
[∵ (a + b)2 - (a - b)2 = 4ab]
= 1012 × 100
= 1012 × 102
= 1014
Hence, n = 14

Let the required prime number be p.
Let p when divided by 6 give n as quotient and r as remainder. Then p = 6n + r, where 0 $$ \leqslant $$ r < 6
Now, r = 0, r = 2, r = 3 and r = 4 do not give p as prime.
∴ r $$ \ne $$ 0, r $$ \ne $$ 2, r $$ \ne $$ 3, and r $$ \ne $$ 4
Hence, r = 1 or r = 5

Given series is 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, .....
∴ 96th term is 4
So, $${T_{97}}$$ = 1 and $${T_{98}}$$ = 2
Hence, the 98th term is 2

We may have (64 and 1), (32 and 2), (16 and 4) and (8 and 8)
In any case, the sum is not 35

(a) In 183654729, 1836547 is not divisible by 7
(b) In 381654729, 38 is divisible by 2,
381 is divisible by 3,
3816 is divisible by 4,
38165 is divisible by 5,
381654 is divisible by 6,
3816547 is divisible by 7
38165472 is divisible by 8 and 381654729 is divisible by 9
(c) In 983654721, 983 is not divisible by 3
(d) In 981654723, 9816547 is not divisible by 7

231
= 2 × 230
= 2 × (22 )15
= 2 × 415
When n is odd, (xn + an) is divisible by (x + a)
∴ (415 + 115) is divisible by (4 + 1)
⇒ (415 + 1) is divisible by 5
⇒ (230 + 1) is divisible by 5
⇒ On dividing 230 by 5, we get (5 - 1) i.e., 4 as remainder.
∴ Remainder obtained on dividing 231 by 5
= Remainder obtained in dividing (2 × 4) i.e., 8 by 5 = 3

$${p^3} - {q^3} = \left( {p - q} \right)$$ &nbsp $$\left\{ {{{\left( {p - q} \right)}^2} - xpq} \right\}$$
$$ \Rightarrow \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right)$$     $$ = \left( {p - q} \right)$$   $$\left\{ {{{\left( {p - q} \right)}^2} - xpq} \right\}$$
$$\left\{ {\because {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right\}$$
By cancelling same terms of both sides
$$ \Rightarrow {p^2} + {q^2} + pq = {p^2} + {q^2}$$     $$ - 2pq - xpq$$   $$\left\{ {{{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right\}$$
$$\eqalign{ & \Rightarrow 3pq = - xpq \cr & \Rightarrow x = - 3 \cr} $$

Since, 653xy is divisible by 80, we must have y = 0
Now, 653x0 must be divisible by both 5 and 16
Clearly, it is divisible by 5 for all values of x
Now, the number 53x0 must be divisible by 16
The least value of x is clearly 6
So, x + y
= 6 + 0
= 6

= (a2 + b2 + c2 - ab - bc - ca)
= (a + b + c)2 - 3 (ab + bc +ca)
= (7 + 5 + 3)2 - 3 (35 + 15 + 21)
= (15)2 - 3 × 71
= (225 - 213)
= 12

Let P = x + $${r_{1}}$$ and Q = y + $${r_{2}}$$, where each of x and y are divisible by the common divisor.
Then, P + Q = (x + $${r_{1}}$$) + (y + $${r_{2}}$$)
                    = (x + y) + ($${r_{1}}$$ + $${r_{2}}$$)
(P + Q) leaves remainder $${r_{3}}$$ when divided by the common divisor.
⇒ [(x + y) + ($${r_{1}}$$ + $${r_{2}}$$) - $${r_{3}}$$] is divisible by the common divisor.
Since (x + y) is divisible by the common divisor, so divisor = $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$

Let the two numbers be are a and b
∴ a + b = 14.....(i)
a - b = 10.....(ii)
By adding equation (i) and (ii) we get
2a = 24
∴ a = 12 and b = 2
∴ Product of these two numbers
= 12 × 2 = 24

n3 is odd ⇒ n is odd and n2 is odd
∴ I and II are true.

Three consecutive multiples of 3 are 3m, 3(m + 1) and 3(m + 2)
Their product = 3m × 3(m + 1) × 3(m + 2)
                      = 27 × m × (m + 1) × (m + 2)
Putting m = 1, this product is (27 × 1 × 2 × 3) = 162
So, this product is always divisible by 162

$$\eqalign{ & \left( {800 \div 64} \right) \times \left( {1296 \div 36} \right) \cr & = \frac{{800}}{{64}} \times \frac{{1296}}{{36}} \cr & = 450 \cr} $$

225 = 9 × 25, where 9 and 25 are co-primes.
So, a number is divisible by 225 if it is divisible by both 9 and 25
Given number is divisible by 25, only if 7q is divisible by 25, i,e, if q = 5
Sum of digits of given number
= (8 + 7 + 6 + p + 3 + 7 + 5)
= 36 + p, which must be divisible by 9
This is possible if p = 0
Hence, p = 0,   q = 5