Remainder = 48, Divisor = (5 × 48) = 240
12 × Quotient = Divisor
⇒ Quotient = $$\frac{240}{12}$$ = 20
Dividend = (240 × 20) + 48 = 4848

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

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 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

The least 6-digit number = 100000
On dividing 100000 by 349, we get 186 as remainder.
Required number :
= 100000 - (349 - 186)
= 100000 + 163
= 100163

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

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 = $$\frac{1}{{\sqrt {27} }} = \frac{1}{{3\sqrt 3 }}$$

Clearly, 2525 is not divisible by any of the numbers 7, 11 and 13
The smallest 3-digit prime number is 101

Hence (D) is true.

Take m = 15 and n = 20
Then, each one of m and n is divisible by 5
But (m + n) is not divisible by 10
Hence, (m + n) is divisible by 10 is not true.

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

= 52 + 62 + .... + 102 + 202
= (12 + 22 + 32 +..... + 102) - (12 + 22 + 32 + 42) + 400
= $$\frac{1}{6}$$ n(n + 1) (2n + 1) - (1 + 4 + 9 + 16) + 400, where n = 10
= $$\left( {\frac{1}{6} \times 10 \times 11 \times 21} \right)$$     - 30 + 400
= (385 - 30 + 400)
= 755

2m > 960 when the least value of m is 10
Then, 210 = 1024 and 1024 - 960 = 64 = 26
∴ m = 10 and n = 6

(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.

Let the number be x.
Let x when divided by 136 give q as quotient and 36 as remainder.
Then, x = 136q + 36
= (17 × 8q) + (17 × 2) + 2
= 17 × (8q + 2) + 2
So, the given number when divided by 17 gives 2 as remainder.

y = 2 × 1 + 1 = 3
x = 3 × y + 1 = 3 × 3 + 1 = 10
Clearly, 10 when divided by 6, leaves a remainder 4

= 136 × 12 × 8
= 136 × 96
= 136 × (100 - 4)
= (136 × 100) - (136 × 4)
= 13600 - 544
= 13056

Sum of the digits = 12, which is divisible by 3. So, the given number is divisible by 3
(Sum of digits at odd places) - (Sum of digits at even places)
= 6 - 6 = 0
So, the given number is divisible by 11
The given number when divided by 37 gives 3003003003
So, the given number is divisible by 37
The given number when divided by 111 gives 1001001001
Clearly, it is divisible by 111 as well as by 1001
Hence, the given number is divisible by each one of 3, 11, 37, 111 and 1001

We keep on dividing 99999..... by 7 till we get 0 as remainder.

∴ Required number = 142857
Number of digit = 6

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

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