If 210 is divisible by 11 then remainder becomes 1.
1 is subtracted from 210 so that the sum completely divisible by 11.

Let N = n2 (n2 - 1) = n2 (n - 1) (n + 1)
Then, n = 2
⇒ N = 22 × (2 - 1) (2 + 1)
        = (4 × 1 × 3) = 12
Hence, the required number is 12

350 + 926 + 2718 + 928 + 929
= 350 + 352 + 354 + 356 + 358
= 350[1 + 32 + 34 + 36 + 38]
= 350[1 + 9 + 81 + 729 + 6561]
= 350 × 7381
This no will be divided by 5, 7, 2, 50 it will be divisible by 11.

$$\eqalign{ & {\text{Given, }}\root 3 \of {12} \cr & {\bf{Concept \,used:}} \cr & {\text{If }}a > b,{\text{ then:}} \cr & {a^{\frac{1}{3}}} > {b^{\frac{1}{3}}} \cr & {\bf{Calculation:}} \cr & {\text{Option}}\left( {\text{A}} \right):\root 9 \of {1500} = {\left( {1500} \right)^{\frac{1}{9}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {9,\,3} \right) = 9 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 9}}{\text{ and }}{\left( {1500} \right)^{\frac{1}{9} \times 9}} \cr & i.e.,{\left( {12} \right)^3}{\text{ and }}{\left( {1500} \right)^1} \cr & i.e.,\left( {1728} \right){\text{ and }}\left( {1500} \right) \cr & {\text{Since, }}1728 > 1500 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {1500} \right)^{\frac{1}{9}}} \cr & {\text{Option}}\left( {\text{B}} \right):\root 5 \of {60} = {\left( {60} \right)^{\frac{1}{5}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {5,\,3} \right) = 15 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 15}}{\text{ and }}{\left( {60} \right)^{\frac{1}{5} \times 15}} \cr & i.e.,{\left( {12} \right)^5}{\text{ and }}{\left( {60} \right)^3} \cr & {12^2} \times {12^3}{\text{ and }}{12^3} \times {5^3} \cr & 144 \times {12^3}{\text{ and }}125 \times {12^3} \cr & {\text{1}}{{\text{2}}^3}{\text{ is common both but }}114 > 125 \cr & \Rightarrow {\text{Hence }}{12^5} > {60^3}{\text{ or }}{12^{\frac{1}{3}}} > {60^{\frac{1}{5}}} \cr & {\text{Option}}\left( {\text{C}} \right):\root 6 \of {121} = {\left( {121} \right)^{\frac{1}{6}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {6,\,3} \right) = 6 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 6}}{\text{ and }}{\left( {121} \right)^{\frac{1}{6} \times 6}} \cr & i.e.,{\left( {12} \right)^2}{\text{ and }}{\left( {121} \right)^1} \cr & i.e.,{\left( {12} \right)^2}{\text{ and }}{\left( {11} \right)^2} \cr & {\text{Since, }}12 > 11 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {121} \right)^{\frac{1}{6}}} \cr & {\text{Option}}\left( {\text{D}} \right):\root {12} \of {33214} = {\left( {33214} \right)^{\frac{1}{{12}}}} \cr & \Rightarrow \root 3 \of {12} = {12^{\frac{1}{3}}} \cr & \Rightarrow {\text{LCM}}\left( {12,\,3} \right) = 12 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3} \times 12}}{\text{ and }}{\left( {33214} \right)^{\frac{1}{{12}} \times 12}} \cr & i.e.,{\left( {12} \right)^4}{\text{ and }}{\left( {33214} \right)^1} \cr & i.e.,{\left( {20736} \right)^1}{\text{ and }}{\left( {33214} \right)^1} \cr & {\text{Since, }}33214 > 20736 \cr & \Rightarrow {\left( {12} \right)^{\frac{1}{3}}} > {\left( {33214} \right)^{\frac{1}{{12}}}} \cr & \therefore {\text{ The required greatest value is }}\root {12} \of {33214} . \cr} $$

Let the three fractions be p, q and r, where p < q < r.
According to the question,
$$\eqalign{ & \frac{r}{p} = \frac{7}{6} \cr & \Rightarrow r = \frac{7}{6}p \cr & {\text{Again, middle fraction}} \cr & = q = \frac{7}{6} - \frac{1}{3} = \frac{{7 - 2}}{6} = \frac{5}{6} \cr & \therefore p + q + r = 2\frac{{11}}{{24}} \cr & \Rightarrow p + \frac{5}{6} + \frac{7}{6}p = \frac{{59}}{{24}} \cr & \Rightarrow p + \frac{{7p}}{6} = \frac{{59}}{{24}} - \frac{5}{6} \cr & \Rightarrow \frac{{6p + 7p}}{6} = \frac{{59 - 20}}{{24}} = \frac{{39}}{{24}} \cr & \Rightarrow 13p = \frac{{39}}{{24}} \times 6 = \frac{{39}}{4} \cr & \Rightarrow p = \frac{{39}}{{4 \times 13}} = \frac{3}{4} \cr} $$

Remaining number LCM (9, 11, 13) + 6 = 1287 + 6 = 1293
Least number = 1294 - 1293 = 1

Let the smaller number be x
Then, larger number = (x + 1365)
∴ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
Hence, the smaller number = 270

3-digit numbers divisible by 6 are 102, 108, 114 ..... , 996
This is an A.P. in which a = 102, d = 6 and Tn = 996
∴ Tn = a + (n - 1) d
⇒ 102 + (n - 1) × 6 = 996
⇒ (n - 1) × 6 = 894
⇒ (n - 1) = 149
⇒ n = 150

Six digit largest number 999999

Sis digit largest square number = 999999 - 1998 = 998001

We know that a number having x,y,z as its digit, is a multiple of 11,
If z + x - y = 0
Hence, y = z + x

$$\eqalign{ & \left( {{7^{19}} + 2} \right) \div 6 \cr & = \frac{{{{\left( {6 + 1} \right)}^{19}} + 2}}{6} \cr & = \frac{{{1^{19}} + 2}}{6} \cr & = 3\,{\text{remainder}} \cr} $$

Alternate
(xn - an) is divisible by (x - a) for all values of n.
∴ (719 - 119) is divisible by (7 - 1)
⇒ (719 - 1) is divisible by 6
⇒ On dividing (719 + 2) by 6, remainder obtained = 3

Divisor = 555 + 445 = 1000
Quotient = 2(555 - 445) = 220
Remainder = 30
As we know,
Dividend = Divisor × Quotient + Remainder
Dividend = 1000 × 220 + 30
Dividend = 220000 + 30
∴ Dividend = 220030

Each prime number have to distinct factor number itself and 1.

Given,
35, 48, 55
24, 35, 49
42, 55, 69
21, 32, 43
Concept used:
Co-prime numbers = Co-prime numbers are ones with just one common element, which is 1.
Calculations:
According to the question,
Factors of (35, 48, 55) = 5, 1
⇒ 35 = 5 × 7 × 1
⇒ 48 = 2 × 2 × 2 × 2 × 3 × 1
⇒ 55 = 5 × 11 × 1
Factors of (24, 35, 49) = 7, 1
⇒ 24 = 2 × 2 × 2 × 3 × 1
⇒ 35 = 5 × 7 × 1
⇒ 43 = 7 × 7 × 1
Factors of (42, 55, 69) = 3, 1
⇒ 42 = 2 × 3 × 7 × 1
⇒ 55 = 5 × 11 × 1
⇒ 69 = 3 × 23 × 1
Factors of (21, 32, 43) = 1
⇒ 21 = 3 × 7 × 1
⇒ 32 = 2 × 2 × 2 × 2 × 2 × 1
⇒ 43 = 43 × 1
∴ The set of number which are co-prime number are (21, 32, 43)

Total number from 500 to 650 = 650 - 500 + 1 = 151
Total number from 500 to 650 which are divisible by 3
$$ = \frac{{650}}{3} - \frac{{500}}{3} = 216 - 166 = 50$$
Total number for 500 to 650 which are divisible by 7
$$ = \frac{{650}}{7} - \frac{{500}}{7} = 92 - 71 = 21$$
Total number from 500 to 650 which are divisible by 21
$$ = \frac{{650}}{{21}} - \frac{{500}}{{21}} = 30 - 23 = 7$$
Number neither divisible by 3 nor 7
= 151 - 50 - 21 + 7
= 151 - 71 + 7
= 80 + 7
= 87

Go through option
[(x + 2) + (x + 1) + x]2 = [(x + 2)2 + (x + 1)2 + x2] + 292
Let x = 3
(5 + 4 + 3)2 = 25 + 16 + 9 + 292
144 < 342
Option D, x = 6
(8 + 7 + 6)2 = 64 + 49 + 36 + 292
441 = 441
∴ Option D satisfy this equation
So, largest is 8.

Two consecutive natural numbers are always co-prime number.

The smallest 4-digit number is 1000
This when divided by 7 leaves 6 as remainder.
∴ 1001 is the smallest 4-digit number exactly divisible by 7

$$\eqalign{ & A = 0.3\overline {12} = \frac{{312 - 3}}{{990}} = \frac{{309}}{{990}} \cr & B = 0.4\overline {15} = \frac{{415 - 4}}{{990}} = \frac{{414}}{{990}} \cr & C = 0.30\overline 9 = \frac{{309 - 30}}{{900}} = \frac{{279}}{{900}} \cr & A + B + C = \frac{{309}}{{990}} + \frac{{414}}{{990}} + \frac{{279}}{{900}} \cr & = \frac{{3090 + 4140 + 3069}}{{9900}} \cr & = \frac{{10269}}{{9900}} \cr & = \frac{{1141}}{{1100}} \cr} $$