(10x + y)(x + y) = 424 . . . . . . (i)
(10y + x)(x + y) = 280 . . . . . . (ii)
Add equation (i) and (ii)
11(x + y)(x + y) = 704
(x + y)2 = 64
x + y = 8

82k + 52k
If K is odd number
Then, K = 1, 3, 5 _ _ _ _
82 × 1 + 52 × 1
= 64 + 25
= 89

If 8764x5, is divisible by 9 then, sum of digit
8 + 7 + 6 + 4 + x + 5 = 0/9/18/36/45
30 + x = 45
x = 15

$$\eqalign{ & \frac{{{{10}^{100}}}}{{{5^{75}}}} \cr & = \frac{{{2^{100}} \times {5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {2^{75}} \times {5^{25}} \cr & = {2^{75}} \times {10^{25}} \cr} $$

$$\eqalign{ & 2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{4} + 5\frac{1}{5} \cr & = \left( {2 + 3 + 4 + 5} \right) + \left( {\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right) \cr & = 14 + \left( {\frac{{30 + 20 + 15 + 12}}{{60}}} \right) \cr & = 14 + \frac{{77}}{{60}} \cr & = 14 + 1 + \frac{{17}}{{60}} \cr & = 15 + \frac{{17}}{{60}} \cr & \therefore {\text{Required answer}} = 1 - \frac{{17}}{{60}} \cr & = \frac{{60 - 17}}{{60}} \cr & = \frac{{43}}{{60}} \cr} $$

Divisibility law of 4: A number divisible by 4 if its last 2 digit is divisible by 4
Divisibility law of 11: If the difference of the alternating sum of digits of the number is a multiple of 11 (e.g. 2343 is divisible by 11 because 2 - 3 + 4 - 3 = 0, which is a multiple of 11).
⇒ 15x1y2 is divisible by 4 if y = 1 or 3
If y = 1, then
⇒ 15x112 is divisible by 11 if x = 6
If y = 3 then
⇒ 15x132 is divisible by 11 if x = 4
If y = 1 and x = 6, then
⇒ x + y = 1 + 6 = 7
If y = 3 and x = 4, then
⇒ x + y = 3 + 4 = 7

$$\eqalign{ & {\text{Let number are }}x{\text{ and }}y \cr & {\text{According to the question,}} \cr & x \times y = 48 \cr & {\text{and }}x = \frac{{2 + \mathop {\mathop 5\limits^ \uparrow }\limits^{{\text{total fingers in one hand}}} \times 2}}{3} = \frac{{12}}{3} = 4 \cr & y = \frac{{48}}{4} = 12 \cr} $$

S1 = 2, 9, 16, . . . . . ., 632
S2 = 7, 11, 15, . . . . . ., 743
S1 = 2, 9, 16, 23, 30, 37, 44, 51 . . . . . .
S2 = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, . . . . . .
Number of terms in $${{\text{S}}_1} = \frac{{632 - 2}}{7} + 1 = 91$$
∵ In S1 there is no further terms beyond 632 therefore there is no common terms.
∴ Number of terms in series S1 = 91
If this series start from the 23 then number of terms = 91 - 3 = 88
Now, number of common terms in both series S1 and S2 = $$\frac{{88}}{7}$$ = 22 = n
Now, common terms 23, 51, . . . . . . 22
$$\eqalign{ & {\text{Sum}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = 11\left[ {46 + 21 \times 28} \right] \cr & = 11\left[ {46 + 588} \right] \cr & = 11 \times 634 \cr & = 6974 \cr} $$

Let the number be = x
Then according to question
21 × x + 100 = x2
x2 - 21x - 100 = 0
x2 - 25x + 4x - 100 = 0
x(x - 25) + 4(x - 25) = 0
(x - 25)(x + 4) = 0
x = -4 is not possible
x = 25

$$\eqalign{ & {\text{Sum of A}}{\text{.P}}{\text{.}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & {n^{{\text{th}}}}{\text{term}} = a + \left( {n - 1} \right)d \cr & {3^{{\text{rd}}}}{\text{term}} = a + \left( {3 - 1} \right)d = 19 \cr & a + 2d = 19\,......\,\left( {\text{i}} \right) \cr & {\text{Sum of 10 term}} = \frac{{10}}{2}\left[ {2a + 9d} \right] = 390 \cr & 2a + 9d = 78\,......\,\left( {{\text{ii}}} \right) \cr & a + 2d = 19\,......\,\left( {\text{i}} \right) \cr & {\text{From equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & \boxed{a = 3}\,\,\boxed{d = 8} \cr} $$

According to the given condition
Denominator = 7 + C [C > -7]
\[\begin{array}{*{20}{c}} {{\text{Denominator}} = }&{{\text{Greator number}}}&{}&{{\text{Smaller number}}} \\ {}&{\left( {7 + C - 4} \right)}& \times &{\left( {7 + C - 4 - 2} \right)} \end{array}\]
Equation $$ = \frac{{\left( {3 + C} \right)\left( {1 + C} \right)}}{{7 + C}}$$
By putting value according to option
C = -2 for minimum value
= -1/5

x = (164)169 + (333)337 - (727)726
x = 4 + 3 - 9
x = . . . . . . 7 - 9
x = 17 - 9
x = 8

On arranging the digit of number 150328 in ascending = 853210
According to the question,
853210 - 13 × 5 = 853210 - 65 = 853145
Clearly it is divisible by 5.

9991 × 10009
= 9991(10000 + 9)
= 99910000 + 89919
= 99999919

Alternate solution
9991 × 10009
= (10000 - 9)(10000 + 9)
= (10000)2 - (9)2
= 100000000 - 81
= 99999919

Let the natural number (n) be 1.
∴ n(n + 1)(n + 2)
= 1(1 + 1)(1 + 2)
= 6

10 digit number 1330x558y2 is divisible by 88, means number is also divisible by 11 and 8
⇒ Sum of digit of the number even place = 3 + 0 + 5 + 8 + 2 = 18
⇒ Sum of digit of the number odd place = 1 + 3 + x + 5 + y = 9 + x + y
As we know,
⇒ 9 + (x + y) – 18 = 0
⇒ x + y = 9

According to the question,
$$\eqalign{ & \,5\,\,\,\,\,E\,\,\,\,\,9 \cr & \,2\,\,\,\,\,F\,\,\,\,\,8 \cr & \,3\,\,\,\,\,G\,\,\,\,\,7 \cr & \underline {\overline {11\,\,\,\,\,1\,\,\,\,\,4\,} } \cr} $$
⇒ Sum of 2, E, F and G must be 11. For maximum F we will have to take E and G zero.
∴ F = 9

Sum of first $$n$$ natural numbers $$\frac{{n\left( {n + 1} \right)}}{2}$$
∴ Required average $$ = \frac{{n\left( {n + 1} \right)}}{{2 \times n}} = \frac{{n + 1}}{2}$$

x = (633)24 - (277)38 + (266)54
x = 1 - 9 + 6
x = 7 - 9
x = . . . . . 7 - 9
x = 17 - 9
x = 8

Since x, y and z are distinct prime number (x < y < z)

y + z = 70 - 2 = 68
31 + 37 = 68
Hence, x = 2, y = 31 and $$\boxed{{\text{z}} = 37}{\text{ Answer}}$$