In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:

Let the two number are a and b
According to the question
sum of squares of two numbers = 97
i.e.., a ² + b ² = 97 .....................(i)
and square of their difference = 25
i.e., (a - b) ² = 25 .................(ii)
? a - b = 5...............(iii)
from Eq. (ii)
a² + b² - 2ab = 25
(a² + b²) - 2ab = 25
Now put the value of a ² + b ² from Eq. (i)
? 97 - 2ab = 25
? 2ab = 72
? ab = 36.................(iv)
Now, we have
(a + b)² = ( a² + b²) + 2ab
Now put the value of a ² + b ² from Eq. (i) and ab from Eq. (iv)
(a + b)² = 97 + 72 = 169
a + b = 13.............(v)
Now add the Eqs. (iii) and (v), we get
a - b + a + b = 5 +13
2a = 18
? a = 9
now put the value of a in Eq. (v)
a + b = 13
9 + b = 13
b = 13 - 9
b = 4
? Product of both the numbers = ab = 9 x 4 = 36

Let sum = P
and given that n = 3 yr
According to the question,
Amount = 27 P
According to the formula,
Amount = P (1 + R/100)ⁿ
? 27 P = P (1 + R/100)³
? 27 = (1 + R/100)³
? (1 + R/100)³ = (3)³
? 1 + R/100 = 3
? R/100 = 3 - 1 = 2
? R = 200%

(4 x 4 x 4 x 4 x 4 x 4)⁵ x (4 x 4 x 4)⁸ ÷ (4)³ = (64)^?
Apply the law of Fractional Exponents and Laws of Exponents
if a multiply n times a x a x a x....up to n times, then
a x a x a x a ......up to n times = aⁿ
By simplifying the equation
? (4⁶)⁵ x (4³)⁸ x 1/(4)³ = (4³)^?
? (4)³⁰ x (4)²⁴/(4)³ = (4)^{3 x ?}
? 4^{30 + 24 - 3} = 4^{3 x 7}
And comparing the exponents both the sides
? 4⁵¹ = 4^{3 x ?} ? 3 x ? = 51
? ? = 51/3 = 17

Since a * b = a + b + a / b
? 12 * 4 = 12 + 4 + 12 / 4
= 12 + 4 + 3 = 19

S.I. = Rs. (600 x 5 x 2)/100 = Rs.60
C.I.= Rs. [600 x (1 + 5/100)² - 600] = Rs. 61.50

? Requred Difference = Rs. (61.50 - 60) = Rs.1.50

Let numbers be 10N, 12N, 15N and 18N.
Then, LCM = 180N and HCF = N
Hence, required LCM = 180 x 3 = 540

a² + 1 = a ? a + 1/a = 1
On squaring both sides, we get
a² + 1/a² + 2 = 1
? a² + 1/a² = -1
On cubing both sides, we get
(a² + 1/a²)³ = (-1)³
? a⁶ + 1/a⁶ + 3a² x 1/a²(a² + 1/a²) = -1
? a⁶ + 1/a⁶ + 3 x (-1) = -1
Now, a⁶ + 1/a⁶ + 1 = 3
As, a¹² + a⁶ + 1 can also be written as a⁶ + 1/a⁶ + 1
? a¹² + a⁶ + 1 = 3

Three vowels (O, E, A) can be arranged in the odd places in 3! ways (1st position, 3rd position, 5th position) and two consonant (M, G) can be arranged in the even place in 2! ways (2nd place and 4th place).
? Total number of ways = 3! X 2! = 12

Let the total amount = x
According to the question.
14 girls are getting x amount.
1 girl will get x/14 amount.
Similarly if
18 girls are getting x amount.
1 girl will get x/18 amount.
according to question,
x/14 - x/18 = 160
? (9x - 7x)/126 = 160
? 2x/126 = 160
? x/126 = 80
? x = 126 x 80 = ? 10080

Now, 1397 = 11 x 127

∴ The required 3-digit number is 127, the sum of whose digits is 10.