Ancient India 10 Q

If n3 is odd, so ‘n’ must be odd. Also, n2 will be odd. So, (I) and (II) must be true.

4 + 32 + 108 + ... + 4000 = 4(1 + 8 + 27 + .... + 1000)

= 4 (13 + 23 + 33 + ….. + 103)

$=4\times {\left(\frac{10\times 11}{2}\right)}^2$

= 4 × 552 = 12100

Required numbers are 509, 519, 529, 539, 549, 559, 569, 579, 589, 590, 591, 592, …, 598. So, there are 18 numbers in all.

Note: In any set of 100 natural numbers digit from 1 to 9 will appear 20 times and will appear in 19 numbers at unitʼs and tenʼs place.

In every 100 numbers a digit appears 20 times at units and tens place. So, in 200 number the digit 6 appears 20 × 2 = 40 times at units and tens place and 100 times in the hundred place.

Total = 40 + 100 = 140 times

Let the digits be a and b such that the number is 10a + b.

 ab = 14 and 10a + b + 45 = 10b + a

i.e., 9a – 9b = –45

i.e., a – b = –5

(a + b)2 = (a – b)2 + 4ab = 25 + 4(14) = 81

 a + b = 9

 a = 2, b = 7

 The number is 27.

Use options & check.

Let digit at tenʼs place be x.

Digit at unitʼs place = x + 1

5 times sum of digit = 5(x + x + 1)

=10x + 5

The number has to be greater than (10x + 5).

Also x can take values from 1 to 8 only.

x = 1, no. = 12

x = 2, no. = 23

x = 3, no. = 34

x = 4, no. = 45

For all these values of ‘x’, number is not more than 5 times the sum of digits.

x = 5, no. = 56

x = 6, no. = 67

x = 7, no. = 78

x = 8, no. = 89

For all these numbers the numbers are greater than 5 times the sum of digits.

So, sum of numbers = 56 + 67 + 78 + 89 = 290

$x=\frac{-1}{2}$

$Option(a):2^{\frac{1}{x}}=2^{\frac{1}{\frac{-1}{2}}}=2^{-2}=\frac{1}{4}$

$Option(b):\frac{1}{x}=\frac{1}{\frac{-1}{2}}=-2$

$Option(c):\frac{1}{x^2}=\frac{1}{{\left(\frac{-1}{2}\right)}^2}=4$

$Option(d):2^x=2^{\frac{-1}{2}}=\frac{1}{2^{\frac{1}{2}}}=\frac{1}{\sqrt{2}}$

$\text{So, smallest is }\frac{1}{x}.$

By looking at the options, we can find that the value given in option (b) will be negative and remaining options will give us a positive value.

So, the smallest value will be $\frac{1}{x}$.

The sum of first ten natural numbers is 55.

Now, the sum of next set of ten natural numbers is 65 (from 2 to 11)

Now, the sum of next set of ten natural numbers is 75 (from 3 to 12)

So, possible values are 55, 65, 75, 85, ........

All values with unit digit 5 are possible starting from 55.

Given surds are 21/2, 41/3 and 61/4.

LCM of 2, 3 and 4 is 12.

21/2 = (26)1/12 = (64)1/12, 41/3 = (44)1/12

2561/12 and 61/4 = (63)1/12 = (216)1/12

$\sqrt[4]{80}+\sqrt[4]{3125}-\sqrt[4]{6480}$

$=\sqrt[4]{2^4\times 5}+\sqrt[4]{5^4\times 5}-\sqrt[4]{2^4\times 3^4\times 5^1}$

$=2\sqrt[4]{5}+5\sqrt[4]{5}-6\sqrt[4]{5}=\sqrt[4]{5}$