Which of the following statements is/are true?
I. $$4^{10} + 6^{10}$$ is divisible by 52.
II. $$7^{15} + 64^{5}$$ is divisible by 11.
III.$$2^{20} - 49^{10}$$ is divisible by 9.
IV. $$3^{15} - 8^{5}$$ is divisible by 5.

I) When a number has to be divisible by 52, basically it has to be divisible by 4 and 13.

Taking $$2^{10\ }$$common from the expression $$2^{10\ }*\left(2^{10}+3^{10}\right)$$. $$2^{10\ }$$ is divisible by 4 but not by 13.

So lets check if $$2^{10\ }+3^{10}$$ is divisible by 13.

Remainder of$$2^{10\ }$$when divided by 13.

$$2^{10\ }=2^5\cdot2^5$$

$$2^5\%13=32\%13=6$$

$$2^{10\ }=2^5\cdot2^5 = \left(13k+6\right)\cdot\left(13k+6\right)=13K+36=13K+10$$

So the remainder is 10.

Remainder of $$3^{10\ }$$when divided by 13.

$$3^{10\ }=3^5\cdot3^5$$

$$3^5\%13=243\%13=9$$

$$3^{10\ }=3^5\cdot3^5 = \left(13k+9\right)\cdot\left(13k+9\right) = 13K+81 = 13K+3$$

The remainder of $$2^{10 }+3^{10}$$when divided by 13 = 10+3 = 13, so it is divisible by 13.

II. $$7^{15} + 64^{5}$$ is divisible by 11.

Check for $$7^3$$ %11

343%11=2

$$7^{15}=\left(7^3\right)^5=\left(11k+2\right)^5=11K+32=11K+10$$

This gives the remainder 10.

Remainder of $$64^5$$ %11

This is same as$$\left(2^6\right)^5\%11=2^{30}\%11$$

$$2^{10}\%11=1024\%11=1$$

$$2^{30}\%11=\left(2^{10}\right)^3\%11=1^3\%11$$

This gives the remainder 1.

$$\left(7^{15}+64^5\right)\%11=\left(10+1\right)\%11=0$$

So, it is divisible by 11.

III.$$2^{20} - 49^{10}$$ is divisible by 9.

Re-write it as $$\left(2^{10}\right)^2-\left(7^{10}\right)^2$$

It is in the form of $$a^2-b^2$$=$$(a+b)\cdot(a-b)$$.

$$\left(2^{10}+7^{10}\right)\cdot\left(2^{10}-7^{10}\right)$$

Apply the same for $$\left(2^{10}-7^{10}\right)$$

$$\left(2^5+7^5\right)\cdot\left(2^5-7^5\right)\cdot\left(2^{10}+7^{10}\right)$$

Now, check each of the terms to see if they are divisible by 9. If atleast one is divisible, then the whole term is divisible by 9.

Check $$\left(2^5+7^5\right)$$

$$\left(32+343\cdot49\right)\%9$$

$$\left(9k_1+5+\left(9k_2+1\right)\cdot\left(9k_3+4\right)\right)\%9$$

where $$k_1,\ k_2,\ k_3\ are\ constant$$

$$\left(9k_1+5+\left(9k_4+4\right)\right)\%9$$

$$\left(9K+5+4\right)\%9$$

$$\left(9K+9\right)\%9$$

Since this term is divisible by 9, statement 3 is true.

IV. $$3^{15} - 8^{5}$$ is divisible by 5.

$$3^{15} - 8^{5}$$ = $$3^{15}-2^{15}$$

Rewriting it as $$\left(3^5\right)^3-\left(2^5\right)^3$$

This is in the form $$a^3-b^3=\left(a-b\right)\cdot\left(a^2+b^2+a\cdot b\right)$$

$$\left(3^5-2^5\right)\cdot\left(3^{10}+2^{10}+6^5\right)$$

$$\left(243-32\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

$$\left(211\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

When divided by 5, the remainder will depend on the unit digit. If it is below 5, that will be the remainder. If it is above 5, subtract the number from 5, the result will be the remainder.

$$\left(1\right)\cdot\left(3\cdot3+2\cdot2+1\right)=\left(1\right)\cdot\left(9+4+1\right)$$

Any power of six will have the unit digit as 6; hence, the remainder will be 1.

14%5=4.

So, the number will not be divisible by 5.

Hence option B is correct.