Question

22 and 24 and 25 plz

Answer 1

(22)

$2222\equiv3\pmod{7}\ \ \longrightarrow\ (1)\\\\(2222)^3\equiv3^3=27\equiv-1\pmod{7}\\\\((2222)^3)^2=(2222)^6\equiv(-1)^2=1\pmod{7}\\\\\therefore\ (2222)^{6n}\equiv1\pmod{7}$

Since 5555 gives quotient 925 and remainder 5 on division by 6,

$((2222)^6)^{925}=(2222)^{5550}\equiv1\pmod{7}\\\\(2222)^{5555}\equiv(2222)^5\pmod{7}$

Now, from (1),

$(2222)^5\equiv3^5\equiv\mathbf{5}\pmod{7}$

Hence the answer is (2) 5.

(24)

First we factorise 15.

15 = 5 × 3

Now, to get the answer, we have only to find the no. of multiples of powers of the highest factor among them, inclusively below 100 . Here it's 5.

So the no. of multiples of powers of 5 below 100 is the answer. Let's find.

Below 100 there are 20 multiples of 5.

Below 100 there are 4 multiples of 5² = 25.

Below 100 there are no multiples of 5³ = 125. So let's finish.

Hence the answer is 20 + 4 = 24, i.e., option (2).

(25)

Here the terms 4!, 5!, 6!,..., 100! are each congruent to 0 modulo 24, so,

$\displaystyle\sum_{n=1}^{100}n!\equiv1!+2!+3!\pmod{24}\\\\\sum_{n=1}^{100}n!\equiv1+2+6\pmod{24}\\\\\sum_{n=1}^{100}n!\equiv\mathbf{9}\pmod{24}$

Hence the answer is (3) 9.