Clearly, the required number divides (129-3) = 126 and (545-5) = 540 exactly.

 Required number = HCF (126,540)

Now,

and,

 HCF(126,540) = product of common terms with lowest power

=

= 18

Hence, the required number is 18.

We have to find the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively.

Required number = HCF { (70 – 5) , (125 – 8 )}

= HCF {65 , 117 }

We have to find factors of 65 & 117 .

So, 65 =

Required number = HCF { 65 , 117 }

= 13

Consider

We are interested in finding remainder when is divisible by 63.

Here, we can see that it is always divisible by 6.

We know

Let the number be N.

When the number is divided by 18, it can be expressed as N = Q × 18 + 11.

Further, we can write N = 6× (3Q) + 6 + 5

When N = 6× (3Q) + 6 + 5 is divided by 6, the remainder will be 5.

Let p(t) = mt³ + 4t² + 3t – 4 and g(t) = t³ – 4t + m

When p(t) and g(t) are divided by (t-3) then p(3) and g(3) are obtained as remainder

According to question

⇒ p(3) = g(3)

⇒ m(3)³ + 4(3)² + 3(3) – 4 = (3)³ – 4(3) + m

⇒ 27m+36+9-4 = 27 – 12 + m

⇒ 26m = -26

⇒ m = -1

Given for , remainder is 3

So, for  (m⁵ +m⁴ + m³ +m² +m +1)/7 ,remainder [ (3⁵ +3⁴ + 3³ +3² +3 +1)/7]

=

Remainder = 0

Hence, option b is the correct answer.

By Euler’s theorem

First, we reduce it to  divided by 7.

Now, we know that,

, remainder is 3

, remainder is 2

, remainder is -1

So, , remainder is -3

Similarly, , remainder is again -3.

So, the remainder of the overall expression is  10 times

So, , remainder is -2 or

Hence, option d is the correct Answer