Let the three odd consecutive natural numbers be P, P + 2 and P + 4.
∴  According to the given question
P + P + 2 + P + 4 = 87

Correct Option: D

Let us assume the 3 odd consecutive natural numbers be P, P + 2 and P + 4.
∴  According to the given question
P + P + 2 + P + 4 = 87
⇒   3P + 6 = 87
⇒   3P = 81
⇒ ∴   P = 27
∴  Smallest number = 27

Ten’s digit of original number = P
∴  Unit’s digit = 2P
According to given question;
∴  Number = 10P + 2P = 12P
According to the question,

Correct Option: C

Ten’s digit of original number = P
∴  Unit’s digit = 2P
According to given question;
∴  Number = 10P + 2P = 12P
According to the question,

Required unit’s digit = Unit’s digit in the product of 7 × 5 × 8 × 3 × 9

Correct Option: A

Required unit’s digit = Unit’s digit in the product of 7 × 5 × 8 × 3 × 9 = 0

As we know that
7¹ = 7;
7² = 49;
7³ = 343;
7⁴ = 2401 ;
7⁵ = 16807
i.e. The unit’s digit repeats itself after power 4.
Remainder after we divide 245 by 4 = 1

Correct Option: D

As we know that
7¹ = 7;
7² = 49;
7³ = 343;
7⁴ = 2401 ;
7⁵ = 16807
i.e. The unit’s digit repeats itself after power 4.
Remainder after we divide 245 by 4 = 1
∴ Unit’s digit in the product of (4387)²⁴⁵ × (621)⁷² = Unit’s digit of (4387)²⁴⁵ × Unit digit of (621)⁷²
(621)⁷² = Unit’s digit of (7)²⁴⁵ × Unit digit of (1)⁷²
(621)⁷² = Unit’s digit of (7)¹ × Unit digit of (1)⁷²
∴ Unit’s digit in the product of (4387)²⁴⁵ × (621)⁷² = 7 × 1 = 7

As per given question,
Last digit of (1001)²⁰⁰⁸ + 1002
As we know that
Last digit of (1001)²⁰⁰⁸ + 1002 = Last digit of (1)²⁰⁰⁸ + last digit of 1002

Correct Option: B

As per given question,
Last digit of (1001)²⁰⁰⁸ + 1002
As we know that
Last digit of (1001)²⁰⁰⁸ + 1002 = Last digit of (1)²⁰⁰⁸ + last digit of 1002
= 1 + 2 = 3