Olympiad Test: Number System - 2 - Class 7 MCQ

Unit digit in (6374)¹⁷⁹³ = Unit digit in (4)¹⁷⁹³
= Unit digit in [(42)⁸⁹⁶ × 4]
= Unit digit in (6 × 4) = 4
Unit digit in (625)³¹⁷ = Unit digit in (5)³¹⁷ = 5
Unit digit in (341)⁴⁹¹ = Unit digit in (1)⁴⁹¹ = 1
Required digit = Unit digit in (4 × 5 × 1) = 0.

The unit digit in the 71ⁿ for any value of n is 1.
Thus the unit digit is 71ⁿ –1 is 0.

look carefully it is written in a specific pattern using a³+b³
Given Expression =

107 × 107 + 93 × 93 = (107)² + (93)²
= (100 + 7)² + (100 – 7)²
= 2 × [(100)² + 7²]       [Ref: (a + b)² + (a – b)²
= 2(a² + b²)] = 20098

(xⁿ + 1) will be divisible by (x + 1) only when n is odd.
(67⁶⁷ + 1) will be divisible by (67 + 1)
(67⁶⁷ + 1) + 66, when divided by 68 will give 66 as remainder.

Required sum = (2 + 3 + 5 + 7 + 11) = 28.

(12345679 × 2 × 6²)  = 12345679 × 72
= 12345679 × (70 +2)
= 12345679 × 70 + 12345679 × 2
= 864197530 + 24691358
= 888888888

20 × n = (64 + 36)(64 – 36) = 100 × 28

(2² + 4² + 6² + ... + 20²) 
= (1 × 2)² + (2 × 2)² + (2 × 3)² + ... + (2 × 10)²
= (2² × 1²) + (2² × 2²) + (2² × 3²) + ... 
+ (2² × 10²)
= 2² × [1² + 2² + 3² + ... + 10²]
[Formula: (1² + 2² + 3² + ... + n²)

= (4 × 5 × 77)
= 2 × 77 × 10

Let S_n = (1 + 2 + 3 + ... + 45)
This is an A.P. in which a = 1, d = 1, n = 45 and l = 45

Required sum = 207 × 5.

Let S_n = (2 + 4 + 6 + ... + 20).
This is an A.P.
in which a = 2, d = 2 and l = 20
Let the number of terms be n. Then,
a + (n – 1)d = 20
2 + (n – 1) × 2 = 20
n = 10.

2056 × 987 = 2056 × (1000 – 13)
= 2056 × 1000 – 2056 × 13
= 2056000 – 26728
= 2029272

Let the number be x.

Let the three integers be x, x + 2 and x + 4.
Then, 3x = 2(x + 4) + 3
x = 11
Third integer = x + 4 = 15

Given number expression = (397)² + (104)² + 2 × 397 × 104
= (397 + 104)²
= (501)2 = (500 + 1)²
= (500)2 + (1)² + (2 × 500 × 1)
= 250000 + 1 + 1000
= 251001

(35423 + 7164 + 41720) – (317 × 89) 
= (35423 + 7164 + 41720) – (28213) 
= 84307 – 28213
= 56094

For every natural number n, (aⁿ – bⁿ) is completely divisible by (a – b).

(3²⁵ + 3²⁶ + 3²⁷ + 3²⁸) = 3²⁵ × (1 + 3 + 32 + 33) 
= 3²⁵ × 40
= 3²⁴ × 3 × 4 × 10
= (3²⁴ × 4 × 30), which is divisible by 30.

Let x = 6q + 3.
Then, x² = (6q + 3)²
= 36q² + 36q + 9
= 6(6q² + 6q + 1) + 3
Thus, when x² is divided by 6, then remainder = 3.

When n is even; (xⁿ- aⁿ) is completely divisible by (x + a)
(17²⁰⁰ – 1²⁰⁰) is completely divisible by (17 + 1), i.e., 18.
(17²⁰⁰ – 1) is completely divisible by 18.
On dividing 17²⁰⁰ by 18, we get 1 as remainder.