Math Olympiad Test: Real Numbers- 2 - Class 10 MCQ

6ⁿ − 5ⁿ
For n = 1
6¹ − 5¹ = 6 − 5 = 01
⇒ Unit digit is 1, i.e., ends with 1.
For n = 2
6² − 5² = 36 − 25 = 11
⇒ Unit digit is 1, i.e., ends with 1.
For n=3
6³ − 5³ = 216 − 125 = 91
⇒ Unit digit is 1, i.e., ends with 1.
By continuity, we can say that (6ⁿ − 5ⁿ) always ends with 1.

Solve for divisor of given expression: 
Given, 9²ⁿ - 4²ⁿ is in the form of a²ⁿ -b²ⁿ where n is a natural number.
Since a²ⁿ - b²ⁿ is always divisible by (a - b) and (a + b),
9²ⁿ - 4²ⁿ is always divisible by:
9 - 4 = 5
9 + 4 = 13
∴ 9²ⁿ - 4²ⁿ is always divisible by both 5 and 13.

• The H.C.F. of 81 and 50 is 1, as they have no common factors other than 1.
• The L.C.M. is found by multiplying the two numbers since H.C.F. is 1: 81 × 50 = 4050.
• Therefore, the product of H.C.F. and L.C.M. is 1 × 4050 = 4050.

L.C.M. × H.C.F. = 306 × 657
⇒ L.C.M. × 9 = 306 × 657
⇒ 

Smallest prime number = 2
Smallest composite number = 4
∴ Required product = 2 × 4 = 8

H.C.F. of 210 and 55 = 5
210 × 5 + 55y = 5
⇒ 55y = 5 - 1050
⇒ 55y = -1045
⇒ y = 1045/55
= -19

13915 = 5 × 11 × 11 × 23
Required difference = 23 - 5 = 18

The LCM of numbers from 1 to 10 need to be found out to find the least number We know that 2, 3, 5 and 7 are prime numbers
Prime factorization of
4 = 2 × 2
6 = 2 × 3
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
So we get required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520
Therefore, the least number that is divisible by all the numbers between 1 and 10 is 2520.

√27 x √3 = √81 = 9

Here 1225 = 5² × 7²
∴ Required difference = 2 - 2 = 0

95 = 5 × 19
and 152 = 23 × 19
H.C.F = 19

576 = 2⁶ × 3⁴
6 + 2 = 8

324 = 2² × 3²