Olympiad Test: Number System - 3 - Free MCQ with solutions Year 7 Mathematics

We know that (1² + 2² + 3² + ... + n²)

Putting n=10, required sum

= 385

This is an A.P. in which a = 51, l = 100 and n = 50.

Unit digit in 7⁹⁵ = Unit digit in [(7⁴)²³ × 7³]
= Unit digit in [(Unit digit in(2401))²³ × (343)]
= Unit digit in (1²³ × 343) = Unit digit in (343) = 3
Unit digit in 358 = Unit digit in [(3⁴)¹⁴ × 3²]
= Unit digit in [Unit digit in (81)¹⁴ × 3²]
= Unit digit in [(1)¹⁴ × 3²]
= Unit digit in (1 × 9)
= Unit digit in (9) = 9
Unit digit in (7⁹⁵ – 3⁵⁸)
= Unit digit in (343 – 9)
= Unit digit in (334) = 4.

904 × 1904 = (1904)²
= (1900 + 4)²
= (1900)² + (4)² + (2 × 1900 × 4)
= 3610000 + 16 + 15200
= 3625216

The least six digit number is 100000.
When 100000 ÷ 111,
Quotient 990 and Remainder = 100
Therefore required number
=100000 + ( 111 – 100)
= 100000 + 11 
= 100011

Here a = 3 and r =6/3 =2.
Let the number of terms be n.
Then, t_n = 384
ar^n-1 = 384
3 × 2^{n – 1} = 384
2^n-1 = 128 = 2⁷
n – 1 = 7
n = 8
Number of terms = 8.

Largest 5-digit number = 99999
When 99999÷91, Quotient = 1098 and
Remainder = 81
Required number = (99999 – 81)
= 99918.

By hit and trial, we find that
47619 × 7 = 333333.

Let the required number be n,
Then 25% × 2/5 × n = 125
By solving the above equation,
∴ n = 1250

The square of a natural number never ends in 7.
42437 is not the square of a natural number.