Write $3$ -digit numbers ending with $0, 1, 4, 5, 6, 9$ one for each digit, but none of them is a perfect square.

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Solution

We will first write the three digit numbers between $14^{2} = 196$ and $15^{2} = 225$ ending with $0, 1, 4, 5, 6, 9$ where none of them will be a perfect square. And those numbers are as follows:

$200, 201, 204, 205, 206$ and $209$

Also we write the three digit numbers between $17^{2} = 289$ and $18^{2} = 324$ ending with $0, 1, 4, 5, 6, 9$ where none of them will be a perfect square. And those numbers are as follows:

$300, 301, 304, 305, 306$ and $309$

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Similar questions

Column I	Column II
Greatest 3 digit perfect square	$1, 024$
Smallest 5 digit perfect square	$9, 801$
Smallest 3 digit perfect square	$961$
Greatest 4 digit perfect square	$10, 000$
Smallest 4 digit perfect square	$100$

State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeroes.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may be a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

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