Find the greatest number of 6 digit which is a perfect square and which is also divisible by 16, 18 and 45
Answers
Answer:
The required number is 921600
Step-by-step explanation:
We find the LCM of 16, 18 and 45 first.
16 = 2 × 2 × 2 × 2
18 = 2 × 3 × 3
45 = 3 × 3 × 5
So LCM = 2 × 2 × 2 × 2 × 3 ×3 × 5
= 720
But 720 is not a perfect square. In order to make it a perfect square, we multiply it by 5 and thus it becomes (720 × 5) = 3600
We have to find the greatest six digit number under the given conditions.
We multiply 3600 by perfect squares to find the 6 digit number.
The least perfect square multiplied by which 3600 becomes a 6 digit number is 100. So we begin with 100.
3600 × 100 = 360000, a perfect square
3600 × 121 = 435600, a perfect square
3600 × 144 = 518400, a perfect square
3600 × 169 = 608400, a perfect square
3600 × 196 = 705600, a perfect square
3600 × 225 = 810000, a perfect square
3600 × 256 = 921600, a perfect square
3600 × 289 = 1040400, a perfect square but a 7 digit number
Therefore the required number is 921600, a perfect square of 960