Question

वह छोटी से छोटी वर्ग संख्या ज्ञात कीजिए जो 9, 15 और 21 में से प्रत्येक से विभाज्य हो। Find the smallest square number which is divisible by each of the numbers 9, 15 and 21.

1) 315
2) 1575
3) 11025
4) 2205​

Answer 1

Answer:

First find the LCM of 9,15 and 20 that is 180.

Now if you see the Prime factorization of 180, it is 2×2×3×3×5

So now to make 180 a perfect square we need to multiply it by 5 so that it's Prime factorization is 2×2×3×3×5×5.

Now 180×5=900

Now the square root of 900 is 30

Therefore, 30 is the smallest square number that is equally divisible by 9,15 and 20.

Answer 2

Solution :-

As we know that, to find the smallest number divisible by given numbers first we have to find the LCM.

So, Lets first find the LCM of given numbers .

Prime Factors of 9, 15 and 21 :-

→ 9 = 3 * 3

→ 15 = 3 * 5

→ 21 = 3 * 7

then,

→ LCM = 3² * 5 * 7 = 315

Now, we need a perfect Square number . As we know that, perfect square numbers are in the pair of two .

So, in order to make LCM as perfect square we have to multiply it with 5 and 7 .

therefore,

→ Required Number = LCM * 5 * 7

→ Required Number = 315 * 35

→ Required Number = 11025 (Ans.)

Hence, the smallest square number which is divisible by each of the numbers 9, 15 and 21 is equal to 11025 .

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