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The least perfect square, which is divisible by each of 21, 36 and 66
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05 Apr 2019, 05:43
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The least perfect square, which is divisible by each of 21, 36 and 66 is A. 213444 B. 214344 C. 214434 D. 231444 E. 231434
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Re: The least perfect square, which is divisible by each of 21, 36 and 66
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05 Apr 2019, 05:56
The number has to be divisible by 2,3,7 and 11 in order to be divisible by 21,36 and 66
Since all now are even  all now are divisible by 2 Since sum of all no's is 18 except E which adds to 17, so we can exclude E Now check divisibility by 7: only A is divisible by 7
Hence answer is A
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Re: The least perfect square, which is divisible by each of 21, 36 and 66
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05 Apr 2019, 16:25
We need the answer to be divisible by (3)(7), by (2^2)(3^2) and by (2)(3)(11), so it needs to be divisible by 2^2, 3^2, 7 and 11. It also needs to be a perfect square, so it must have even exponents in its prime factorization, and the smallest exponents we could use are 2's for everything, so we just want the value of (2^2)(3^2)(7^2)(11^2). I'd never want to calculate that, so we can just use some divisibility tests to find the right answer. The answer is divisible by 4, so the last two digits of the right answer must form a multiple of 4. Only answers A, B and D are possible. The answer is also divisible by 9, so the sum of the digits must be a multiple of 9. That leaves only A and B. Finally we can check for divisibility by 7. We can take the wrong answer first: 214,344. To check if this is a multiple of 7, we can just start subtracting large multiples of 7 from it. So we can take away 210,000, leaving us with 4,344. Now we can take away 4200, leaving us with 144. That's not divisible by 7, so 214,344 can't be either. That leaves only answer A, so that must be right.
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Re: The least perfect square, which is divisible by each of 21, 36 and 66
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05 Apr 2019, 20:49
We have to find a number which is divisible by 7,9,4,11
for divisibility test of 9 , sum of all digits should be divisible by 9
for 4, last two digits divisible by 4
for 11 , difference of sum.of digits at even places and odd places should be divisible by 11
This way we can eliminate b,c,d,e options
So IMO A
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The least perfect square, which is divisible by each of 21, 36 and 66
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05 Apr 2019, 22:01
AkshdeepS wrote: The least perfect square, which is divisible by each of 21, 36 and 66 is
A. 213444 B. 214344
C. 214434 D. 231444
E. 231434 Step 1: Prime Factorization :
21 = 3 * 7
36 = 2 * 2 * 3 * 3
66 = 2 * 3 * 11
Rule : A number is a perfect square is it has all its prime factors in pairs or it has even powers prime factors
As we are looking for least value of perfect square, we will take only one pair of each prime combinedly from all numbers.
So, 2 * 2 * 3 * 3 * 7 * 11 (As we have only one 7 and one 11)
To make this number a perfect square we need a 7 and an 11 for pairing.
Least perfect square = 2*2*3*3*7*7*11*11
Least perfect square = 4*9*49*121
Least perfect square = 213444
This multiplication is not that time consuming (A)
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Re: The least perfect square, which is divisible by each of 21, 36 and 66
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06 Apr 2019, 09:08
m1033512 wrote: We have to find a number which is divisible by 7,9,4,11
for divisibility test of 9 , sum of all digits should be divisible by 9
for 4, last two digits divisible by 4
for 11 , difference of sum.of digits at even places and odd places should be divisible by 11
This way we can eliminate b,c,d,e options
So IMO A
award kudos if helpful
Posted from my mobile device <<<for 11 , difference of sum.of digits at even places and odd places should be divisible by 11 >>> or difference is ZERO ..right?




Re: The least perfect square, which is divisible by each of 21, 36 and 66
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06 Apr 2019, 09:08






