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Which of the following is a multiple of 6^2?
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31 Jul 2017, 23:55
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75% (02:04) correct 25% (01:26) wrong based on 84 sessions
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Which of the following is a multiple of 6^2? (A) 36,045 (B) 43,623 (C) 52,224 (D) 89,892 (E) 92,044
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Which of the following is a multiple of 6^2?
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Updated on: 01 Aug 2017, 00:19
Bunuel wrote: Which of the following is a multiple of 6^2?
(A) 36,045
(B) 43,623
(C) 52,224
(D) 89,892
(E) 92,044 A multiple of 36 should be a multiple of 2 and 3 A & B out (Odd) E out (Sum of digits not divisible by 3) Between C & D 52224 sum of digits is 15 (not divisible by 9) 89892 sum of digits is 36 (Divisible by 9 and 4) Any number divisible by 3 (sum of digits should be divisible by 3) Any number divisible by 9 (sum of digits should be divisible by 9) D is divisible by 9 as well as 4 (last two digits divisible by 4) Hence D
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Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:09
Bunuel wrote: Which of the following is a multiple of 6^2?
(A) 36,045
(B) 43,623
(C) 52,224
(D) 89,892
(E) 92,044 Prime factorization of \(6^2\) ie; \(36 = 2^2 * 3^2\)
Multiple of \(36\) should be divisible by both \(2\) and \(3\) and should have at least \(2^2\) and \(3^2\) as factors.
Checking options;
(A) \(36,045 \) Not divisible by \(2\)  Hence Not multiple of \(36\)
(B) \(43,623 \) Not divisible by \(2\)  Hence Not multiple of \(36\)
(C) \(52,224 \) Has \(2^2\) and only one \(3\) as factor  Hence Not multiple of \(36\)
(D) \(89,892 \) Has \(2^2\) and \(3^2\) as factor  Hence Is Multiple of \(36\)
(E) \(92,044 \) Not divisible by \(3\)  Hence Not multiple of \(36\)
Answer (D)...



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Re: Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:17
For a number to be a multiple of 6^2 it has to divide by the factors of 6^2. Accordingly we can immediately eliminate options A and B as they are odd numbers (and thus clearly don't divide by 2, 6 etc). We can also eliminate option E as it doesn't divide by 6 (the sum of the number is even but it doesn't divide by 3). So we left with options C and D. As option C doesn't divide by 9 up to the some of its digit number, D is the correct answer as the sum of its digit is 36 and thus divide by 9.
Answer: D



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Re: Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:20
sashiim20  How did you figure out quickly how many 2's and 3's each of those big numbers has?



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Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:22
oryahalom wrote: sashiim20  How did you figure out quickly how many 2's and 3's each of those big numbers has? Just check if number is divisible by 2 and 3 and also divisible by 4 and 9... We just need to find if the number has at least two 2 (ie;4) and two 3 (ie; 9)... this is enough to prove number is divisible by 36... we don't need to do full division...



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Re: Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:25
sashiim20 wrote: oryahalom wrote: sashiim20  How did you figure out quickly how many 2's and 3's each of those big numbers has? Just check if number is divisible by 2 and 3... we can also check the divisibility of 4 (last 2 digits divisible by 4 which is same as 2^2) and 9 (sum of digits divisible by 9 which is same as 3^2)
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Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:26
sashiim20 wrote: oryahalom wrote: sashiim20  How did you figure out quickly how many 2's and 3's each of those big numbers has? Just check if number is divisible by 2 and 3... My question is not whether the number is divisible by two or three but how many 2's and 3's each of the numbers has (up to prime factorization). In big numbers it seems to be time consuming.



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Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:31
oryahalom wrote: sashiim20 wrote: oryahalom wrote: sashiim20  How did you figure out quickly how many 2's and 3's each of those big numbers has? Just check if number is divisible by 2 and 3... My question is not whether the number is divisible by two or threes but how many 2's and 3's each of the numbers has (up to prime factorization). In big numbers it seems to be time consuming. You don't need to do prime factorization... Please check the post above i have mentioned, divisibility of 2,3, 4 and 9 is enough (Or in other words, check if 2,3,4 and 9 are factors of number)... divisibility of 4 will automatically prove number is divisible by 2. Divisibility of 3  sum of numbers are divisible by 3 Divisibility of 4  last 2 digits are divisible by 4 Divisibility of 9  sum of numbers are divisible by 9 From the given options  A, B, E are out. Only left options are C and D... Following the divisibility rules we can figure out which is divisible by 36. C is divisible by 3 but not by 9. Hence Eliminated. Answer D... Hope its clear now.. Let me know if you have any other doubts... Thank you



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Re: Which of the following is a multiple of 6^2?
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01 Aug 2017, 00:45
\(6^2 or 36\) when multiplied by any number, cannot end with a 3 or 5. So straight away A and B are out. Of Options C,D and E  numbers 52224 and 92004 are not divisible by 9, which is a necessary condition for a number to be divisible by \(36(2^2 * 3^2)\) We are left with Option D(89892), which is the correct answer.
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Re: Which of the following is a multiple of 6^2?
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01 Aug 2017, 21:57
Options A, B, E fail the divisibility test of numbers 2 and 3, and hence are out of contention. Out of option C and D, only D is divisible by 9 (because \(6^2\) can also be written as \(9 * 4\)). Answer is Option D.
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Re: Which of the following is a multiple of 6^2?
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04 Aug 2017, 05:37
Bunuel wrote: Which of the following is a multiple of 6^2?
(A) 36,045
(B) 43,623
(C) 52,224
(D) 89,892
(E) 92,044 6^2 = 3^2 * 2^2 = 9 * 4 So to check which of the following is multiple of 6^2 ., we need to check which one is divisible by both 4 and 9. Divisibility by 4 is easier than 9 , So first eliminate the choices which are not divisible by 4. Check the last 2 terms of the numbers. Clearly A and B are not divisible by 4 Now lets check divisibility of C D E by 9 C. 5+2+2+2+4 = 15 not divisible by 9. D. 8+9+8+9+2 = 36 is divisible by 9. E. 9+2+0+4+4 = 19 is not divisible by 9. So, Answer D
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Re: Which of the following is a multiple of 6^2?
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