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Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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Updated on: 28 Oct 2014, 05:16
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62% (01:14) correct 38% (01:35) wrong based on 194 sessions
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Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287 is divided by 16 A. 1 B. 7. C. 2. D. 3 E. 9
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Originally posted by dimri10 on 23 Jun 2011, 02:45.
Last edited by Bunuel on 28 Oct 2014, 05:16, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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17 Nov 2014, 23:20
dimri10 wrote: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287 is divided by 16
A. 1 B. 7. C. 2. D. 3 E. 9 A number ending in a 0 is divisible by 2. A number ending in 2 zeroes is divisible by 4. A number ending in 3 zeroes is divisible by 8. A number ending in 4 zeroes in divisible by 16. Given the obscene number, you should immediately be convinced that you will need to focus on a very small part of it. 42,527,152,653,425,416,242,624,272,427,215,287 = 42,527,152,653,425,416,242,624,272,427,210000 + 5287 The first number is divisible by 16. You just have to find the remainder when you divide 5287 by 16. That will be the remainder when you divide the original number by 16. 5287/16 gives remainder 7. Answer (B)
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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 03:59
dimri10 wrote: whats the reminder when 42527152653425416242624272427215287 is dividd by 16
a.1 b.7. c.2. d.3 e.9 divisibility by 16 .. take last 4 digits 5287 mod 16 = 7 hence B




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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 05:43
I tried the following approach: 42527152653425416242624272427215287/16 = 42527152653425416242624272427215280/16 + 7/16 Now the last two digits of first term, 80 is divisible by 4 and so is the denominator 16. Again, 80/16 = 20/4 also divides , so remainder = 7 Answer  B But I'm not 100% convinced with my approach.
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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 05:53
subhashghosh wrote: I tried the following approach:
42527152653425416242624272427215287/16 = 42527152653425416242624272427215280/16 + 7/16
Now the last two digits of first term, 80 is divisible by 4 and so is the denominator 16.
Again, 80/16 = 20/4 also divides , so remainder = 7
Answer  B
But I'm not 100% convinced with my approach. Hi Subhash, even i doubt your approach.. I tried randomly sum numbers using ur approach, but it doesnt work.. http://staff.argyll.epsb.ca/jreed/math7 ... 1/1104.htm



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 05:55
Yup.. I'll wait for experts to opine on this.
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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 10:22
sudhir18n wrote: dimri10 wrote: whats the reminder when 42527152653425416242624272427215287 is dividd by 16
a.1 b.7. c.2. d.3 e.9 divisibility by 16 .. take last 4 digits 5287 mod 16 = 7 hence B it seems like a wise mehod.kudos



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 21:06
@sudhir18n even though your approach worked here , i dont think it can be generalized that way. It may not yield the right answer all the time. eg : 2782 when divided by 16 leaves a remainder of 6 but 12782 when divided by 16 leaves a remainder of 14 so if we you get 12782 and were asked to find remainder when divided by 16 if you go by 4 digit approach then that would be wrong.. sudhir18n wrote: dimri10 wrote: whats the reminder when 42527152653425416242624272427215287 is dividd by 16
a.1 b.7. c.2. d.3 e.9 divisibility by 16 .. take last 4 digits 5287 mod 16 = 7 hence B



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 23:18
Spidy001 wrote: @sudhir18n even though your approach worked here , i dont think it can be generalized that way. It may not yield the right answer all the time. eg : 2782 when divided by 16 leaves a remainder of 6 but 12782 when divided by 16 leaves a remainder of 14 so if we you get 12782 and were asked to find remainder when divided by 16 if you go by 4 digit approach then that would be wrong.. sudhir18n wrote: dimri10 wrote: whats the reminder when 42527152653425416242624272427215287 is dividd by 16
a.1 b.7. c.2. d.3 e.9 divisibility by 16 .. take last 4 digits 5287 mod 16 = 7 hence B are you sure 2782 mod 16 = 6 im getting 14 ..



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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23 Jun 2011, 23:51
Yeah u r right. I mis calculated that one .
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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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24 Jun 2011, 01:28
Can someone pls explain what the approach should be for such a question



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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24 Jun 2011, 09:52
We all know 100 is divisible by 4 so 100x100 is divisible by 4x4. To find the remainder of A divided by 16, we just need the last 4 digits.



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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17 Nov 2014, 16:57
16 is a multiple of 8 and 2. A number can be checked for 8, by dividing the last 3 digits and for 2 by dividing the last digit. Here 287 will result in 7 as a remainder when divided by 8 and 1 when divided by 2. Therefore the remainder is 7*1 = 7. Break down the divisor into smaller factors and perform the division check. dimri10 wrote: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287 is divided by 16
A. 1 B. 7. C. 2. D. 3 E. 9



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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05 Dec 2015, 08:25
can we follow this approach every time ? for last 2,3,4 digits and check the divisibility test ?



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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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01 Mar 2017, 14:21
Amit0507 wrote: 16 is a multiple of 8 and 2. A number can be checked for 8, by dividing the last 3 digits and for 2 by dividing the last digit. Here 287 will result in 7 as a remainder when divided by 8 and 1 when divided by 2. Therefore the remainder is 7*1 = 7. Break down the divisor into smaller factors and perform the division check. dimri10 wrote: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287 is divided by 16
A. 1 B. 7. C. 2. D. 3 E. 9 I used the same approach of 287 div by 8 gives remainder as 7. Now I divided 7 by 2 to get remainder as 1. So my Answer was A, which i now know is wrong. But I dont understand your last step of multiplying 7*1 = 7 as remainder. What's your thought process doing that step ? Any help appreciated.
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Re: Whats the reminder when 42,527,152,653,425,416,242,624,272,427,215,287
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