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Two consecutive positive integers, each greater than 9
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20 Aug 2018, 18:19
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Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders? (1) The sum of the remainders is even. (2) The sum of the units digits of the two original integers is 9.
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Two consecutive positive integers, each greater than 9
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20 Aug 2018, 19:54
From statement 1: let's assume two digit numbers 13 and 14 then the remainder when divided by 5 would be 2. If we consider 14 and 15, the remainder would be 4. So we are getting two answers which are even. Hence 1 is insufficient. From statement 2: The only possible unit digits are 4 and 5. Irrespective of the tens digit, the remainder will be 4. Hence 2 is sufficient. B should be the answer. Posted from my mobile device
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Re: Two consecutive positive integers, each greater than 9
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21 Aug 2018, 05:05
Afc0892 wrote: From statement 1: let's assume two digit numbers 13 and 14 then the remainder when divided by 5 would be 2. If we consider 14 and 15, the remainder would be 4. So we are getting two answers which are even. Hence 1 is insufficient.
From statement 2: The only possible unit digits are 4 and 5. Irrespective of the tens digit, the remainder will be 4. Hence 2 is sufficient.
B should be the answer.
Posted from my mobile device The sum of the remainders: the remainder of 13 when divided by 5 is 3. the remainder of 14 when divided by 5 is 4. 3 + 4 = 7 (odd) The only possible way is when the remainder is 0 and when the other remainder is 4. Ex: 15 and 14... 20 and 19 ... 25 and 24... etcc but in all this cases, the sum of the remainders will be 4. So, sufficient. Option D



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Two consecutive positive integers, each greater than 9
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26 Aug 2018, 09:55
Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?
(1) The sum of the remainders is even.
When we divide by 5 the only remainders are possible: 0,1,2,3,4. Because integers are consecutive (a and a+1) the reminders should stay together in the cyclicity. Hence only 0 + 4 is possible. Check 29 and 30, they will give you remainders 0 and 4.  Sufficient
(2) The sum of the units digits of the two original integers is 9
0,1,2,3,4,5,6,7,8,9...  consecutive only 4,5 and 0,9.
Case 1: 29 and 30 (sum of units is 9). Remainders 0+4=4 Case 2: 24 and 25 (sum of units is 9). Remainders 0+4=4  Sufficient
Answer D



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Two consecutive positive integers, each greater than 9
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29 Aug 2019, 10:55
LordStark wrote: Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?
(1) The sum of the remainders is even.
(2) The sum of the units digits of the two original integers is 9. Analyzing the question:We are dividing both numbers by 5 first to find the remainders, then we add the remainders in an attempt to answer the question. Since they are consecutive integers, the remainders will likely be consecutive as well. For example, for 11 and 12 we have remainders 1 and 2, for 13 and 14 we have remainders 3 and 4. This breaks down whenever we hit the “reset” at “remainder of 4” because the highest remainder we can get when dividing by 5, is 4 and the next numbers have a remainder of 0. Statement 1:As explained above the remainders are likely consecutive, in that case consecutive numbers must be either odd + even or even + odd, therefore the sum must be odd. In order for the sum to not be odd, it must be the special case of remainder of 4 plus remainder of 0. Hence the sum must be 4, sufficient. Statement 2:Because multiples of 10's are divisible by 5, you only need to pay attention to the units digits. Since the numbers are consecutive we can have either 4 + 5 or 9 + 0 as the consecutive digits (19 + 20 for the latter as an example). Both of them have remainders with a sum of 4 so this is also sufficient. Answer: D Also note: for statement 2 even if the two summed numbers were not consecutive, the sum of remainders would still be 4.
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Two consecutive positive integers, each greater than 9
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29 Aug 2019, 10:55






