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Two consecutive positive integers, each greater than 9

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Two consecutive positive integers, each greater than 9  [#permalink]

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New post 20 Aug 2018, 18:19
2
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

47% (01:55) correct 53% (01:53) wrong based on 64 sessions

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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  [#permalink]

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New post 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.

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Re: Two consecutive positive integers, each greater than 9  [#permalink]

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New post 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.

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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  [#permalink]

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New post 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 &nbs [#permalink] 26 Aug 2018, 09:55
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