Two consecutive positive integers, each greater than 9, are divided by

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Competition Mode Question

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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unraveled · Answer

Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?

Let two consecutive integers are x and x+1.
 Re  + Re  = ?

Since the two integers would be odd and even or even or odd, the remainder would be odd and even or even and odd. Thus sum would always be odd except when the 1st integer is an even integer and 2nd is a multiple of 5. 
Example: For 10 and 11, 
 Re  + Re  = 0 + 1 = 1

For 11 and 12, 
 Re  + Re  = 1+2 = 3

....

For 14 and 15, 
 Re  + Re  = 4 + 0 = 4

(1) The sum of the remainders is even.
For 14 and 15, 
 Re  + Re  = 4 + 0 = 4

SUFFICIENT.

(2) The sum of the units digits of the two original integers is 9.
Let x = 10a + b, where and b positive integer. a>0 & b≥0 
x + 1 = 10a + b + 1
So units digit of integers = b + b + 1 = 9
b = 4 
Hence possible integers are 14 and 15, 24 and 25, 34 and 35 ......

Therefore, 
 Re  + Re  = 4 + 0 = 4  always.

SUFFICIENT.

Answer D.

ArunSharma12 · Answer

statement 1:
possible sum of the remainders can be  
sum of the remainders can be 0,2
not sufficient

statement 2:
if unit digit of the two numbers is 9, then remainder will be 4.
sufficient

Ans: B

AlexCC · Answer

If you list both numbers, to see the reminders of each when divided by 5:
n. n+1. reminder
10. 11. -->0 & 1
11. 12. -->1 & 2
...
14 15 --> 4 & 0
15 16. --> 0 & 1
 1)  Only when the "n+1" is divisible by 5, will "n" has a reminder of 4, the sum of both reminders will be 4, even. All the other cases give a sum of reminders of odd numbers (1,3,5)
both 14 & 15 and 19&20 comply with the rule giving the sum of reminder 4, even. so statement  #1 SUFFICIENT

2)   As two numbers are consecutive, in order to have a sum of unit digits to be 9, there is only 4+5 (5+4 not possible for consecutive integers n and n+1), and 9+0. And for both cases, unit digits 4&5, and unit digits 9&0 have "n+1" number divisible by 5, the other one leaves reminder 4, the sum of the reminders will always be 4, statement  #2 SUFFICIENT

So the answer is D

MayankSingh · Answer

IMO D

Let the numbers be  -  x & (x+1) 
 Take x>9

Possible remainder when x is divided by 5 = { 0,1,2,3,4 } .............(I)
Corresponding to these remainder left by (x+1) when divided by 5 = {1,2,3,4,0 }................(II)

Sum of remainder = (I) + (II) = {1,3,5,7,4}

(1) The sum of the remainders is even.

Only one possible case when sum of remainder = 4 
 Sufficient

(2) The sum of the units digits of the two original integers is 9.

Two consecutive numbers with sum of unit digit as 9 , So number should be of the form X4 & X5
Now, Rem (X4/5) = 4 & Rem (X5/5) = 0
Sum of remainder = 4+0 = 4 
 Sufficient

GMATWhizTeam · Answer

Solution 
 Step 1: Analyse Question Stem 
 • Let us assume that the two consecutive positive integers are n and n+1.
 o  n > 9  and  n+1 > 9  
• We need to find the value of  (\frac{n}{5})_R + (\frac{n+1}{5})_R 
 o Where  ()_R  denotes remainder of.

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE 
 Statement 1:  The sum of the remainders is even.
 • When a number is divided by 5, the possible remainders can be (0, 1, 2, 3, 4)
• Now, when two consecutive integers (or n and n+1) are divided by 5, the possible values of  ((\frac{n}{5})_R, (\frac{n+1}{5})_R )  can be  (0, 1), (1, 2), (2, 3), (3, 4),  and  (4, 0) .
 o And only possible case, when the sum of the remainders is even, is  ((\frac{n}{5})_R, (\frac{n+1}{5})_R ) = (4, 0)  
• Thus,  (\frac{n}{5})_R + (\frac{n+1}{5})_R = 4  
Hence, statement 1 is sufficient and we can eliminate answer Options B, C and E.

Statement 2:  The sum of the units digits of the two original integers is 9.
 • According to this statement, the unit’s digit of n + the units’ digit of n+1 = 9
 o This is only possible if the unit’s digit of n = 4, and the unit’s digit of n+1 = 5.
  Therefore,  (\frac{n}{5})_R = 4  and
  (\frac{n+1}{5})_R = 0   
• So,  (\frac{n}{5})_R + (\frac{n+1}{5})_R = 4  
Hence, statement 2 is also sufficient and the correct answer is  Option D.

yashikaaggarwal · Answer

Two consecutive positive integers, each greater than 9, are divided by 5.
Pairs will be in the form if 10a+b, 10a+b+1

Statement 1. The sum of remainder is even 
That means 10a+b/5 10a+b+1/5 should be even.
=> Let say no. Is 10&11 sum of remainder is =0+1 = 1,(same will be for 20&21, 30&31)
=> Let say no. Is 11&12 sum of remainder is =2+1 = 3,(same will be for 22&21, 32&31)
=> Let say no. Is 12&13 sum of remainder is =2+3 = 5,(same will be for 22&23, 33&32)
Similarly = 13/5+14/5 = 7
14/5+15/5 = 4+0 = 4 (even) ------(1)
15/5+16/5 = 0+1 = 1 
16/5+17/5 = 1+2 = 3
17/5+18/5 = 2+3 = 5
18/5+19/5 = 3+4 = 7
19/5+20/5 = 4+0 = 4 
Therefore the sum of remainder = 4 (sufficient)

Statement 2: The sum of the unit digits of the two integers = 9
That means the no. Will either end from 4&5 or 9&0
Let say the no. is 14&15
=> Sum of remainder = 14/5+15/5 = 4+0 = 4
=> Sum of remainder = 19/5+30/5 = 4+0 = 4 
Therefore the sum of remainder= 4 (sufficient)

IMO D

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Archit3110 · Answer

target find a/5 +b/5 = sum of remainders 
where a , b>9
#1
The sum of the remainders is even.
possible contenders are 14,15 ; (24,25) ; ( 19,20) ; where we get remainder as 4,0 always 4 ; so sufficient
#2
The sum of the units digits of the two original integers is 9.
possible set 
14,15 ; 19,20 ; 24,25...
sufficient
remainder sum is 4
OPTION D ; sufficient

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.

Abi1995 · Answer

Ans - D

Check statement 1- 
Option 
10,11- R=1
12,13-R= 5
14,15, R= 4
4 will always be even remainder.We cannot get 0 or 2 as remainder for consecutive number and remainder cannot be more than 5
This is Sufficient

Check statement 2-
Only way to get sum if unit digit 9 is 4 and 5 in consecutive integers, bow for any such consecutive pair the remainder will always be 4
So this is sufficient too.

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Saasingh · Answer

IMO D

We know numbers are consecutive and each greater than 9. 
So numbers can be (10,11),(11,12),(12,13) ...so on.

(1) Sum of remainders is even.

When division is by 5, remainders can be 0,1,2,3,4.

We get sum of remainders even by choosing 2 even or 2 odd remainders from 0,1,2,3,4.

Its important to understand that 2 consective numbers can never yield same remainder on division with 5. Also the remainders for consective numbers will differ by 1.

Thus, only 0 and 4 satisfy this. (0 is equivalent to 5 and 4,5 are consecutive).

Sum = 0+4=4

In another way, if this logic seems difficult, one may take examples. All consecutive numbers yield odd sum EXCEPT numbers that yield remainder 4 and 0.

Satisfies.

(2) Sum of units digit is 9.

Consective numbers ending in 4 and 5 give 9. Example: 14,15 
Or consecutive numbers ending with 9 and 10. Example: 19,20.

In either case, on division with 5 yields sum of remainders as 4+0 = 0.

Satisfies.

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minustark · Answer

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.

1) When an integer is divided by 5, the remainder can take values: 0, 1, 2, 3, 4. Since the dividends are consecutive, so we can infer one of them is divisible by 5, and another one is the prior integer, which will have remainder of 4. Sum of the remainder is 4. Sufficient

2) Two consecutive integer have to be ended up in 4 and 5 or 9 and 0 to get 9 as units digits. So, one of them will have remainder 4 and another will be divisible by 5. Sum will be 4 in both cases. Sufficient
D is the answer.

exc4libur · Answer

(1) sufic
possible remainders: 1,2,3,4,0
consecutive integers: 9,10/5 or 4,5
remainders: 4,0=4=even

(2) sufic
consec integers: 14,15,19,20
units: 4+5=9, 9,0=9
remainders: 4+0=4

Ans (D)

monikakumar · Answer

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.

1) 1,2,3,4 are remianders when divided by 5
2,4, are even
2 is not possible,
14,15 => remainder sum is 4
sufficient

2)sum of units digit is 9
as it is consecutive integers, it must be 4 and 5
so sum of remainders is 4
suffcient

Ans D

chetan2u · Answer

Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?

Consecutive numbers on division will give Remainders as consecutive. In this case, the pattern of remainder will be 1,2,3,4,0,1,2  
So Sum of the remainders can be ONLY one of 1+2, 2+3, 3+4, 4+0, 0+1.

(1) The sum of the remainders is even.
Now If the remainders Are consecutive, so E+O, the sum will always be O. 
Here only when one number is a multiple of 5 and other leaves a remainder 4, the sum is even. 
Sum =4+0=4.
Numbers could be (19,20); (14,15); (199,200) and so on

(2) The sum of the units digits of the two original integers is 9.
Property of divisibility by 5 is that the units digit should be 5 or 0. 
So remainder =9-5=4.

D

MBAApp19 · Answer

B 
Any combination that adds to 9 when divided by 5 results 4 as sum of remainder

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HoneyLemon · Answer

Is it D ?

lets take 2 int = x and x+1 ... x > 9

If a number is divided by 5 .. then reminder can be one of 0,1....4 ..

(1) The sum of the remainders is even.

Sum of reminder is even that means : EVEN + EVEN OR ODD + ODD ..

ODD + ODD reminder is not possible .. Since the numbers are consecutive ..

So EVEN + EVEN = 4 +0 = 4 . So sum of reminders is 4 .. SUFFICIENT

(2) The sum of the units digits of the two original integers is 9.

2 Consecutive integers in which sum of the units digits is 9. Only possibility is unit digits are 4 and 5 .
so sum of reminders are 4 + 0 = 4 .. 
.. SUFFICIENT

Hence D ..

lacktutor · Answer

Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?

( Statement1 ): The sum of the remainders is even.

(10,11) - odd
(11,12) - odd
(12,13) - odd
(13,14) - odd
(14,15) - EVEN (The sum of remainders is 4+0 =4)
(15,16) - odd
(16,17) - odd
(17,18) - odd
(18,19) - odd
(19,20) - EVEN (The sum of remainders is 4+0 =4)
.....

Sufficient

( Statement2 ): The sum of the units digits of the two original integers is 9.

TWO CONSECUTIVE INTEGERS:
10x +y and 10x+y+1
---> y + y+1 = 9
2y =8 
y = 4
that means two consecutive integers are (14,15), (19,20), (24,25).....
The sum of the remainders is 4+0 = 4 (Always)
Sufficient

Answer (D).

Lipun · Answer

Hi  ArunSharma12 ,

Both of us misinterpreted the Q. It only says ' sum of remainders ' and  NOT  the ' overall remainder '. The cycle of sum of remainders is {1,3,5,7,4}.
The only even sum is 4. Thus, ST1 by itself is sufficient.

Correct Ans: D

Thanks
Lipun

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Two consecutive positive integers, each greater than 9, are divided by

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