Around the World in 80 Questions (Day 7): When positive integer x is

When positive integer x is divided by 9, the remainder is m and when positive integer y is divided by 9, the remainder is n. If m > n, what is the value of m + n ?

(1) x + y is a multiple of 9.
Remainder of (x+y)/9 = remainder of x/9+ remainder of y/9 = m+n. We know that 0<m<9 and 0<n<9, so 0<m+n<18. Since x+y is divisible by 9, that means m+n is multiple of 9 so that the remainder of (x+y)/9 is zero. m+n can only be 9. Sufficient.

(2) x*y divided by 9, the remainder is 5
Remainder of xy/9 = remainder of x/9 * remainder y/9 = m*n. We know the remainder is 5, let's say m*n = 9k+5( k is an integer >=0), so mn can be 5, mn can be 14, mn can be 32... therefore m+n is not certain. Not sufficient.

Answer is A

-We have some prelinimary conditions:
-m,n can be positive integers or equal to 0 (since x and y are positive integers)
-9>m>n
(1) x+y is a multiple of 9=> m+n=0 or m+n=9. But since m>n , n can't be 0 (if n=0, m must be 0 to make the satement hold, and thus making m=n)=>m+n=9=>SUF
(2) x*y divided by 9, the remainder is 5=>m*n=9k+5 (k is integer)=> We have m=5, n=1, m+n=6 or m=7, n=2, m+n=9=>We don't know what exactly m+n=>INS
The answer is A

To find: The value of m + n where m > n.

Statement 1: x + y is a multiple of 9.

Combination 1:

If x=14 and y=22.
14 + 22 = 36 ( multiple of 9) .
14/9, m=5
22/9, n=4.
m + n = 9.

Combination 2:

x = 15 ; y = 30
x+y=45 ( multiple of 9).
15/9, m = 6
30/9, n = 3
m + n = 6+3 = 9.

Combination 3:

x = 24;y = 30
x+y=54 (multiple of 9).
24/9,m=6
30/9,n=3
m+n=9.

Irrespective of the values of m and n, the value of m+n =9.
We have a definite value so statement 1 is sufficient.

Statement 2:

x×y, divided by 9, remainder is 5.

Case 1: x=22;y=17
22/9,m=4
17/9,n=8
m+n=12.

Case 2: x=10;y=14
10/9,m=1
14/9,n=5
m+n=6.
We don't have a definite value of m+n. So statement 2 is not sufficient.

Answer is A, statement 1 alone is sufficient.

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Let x=9k+m & y=9j+n

S1: x+y is divisible by 9.
So, x+y = 9r
Putting in values of x & y gives,
9k + 9j + m+n = 9r
Hence, m+n must be a multiple of 9. But m & n are remainders so their individual values cannot be more than 9. Hence, m+n must be equal to 9.
Sufficient.

S2: x*y/9 gives remainder 5.
x*y = 81kj + 9kn + 9jm + mn
Here all terms except mn are divisible by 9.
Hence, mn/9 must give remainder = 5.
m*n = 9q + 5
Possible values (less than 9) of (m,n) which satisfies above equation are (5,1) & (7,2).
Hence, m+n = 6 or 9.
Not Sufficient.

IMO: A

answer is D.

let x = 9a + m, y = 9b + n where a and b are positive integers.
given m>n. since m and n are reminders when x,y are divided by 9. thus \(0 <= n < m < 9 \) (if in case math formatting is not visible - i have written that 0 <= n < m < 9)

STATEMENT 1
x+y is a multiple of 9. This implies that x+y can be expressed as 9*(some positive integer)
now x+y = 9a+m+9b+n = 9(a+b) + (m+n) . Now m+n is the reminder when x+y is divided by 9. if n is 0, then m must be 9 but this doesn't make sense. so n>0. since x+y is a multiple of 9 and m > n > 0 (so m+n can't be 0), m+n should be 9. hence sufficient

STATEMENT 2
xy = 9c + 5 where c is a positive integer.
Now xy = (9a+m)(9b+n) = 9(9ab) + 9(an) + 9(mb) + mn = 9(9ab + an +mb) + mn. comparing this with xy = 9c + 5, we see that mn=5.
given m>n. so m = 5 and n = 1 is the only possible set of values. so m+n = 6. hence sufficient

SINCE EACH OPTION ALONE IS SUFFICIENT, ANSWER IS D

We can write "x" as:- x = 9a + m & "y" as:- y = 9b + n. Here "a" and "b" are some integers. As "x" and "y" are positive integers, we know that "a" and "b" are non negative.

(1) x + y is a multiple of 9:-

As (x + y) is a multiple of 9, (9a + m + 9b + n) should also be multiples of 9 and the remainder should be 0.

Terms "9a" and "9b" are already multiples of 9, so (m + n) should also be multiples of 9.

As "m" and "n" are remainders after division by 9, their maximum values can only be "8".

So the only multiple of 9 (m + n) can be is "0" or "9". But (m + n) cannot be "0", as it is given m>n and so even if "n" is "0", "m" would have to be at least "1".

So the only possibility for (m + n) is 9.

Hence statement (1) alone is sufficient.

(2) x*y divided by 9, the remainder is 5:-

So, (9a + m)*(9b + n) leaves remainder "5" when divided by "9".
(9a + m)*(9b + n) = 81ab + 9an + 9bm + mn. All the terms of this expression will leave no remainder when divided by "9", except the term "mn".

So, "mn" leaves remainder "5" when divided by "9",

We will check all possible cases of "mn" considering the given condition of m>n
"m" cannot be "0"
For m = 1, "n" can only be "0", which will not leave remainder "5".

For m = 2, "n" can be 0 or 1. Both the cases do not satisfy our condition.

For m = 3, "n" can be 0 or 1 or 2. All of the cases do not satisfy our condition.

For m = 4, "n" can be 0 or 1 or 2 or 3. All of the cases do not satisfy our condition.

For m = 5, "n" can be 0 or 1 or 2 or 3 or 4. Only for n = 1, we get remainder "5"

For m = 6, "n" can be 0 or 1 or 2 or 3 or 4 or 5. All of the cases do not satisfy our condition.

For m = 7, "n" can be 0 or 1 or 2 or 3 or 4 or 5 or 6. Only for n = 2, we get remainder "5"

For m = 8, "n" can be 0 or 1 or 2 or 3 or 4 or 5 or 6. Only for n = 4, we get remainder "5"

So we have three possible combinations of (m,n) and all three give different values for (m + n).

Hence statement (2) alone is NOT sufficient.

Hence correct answer is option A

x & y > 0 & m > n
x = 9q + m; 0< m< x ----a

y = 9q' + n; 0 < n< y ----b


=> x+y = 9k

Additing a+b
x+y = 9(q+q') + (m+n) = 9k
This is only possible if m + n is multiple of 9, therefore m + n = 9 or 18 or 27 or any other multiple of 9.

Statement 1 is insufficient.


=> x*y = 9Q + 5; here 5 < x*y

Substituting values of x & y
(9q + m)(9q' + n) = 81qq' + 9qn + 9q'm + mn = 9(9qq' + qn + q'm) + mn = 9Q + 5
Only possible if mn = 5 & 9qq' + qn + q'm = Q

Therefore, m = 5 & n = 1.
Statement 2 is sufficient.

IMO OA should be B.

When positive integer x is divided by 9, the remainder is m and when positive integer y is divided by 9, the remainder is n. If m > n, what is the value of m + n ?

(1) x + y is a multiple of 9.
Let's say if the two numbers are 16 and 11.
The remainder will be 2 and 7 = 9
if the two numbers are 5 and 4. Their remainders will add upto 9.
The numbers could be 33 + 12
The remainders will be 6 + 3 = 9.
Thus m + n = 9.
This is sufficient.

(2) x*y divided by 9, the remainder is 5
This is insufficient.
Answe = A

When positive integer x is divided by 9, the remainder is m and when positive integer y is divided by 9, the remainder is n. If m > n, what is the value of m + n ?

(1) x + y is a multiple of 9.

x = 9s + m
y = 9t + n

x + y = 9(s+t) + m + n (is a multiple of 9)

m + n = 0 or 9

Since, m>n, and m+n = 0 is not possible,
so, m + n = 9

SUFFICIENT



(2) x*y divided by 9, the remainder is 5

say, when x = 5, then m =5; and, y = 1, then n =1 ; --------> 5*1 mod 9 = 5 ; so m + n =6
possible, also when x = 14, then m =5; and, y = 1, then n =1 ; --------> 14*1 mod 9 = 5 ; so m + n =6
and, so on.

m=5 and y =1 is the only combination possible to get remainder as 5

so, m + n = 6

SUFFICIENT



(D) is the CORRECT answer

Q: What is the value of m+n?

Given, 1. (x+m)/9 = integer and (y+n)/9 = integer
2. M>n

Statement 1: (x + y)/9 = integer

Adding, (x+m)/9 + (y+n)/9 = integer
...(x+y+m+n)/9 = integer
...(x+y)/9 + (m+n)/9 = integer
...we know, (x+y)/9 = integer
...then (m+n)/9 must also be an integer to match RHS
...that is (m+n) = 9 or multiple of 9
...because m & n are remainders, value of m & n each must be than divisor 9
...that is m <=8 and n<=8
...possible values of m & n = 1,2,3,4,5,6,7,8
...we need m + n = 9 or multiple of 9
...but m>n so m & n cannot be same number
...possible values of m+n = 8+1, 7+2, 6+3, 5+4
...no unique value of m+n

Statement 1 is not sufficient.

Statement 2: (xy + 5)/9 = integer

We know, x = 9a+m, y = 9b+n where a & b are quotients, m & n are remainders
...substituting ((9a+m)(9b+n)+5)/9 = integer
...multiplying (9a*9b + 9an + 9bm + mn + 5)/9 = integer
...because (9a*9b)/9 = integer, (9an)/9 = integer, (9bm)/9 = integer, (mn+5)/9 must also be an integer to satisfy the equation
...that is (mn + 5) = 9 or multiple of 9
...we know m>n & m<=8, and n<=8
...then possible values of m & n are
...mn = 9-5
...mn = 4
...if m = 2, n = 2, then mn = 4 but m cannot be equal to n
...if m = 4, n = 1, then mn = 4 which works because m>n
...only unique value of m & n possible
...m + n = 4+1 = 5 unique value

Statement 2 is sufficient.

Therefore, option B is the correct answer.

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S1:- x + y is a multiple of 9 => x + y will leave no remainder => m + n = 9

Sufficient.

S2:- x*y divided by 9, the remainder is 5 => m*n = 9k + 5

We can't determine the value of m and n, so S2 is not sufficient.

A is correct choice.

Stem, x/9 = k1 +m
y/9 = k2 +n
m>n, m+n = ?

(1) x + y is a multiple of 9.
the only way x+y can be a multiple of 9 is if their respective remainders when divided by 9 sum up to 9
Sufficient

(2) x*y divided by 9, the remainder is 5
=> m*n / 9 = r +5
the product here can be 5 or 9+5 = 14 and so on
the possible values for m and n can be 5,1 or 7,2 etc
this solution is not unique
Not suffficient

Option A

When positive integer x is divided by 9, the remainder is m and when positive integer y is divided by 9, the remainder is n. If m > n, what is the value of m + n ?

(1) x + y is a multiple of 9.
Let X= 9a+m
Y= 9b+n
X+Y= 9(a+b) + (m+n)
Since x + y is a multiple of 9 , m+n should also be multiple of 9
(m=5, n= 4)
m+n = 9

(2) x*y divided by 9, the remainder is 5
Let X= 9a+m
Y= 9b+n
X*Y= 9ab + 9an+9bn+mn

Case 1: when mn =5
m=5, n=1 (Since m>n)
m+n=6

Case 2: when mn = 14
m=7,n=2 (Since m>n)
m+n=9

IMO Choice A.

x= 9k+m ------a)
y=9p+n --------b)

Also m>n

we need to calculate value of m+n

st.1)
x+y is a multiple of 9
adding equations a, and b we have

9k+m+9p+n
now m+n has to be multiple of 9, as 9k, and 9p are already a multiple of 9

Additionally since m, and n are remainders we know that their value will be always less than 9. now the sum of two quantities less than 9 will be divisible by 9 if there sum is divisible by 9, hence m+n=9

Sufficient

st.2)

to calculate the value of x*y ;

we will multiply equation a, and b together we have
(9k+m)(9p+n) = 81kp+ 9kn+9pm+mn
all the terms except mn will be divisible by 9, hence remainder when mn is divisible by 9 must be 5
now possible values of m, and n which results in the remainder of 5 are
7,3 :for this pair m+n = 10
6,2 : for this pair m+n= 8

since multiple values of m+n are possible hence st. 2 alone is not sufficient

IMO A

(1) x + y is a multiple of 9.

Let the quotient of \(\frac{x}{9} = a\) and the quotient of \(\frac{y}{9} = b\).

\(\frac{x}{9} = a\) leaves a remainder of \(m\), which can be written as \(x = 9a + m\).

And \(\frac{y}{9} = b\) with a remainder of \(n\) can be rewritten as \(y = 9b + n\).

We are told that \(x + y = 9e\), where \(e\) is an integer, so we can rewrite that as \(9a + m + 9b + n = 9e\). For the left hand side to equal the right hand side, \(m+n\) must sum to a multiple of 9. As, neither \(m\) nor \(n\) can ever be 9 or more, their sum can be \(x+y ≤ 16\), the only multiple of 9 which fits there is 9. Therefore, \(m+n = 9\)

Testing it out, let \(x = 7\) and \(y = 2\), then \(\frac{7}{9} = 0\) with a remainder of \(7\), while \(\frac{2}{9} = 0\) with a remainder of \(2\). In this case \(m+n = 9\).

when \(x = 17\) and \(y = 1\): \(\frac{17}{9} = 1\) with a remainder of \(8\) and \(\frac{1}{9} = 0\) with a remainder of \(1\), so \(n+m = 9\)

when \(x = 52\) and \(y = 11\): \(\frac{52}{9} = 5\) with a remainder of \(7\) and \(\frac{11}{9}= 1\) with a remainder of \(2\). \(n+m = 9\)

There are exceptions to this such as: \(x = 59\) and \(y = 5\) one finds that \(\frac{59}{9}= 6\) with a remainder of \(5\) and \(\frac{5}{9}= 0\) with a remainder of \(5\). However, given that the stem states that \(m>n\) and here \(m = n\), this can be discarded.

\(m+n\) will always sum to \(9\)

SUFFICIENT


(2) x*y divided by 9, the remainder is 5

\(\frac{xy}{9} = e + 5\), where \(e\) is an integer, can be rewritten as \(xy = 9e + 5 \).

Let \(x = 7\) and \(y = 2\): \(7*2= 14\), which fits the above equation. \(\frac{7}{9}= 0\) with a remainder of \(7\) and \(\frac{2}{9}= 0\) with a remainder of \(2\), which makes \(m+n = 9\).

Let \(x = 23\) and \(y = 1\): \(23*1= 23\). \(\frac{23}{9}= 2\) with a remainder of \(5\) and \(\frac{1}{9}= 0\) with a remainder of \(1\), which makes \(m+n = 6\).

This yields two different possible answers for \(m+n\)

INSUFFICIENT


\(ANSWER A\)

X=9a+m, 0 ≤ m ≤8
y=9b+n, 0 ≤ n <=8
m>n


S1)x+y=9(a+b)+m+n clearly m+n=9, as m>n
(m,n) = (8,1), (7,2), (6,3), (5,4)
Suff

S2)
xy=multiple of 9+mn , mn=5; m=5,n=1. m+n=6 (Suff)

We get controversial answers from S1 & 52

Ans D

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From question stem

X=9q+ m
Y=9t+n

We are to find m+n given m>n

Statement 1
X+Y =9k
The multiples of 9 are 9, 18,27,36,45,54, infinite values can be obtained

Not sufficient

Statement 2

XY=9S+5
XY can be 14,23,32,41, etc
Not sufficient


Combining

Even after combining we can't be sure of the value we need to merge both statements.

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