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

x = 9j + m
y = 9k + n
m>n are given

1. x + y is a multiple of 9.
x+y = 9*(j+k) + (m+n)
For x+y to be multiple of 9, m+n must be divisible by 9.
1<=m<=8
1<=n<=8
3<=m+n<=15 because m>n
The only multiple of 9 in this range is 9 itself. So, m+n is 9.
Hence, 1 is sufficient.

2. x*y divided by 9, the remainder is 5
(9j + m)(9k+n) = 9p+5
The only term with no 9 is mn
mn must be of the form 9a+5, where a is a non-negative integer (including 0).
m= 5, n =1 is a possible solution, which gives m+n = 6
m=7, n=2 is another solution, which gives m+n = 9
So, a unique solution is not possible.
Hence 2 is not sufficient.

A is the answer.

x = 9p + m
y = 9q + n
m>n
x & y are +ve integers

Value of m+n?

1) x + y is a multiple of 9

Adding both the equations mentioned above

x+y = 9p + 9q + m + n
= 9(some integer) + m + n

If x+y is a multiple of 9, that means m+n is also a multiple of 9

Since m & n < 9

That means m+n = 9

Sufficient

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

xy = (9p+m) * (9q+n)
= 9(some integer) + mn

Now the remainder is 5 but it doesn't mean that mn = 5

Since m>n, m can be 5 and n can be 1 making mn = 5 and m+n = 6

m can also be 7 and n can be 2 making mn = 14 which when divided by 9 gives remainder of 5. In this case m+n = 9

Not sufficient

Ans A

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given
x=9a+m
y=9b+n
if m>n , value of m+n?
#1
x + y is a multiple of 9.
x+y can be 9,18,27,36,45

17=9*1+8
1=9*0 +1
m+n=9
x+y=27

6=9*(0)+6
21=9*+3
m+n=9

7=9*0+7
20=9*2+2
m+n=9

x+y=36
30=9*3+3
6=9*0+6
m+n=9

15=9*1+6
21=9*2+3

sufficient that m+n is 9

#2
x*y divided by 9, the remainder is 5
possible values of x*y will be
1*5 ; 2*7 ; 1*23 ; 2*16 ; 1*41 ; 2*25
5=9*0+5
1=9*0+1
m+n=6

16=9*1+7
2=9*0+2
m+n=9

41=9*4+5
1=9*0+1

25=9*2+7
2=9*0+2

we get two values of m+n i.e. 6 & 9

insufficient



[b]OPTION A is correct[/b]

we know that x=9a+m and y=9b+n. And m>n.
Value of m and could be (0,1,2,3,4,5,6,7,8)

Taking statement 1.

We are given that x + y is a multiple of 9.. And we know that x=9a+m and y=9b+n. Therefore x+y= 9a+m+9b+n. Since we know that x+y is a multiple of 9. That means m+n should be a multiply of 9. We can also conclude that value of m or n can be any value among (0,1,2,3,4,5,6,7,8). Given that m>n. m can never be 0. That means n has to be any value from (0,1,2,3,4,5,6,7,8). But for m+n should be a multply of 9, meaning that n also can be zero. Thus the value of m+n will be 9. for x+y to be a multiple of 9. Since we are getting a unique value. We can eliminate options. B,C and E.

Taking statement 2.
According to statement x*y= (9a+m)*(9b+n)=9c+5, On solving this we get 81ab+9an+9bn-9c=5-mn on LHS and RHS respectively. We can see that LHS is a multiple of 9. And the absolute value of 5-mn has to be multiple of 9. Since we know that m>n. m cannot be zero. and 5-mn has to be mulptiple of 9 therefore n also cannot be 0. The only values satisfying the condition are m=5 and n=1; m=7 and n=2. Since we get two different value of m+n. We can eliminate option D.

The correct answer is 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 ?

x = 9p+ m
y= 9q +n
but m>n hence m cannot be 1 and n cannot be 8
hence m = 2,3,4,...,8
n = 1,2,3,...,7

(1) x + y is a multiple of 9.
multiple of 9 = 9,18,27
suppose x+y = 18
x=15,y=3 then m=6, n=3 hence m+n = 9
suppose x+y = 27
x=14, y = 13 then m=5,n = 4 m+n =9
suppose x=15, n = 12; m=6,n=3 m+n = 9

hence 1 is enough then the answer could be A or D
(2) x*y divided by 9, the remainder is 5
x*y= (9p+m)*(9q+n)=81pq+9pn+9qn+mn from this we know that m*n is 5
so m must be 5 and n must be 1 only then m+n =6 enough
answer is D