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When 3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9, is a+b=1?

1) The units of all the digits’ sum of 6a3+3b5 is 8
2) a+b<10


* A solution will be posted in two days.

Hi,

Info from Q stem


3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9..
this means 6+a+3+3+b+5 should be a multiple of 9..

6+a+3+3+b+5 =17+a+b
for 17+a+b to be multiple of 9, either 17+a+b=18 or 27 or 36..
A)so a+b=1. possible
B)a+b=10 possible
C)a+b=19.. not possible as one of the two a or b will be two digit number..
we have to answer if a+b=1..

lets see the statements
1) The units of all the digits’ sum of 6a3+3b5 is 8
since units digit is 18 .. Sum is 18 and not 27..
so a+b=1
Suff..

2) a+b<10
We have only two possiblities taht a+b is 1 or 10..
Since it is given a+b<10, a+b=1
Suff

D


I am afraid I do not understand

We have 6+a+3+3+b+5 = 9k , k being an positive integer
so 17 + a+ b = 9k
a+b = 9k-17

If k = 1 a+b negative => impssible
If k =2 a+b =1 Ok
If k =3 a+b =10 Ok
If k =3 a+b =19 Impossible because a and b less than 9 both

So try the two sentences

1) The units of all the digits’ sum of 6a3+3b5 is 8

6a3
+3b5
______
xyz

so it means that z= 3+5 = 8 that's ok and then ???
we also can have a+b =10

1
6a3
+3b5
______
1008

and 1008 is divisible by 9 !
so for me (1) is INSUFF

am I right ?
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chetan2u
MathRevolution
When 3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9, is a+b=1?

1) The units of all the digits’ sum of 6a3+3b5 is 8
2) a+b<10


* A solution will be posted in two days.

Hi,

Info from Q stem


3-digit integer 6a3 plus 3-digit integer 3b5 is divisible by 9..
this means 6+a+3+3+b+5 should be a multiple of 9..

6+a+3+3+b+5 =17+a+b
for 17+a+b to be multiple of 9, either 17+a+b=18 or 27 or 36..
A)so a+b=1. possible
B)a+b=10 possible
C)a+b=19.. not possible as one of the two a or b will be two digit number..
we have to answer if a+b=1..

lets see the statements
1) The units of all the digits’ sum of 6a3+3b5 is 8
since units digit is 18 .. Sum is 18 and not 27..
so a+b=1
Suff..

2) a+b<10
We have only two possiblities taht a+b is 1 or 10..
Since it is given a+b<10, a+b=1
Suff

D


I am afraid I do not understand

We have 6+a+3+3+b+5 = 9k , k being an positive integer
so 17 + a+ b = 9k
a+b = 9k-17

If k = 1 a+b negative => impssible
If k =2 a+b =1 Ok
If k =3 a+b =10 Ok
If k =3 a+b =19 Impossible because a and b less than 9 both

So try the two sentences

1) The units of all the digits’ sum of 6a3+3b5 is 8

6a3
+3b5
______
xyz

so it means that z= 3+5 = 8 that's ok and then ???
we also can have a+b =10

1
6a3
+3b5
______
1008

and 1008 is divisible by 9 !
so for me (1) is INSUFF

am I right ?

Hi,
What you are reading statement I as, is not correct..

You are reading it as Unit's digit of SUM of 6a3 and 3b5 is 8...NO

It says
The units of all the digits’ sum of 6a3+3b5 is 8..
your SUM is 1008 .. so the sum is 1+0+0+8 =9..... NOT 8..

But the other option is when a+b=1..
in that case 6a3+3b5 = 918.....
sum of digits = 9+1+8 = 18..

and here the units digit is 8... CORRECT
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chetan2u

Hi,
What you are reading statement I as, is not correct..

You are reading it as Unit's digit of SUM of 6a3 and 3b5 is 8...NO

It says
The units of all the digits’ sum of 6a3+3b5 is 8..
your SUM is 1008 .. so the sum is 1+0+0+8 =9..... NOT 8..

But the other option is when a+b=1..
in that case 6a3+3b5 = 918.....
sum of digits = 9+1+8 = 18..

and here the units digit is 8... CORRECT


chetan2u

Can you pls explain the sum part and its use in the first statement.

Are you saying that the first statement is telling about sum of all digits in both the numbers?

Also, can you pls explain this part:

1) The units of all the digits’ sum of 6a3+3b5 is 8
since units digit is 18 .. Sum is 18 and not 27..
so a+b=1
Suff..
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gauravk
chetan2u

Hi,
What you are reading statement I as, is not correct..

You are reading it as Unit's digit of SUM of 6a3 and 3b5 is 8...NO

It says
The units of all the digits’ sum of 6a3+3b5 is 8..
your SUM is 1008 .. so the sum is 1+0+0+8 =9..... NOT 8..

But the other option is when a+b=1..
in that case 6a3+3b5 = 918.....
sum of digits = 9+1+8 = 18..

and here the units digit is 8... CORRECT

chetan2u

Can you pls explain the sum part and its use in the first statement.

Are you saying that the first statement is telling about sum of all digits in both the numbers?

Also, can you pls explain this part:

1) The units of all the digits’ sum of 6a3+3b5 is 8
since units digit is 18 .. Sum is 18 and not 27..
so a+b=1
Suff..


Hi Gaurav,

Let me try to explain you and also others will get to know more.

Now the S1 says the units of all the digits’ sum of 6a3+3b5 is 8. Here we need to sum of all numbers i.e. we get 17+a+b (6+a+3+3+b+5)

We have two digit number 17 and a+b, now here we need to get 8 as the units digit for this sum then a+b has to be equal to 1...then 17+1= 18 and 8 is the units digit.

Hope this clears.
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