chetan2u wrote:
MathRevolution wrote:
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
DI 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 ?