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07 Mar 2016, 16:59
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D
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Difficulty:
95%
(hard)
Question Stats:
34% (02:24) correct
66%
(02:24)
wrong
based on 244
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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.
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.
09 Mar 2016, 00:47
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MathRevolution
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
11 Mar 2016, 05:59
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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
In the original condition, in order to satisfy 6a3+3b5=9integer, 6+a+3+3+b+5=17+a+b should be divided by 9 as well. Then, a+b should be 1, 10. Since 1)=2) a+b=1, it is yes and sufficient.
Therefore, the answer is D.
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
In the original condition, in order to satisfy 6a3+3b5=9integer, 6+a+3+3+b+5=17+a+b should be divided by 9 as well. Then, a+b should be 1, 10. Since 1)=2) a+b=1, it is yes and sufficient.
Therefore, the answer is D.
