Aptitude - Numbers - Discussion

@Uma.

Super, Thanks for explaining.

To know the number is divisible by 11, the rule is,
Sum of all even places=sum of all odd places.
In 13b7,
3 + 7 = 1 + b.
b=9.

Then a+8 = 9.
a = 1.
a+b = 1 + 9 = 10.

Nice explain @Samyuktha.

Guys, 11 divisibility rule is applicable here so solve it by that rule. It will be more easy.

4 digit number 13b7 i. e divided 11.
So even-odd then :(3+7)-(1+b )apply b=9.
10-10=0.
a+b=1+9=10.

But what about condition that 13b7 is divisible by 11.

Here 4 a 3
And 9 8 4
Gives 1 3 b 7
Since 4+9=13
and at last, we are getting 13 it means the previous no has not generated any carry which can only be possible when a=1. as then we will get 8+1=9.

Therefor a=1, b=9.
a+b=10.

+984
-------
13b7

The first thing is to figure out the possible values of the sum. The missing digit can be 0 to 9:
1307, 1317, 1327, 1337, 1347, 1357, 1367, 1377, 1387, 1397.

Only one of these is evenly divisible by 11, namely 1397, so b = 9.

Filling that in, we have:
. 4a3
+984
-------
1397

From that, it is easy to figure the top number is 413 and thus a = 1.

a=1
b=9
a+b = 10
Answer:
10

Thank you @Uma.

By placing b=9, yes we can divide the number by 11. In other question, that I have come across, another way to find whether a number is divisible by a given number is by this method: eg: 517*324=5+1+7+x+3+2+4=22+x; To let the number be divisible by 3, we simply put x=2 to give 24 which is divisible by 3.

Now, for this 1397=1+3+9+7=20; 20 is not divisible by 11. Is this method applicable for only certain questions? Because this clearly doesn't work for this one.