Aptitude - Numbers - Discussion

For this you see the question as a primary school student:

Remember one rule just:

For a number to be divisible by 11 the sum of odd placed minus the sum of even places should either be zero (0) or 11.

Example:

abcdef - is divisible by 11 or not. So, check (a+c+e) - (b+d+f) = so what ever you will get should either be zero (0) or 11.

Now, the number was.

4 (hundreds) a (tens) 3 (unit) since 4a3.
9 (hundreds) 8 (tens) 4 (unit) 984.
__________________________________ ______.

13 (hundreds) b (tens) 7 (unit) 13b7.

Look- 3+4 in unit place gives no carry.

Now forget middle place and see.

4+9 =13 (no carry came from tens place its visible).

So 8 in the tens place can add only 2 numbers which can't give carry they are:.

0 as 0+8=8.
1 as 1+8=9.

Rest all number will give carry.

Sp finally at last you have two numbers for 4a3.

Those are 4{0) 3 or 4{1}3 by adding them both with 984 seperately you will get 1387 and 1397.

Check 1387 = abcd = (a+c) - (b+d) => (1+8) - (3+7) => 9-10 = -1 (not the answer).

1397 = abcd = (a+c) - (b+d) => (1+9) - (3+7) => 10-10 = 0 (we needed either zero (0) or 11 we got zero (0). Hence a is 1, so b will be 1+8=9.

So answer is a+b = (1+9) =10.

13b7 must be divisible by 11.

First, divide 13 by 11 we get remainder 2.
Now divide 2b7 by 11. As the value of b is between 0 to 9 quotient must be 2.

Now We want such no which must be 77 which is divisible by 11. Hence the value of b = 9.

As b = a + 8, We get the value of a is 1.

Hence a + b = 1 + 9 = 10.

If the number 5728abc (0<=a, b, c<=9) is exactly divisible by 12, then find the maximum value of a-b+c?

Ans: 4.

I didn't understand. Will anyone explain me?

b - a = 8. Both a and b must be a single digit(they must come from1 to 9).

From 1 to 9 if we place a= 9 and b= 1 then the outcome will be 8. So a+b= 10.

Thanks @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.

Thank you @Uma.

+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

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.

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