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.

If we add 4a3 to 984 then we can clearly see that the result is such a way that the digits in 100'ths place have no carry.

That means a+8 is always less than 9 in order to get 13b7 by comparing the places in addition the 10'ths digits underwent.

a+8=b --- (1).

Then 13b7 is divisible by 11 so the difference of sets, sum of even places and odd places in 13b7 is.

1+b-10=0 or multiple of 11.

Try 0.

Then b=9 which is less or equal to 9 hence by eq (1).

We can get answer.

If you try other than 11 then this adds up a carry in first addition of two number.

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.

4 a 3
+9 8 4
----------
1 3 b 7
----------
If any any number added to 8 which is greater than 2 or equals to 2 gives a carry so choose the number in place of a which is less than 2.

8+a = 8+1 = 9 which doesn't give a carry so a=1.

4 1 3
+9 8 4
---------
1 3 9 7 -----> which is divisible by 11.
-----------
a=1, b=9.
a+b=1+9=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

By divisibility rule of 11, we know that it's the difference between sum of number at odd places and sum of no. at even places must be zero or divisible by 11.

Hence we have,

7+3-(b+1), if we put 9 in place of b we get zero hence the condition is satisfied and the no. Is divisible by 11.

So the no. is 1397.

Now subtract 984 from 1397, we get 1397-984 = 413.

Hence a=1 and b=9.

And a+b = 10.

Soma-->
the number is divisible by 11 if the difference between the sum of its digit at odd places and the sum of its digit at even places is either 0 or number divisible by 11.
4 a 3 | ==> here a can either be 0 or 1
9 8 4 }
13 b 7 |==> so b can be either 8 or 9

but (7+3) - (b+1) should be divisible by 11 or 0
so b can't be 8. so b is 9 and a is 1.
hence (a+b) is 10.

Here they give two numbers 4a3 and 983.
And they said when they are added together, we get 13b7.
It means.

4 a 3.
9 8 4.
------
13 b 7.
--------

When we observe this.
If 3+4=7.
4+9=13 then a = 0 or 1.

We get b = 8 or 9.
And also they said that 13 b 7 is divisible by 11.

If we take b = 9 i.e.,
1397 it is divisible by 11.

So b = 9, a = 1.

Now a+b = 9+1 = 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.

Luckily in this prob, no formula/logic is required.

4+3=7.
8+a=b.
9+4=13.

Since 9+4 = 13 which means there was no carry from previous.

A carry will happen only if 8+ a >= 10.

i.e a = or > 2.

So a has to be 1!

When a is 1.8+1 = 9!

a=1.
b=9.