Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 37)
37.
A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?
Answer: Option
Explanation:
4 a 3 | 9 8 4 } ==> a + 8 = b ==> b - a = 8 13 b 7 |
Also, 13 b7 is divisible by 11 (7 + 3) - (b + 1) = (9 - b)
(9 - b) = 0
b = 9
(b = 9 and a = 1)
(a + b) = 10.
Discussion:
35 comments Page 2 of 4.
Vishal Chauhan said:
8 years ago
Thanks @Uma.
Mir Rokon Uddin said:
9 years ago
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.
From 1 to 9 if we place a= 9 and b= 1 then the outcome will be 8. So a+b= 10.
Ancy said:
9 years ago
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?
Ans: 4.
I didn't understand. Will anyone explain me?
(1)
Swati said:
9 years ago
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.
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.
Mohit Bansal said:
1 decade ago
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.
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.
M.devi said:
1 decade ago
I couldn't understand this problem please say it in another method.
(1)
Samyuktha said:
1 decade ago
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.
+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.
Yash said:
1 decade ago
Nice explanation dudes.
Aishu said:
1 decade ago
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.
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.
Raja said:
1 decade ago
Any one please explain the logic.
(1)
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