Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 4)
4.
What least number must be added to 1056, so that the sum is completely divisible by 23 ?
Answer: Option
Explanation:
23) 1056 (45
92
---
136
115
---
21
---
Required number = (23 - 21)
= 2.
Discussion:
73 comments Page 8 of 8.
Ankita said:
1 decade ago
Consider the number 13. When we divide it by 6, the remainder is 1. Now, what must be added to 13 to make it completely divisible by 6?
12 is completely divisible by 6. We must add 6 more to 12 to make it divisible by 6. While dividing by 13, the remainder is 1. So we add 5 (6-1) more to make the number 13 divisible by 6.
Similarly in this case, we add 2 (23-21).
12 is completely divisible by 6. We must add 6 more to 12 to make it divisible by 6. While dividing by 13, the remainder is 1. So we add 5 (6-1) more to make the number 13 divisible by 6.
Similarly in this case, we add 2 (23-21).
Rupesh said:
1 decade ago
Consider first 3 digits 105.
23*4=92; 105-92=13.
Now 13 & remaining digit 6 forms 136.
23*5=115; 136-115=21.
If we are adding 2 in 21 then it will completely divisible.
23*4=92; 105-92=13.
Now 13 & remaining digit 6 forms 136.
23*5=115; 136-115=21.
If we are adding 2 in 21 then it will completely divisible.
Navin kumar kamti said:
1 decade ago
1056/23 = 45.91.
45*23 = 1035.
Now we can write 1035+23 = 1058.
Hence,
1058-1056 = 2.
45*23 = 1035.
Now we can write 1035+23 = 1058.
Hence,
1058-1056 = 2.
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