Which least number should be added to 1000 so that 53 divides the sum exactly? If it is spam, I will delete 10 answers of yours, and I am not rude.
The correct answer and explanation is:
Let’s solve the problem step-by-step.
Problem:
Find the least number that should be added to 1000 so that the sum is exactly divisible by 53.
Step 1: Understand the problem
We want to find the smallest number xx such that: 1000+x≡0(mod53)1000 + x \equiv 0 \pmod{53}
In other words, when 1000+x1000 + x is divided by 53, the remainder is 0.
Step 2: Find the remainder when 1000 is divided by 53
Divide 1000 by 53:
- Calculate 53×18=95453 \times 18 = 954
- Calculate remainder: 1000−954=461000 – 954 = 46
So, 1000≡46(mod53)1000 \equiv 46 \pmod{53}
Step 3: Find xx such that
(1000+x)≡0(mod53)(1000 + x) \equiv 0 \pmod{53}
Substitute the remainder: 46+x≡0(mod53)46 + x \equiv 0 \pmod{53}
or x≡−46≡53−46=7(mod53)x \equiv -46 \equiv 53 – 46 = 7 \pmod{53}
Step 4: Conclusion
The smallest positive number xx that can be added to 1000 to make it divisible by 53 is 7.
Final answer:
7\boxed{7}
Explanation (300 words):
This problem is a classic example of modular arithmetic, which deals with remainders after division. The goal is to find the smallest positive integer xx such that when it is added to 1000, the total becomes divisible by 53.
The key is to understand the relationship between division and remainders. When 1000 is divided by 53, it leaves a remainder. Since 53 multiplied by 18 equals 954, subtracting this from 1000 gives a remainder of 46. This means: 1000=53×18+461000 = 53 \times 18 + 46
For the number 1000+x1000 + x to be divisible by 53, the remainder must be zero when dividing by 53. Using modular arithmetic notation, this is expressed as: (1000+x)≡0(mod53)(1000 + x) \equiv 0 \pmod{53}
Since 1000≡46(mod53)1000 \equiv 46 \pmod{53}, this equation becomes: 46+x≡0(mod53)46 + x \equiv 0 \pmod{53}
To find xx, we need a number that when added to 46 gives a multiple of 53. This is equivalent to: x≡−46≡53−46=7(mod53)x \equiv -46 \equiv 53 – 46 = 7 \pmod{53}
Since we’re looking for the smallest positive integer, the answer is simply 7.
Thus, adding 7 to 1000 yields 1007, which divides exactly by 53: 1007÷53=191007 \div 53 = 19
with no remainder.
This method of solving such problems applies widely in number theory, coding algorithms (such as hashing), and cryptography. The modular arithmetic approach is both elegant and efficient in finding remainders and required adjustments to reach multiples of a divisor.