What is the smallest number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7 & so on until when divided by 2 leaves a remainder of 1?
the smallest number I got was:
2519
I don't believe there's a smaller one than that (I checked it in Excel). The next one is 5039, which is twice the last plus 1. (Incidentally, the one before 2519 corresponds to N=-1.) To work out which number gives a remainder of 1 for all the integers from 1 to P, work out the least common multiple of all the numbers from 2 to P, and subtract 1. So for P=10, the LCM of 2, 3, 4, 5, 6, 7, 8, 9 and 10 is 2x3x2x5x7x2x3=2520; subtract 1 and you get your 2519.
To find the smallest number that satisfies the given conditions, we can start by finding the least common multiple (LCM) of the divisors.
1. Begin by listing out the divisors from 2 to 10: 2, 3, 4, 5, 6, 7, 8, 9, 10.
2. Find the LCM of these numbers. The LCM is the smallest multiple that all the numbers divide into evenly.
- Start by listing the multiples of the largest number, 10: 10, 20, 30, 40, ...
- Check if any of these multiples satisfy the remainder conditions for all the other divisors.
- If a multiple leaves the correct remainder when divided by a divisor, move on to the next divisor in the list and repeat the process.
- If a multiple does not leave the correct remainder when divided by a divisor, move on to the next multiple and try again.
3. Continue this process until you find the smallest number that satisfies the conditions for all the divisors.
Let's work through an example:
Starting with the largest divisor, 10:
- We need to find a multiple that leaves a remainder of 9 when divided by 10. The closest multiple to this condition is 19 (10 x 1 + 9).
Moving on to the next divisor, 9:
- We need to find a multiple that leaves a remainder of 8 when divided by 9. The closest multiple to this condition is 19 (9 x 2 + 1). Since we already know that dividing 19 by 10 leaves a remainder of 9, it satisfies this condition as well.
Continuing with the remaining divisors:
- Divisor 8: The closest multiple to the condition is 19 (8 x 2 + 3).
- Divisor 7: The closest multiple to the condition is 19 (7 x 2 + 5).
- Divisor 6: The closest multiple to the condition is 19 (6 x 3 + 1).
- Divisor 5: The closest multiple to the condition is 19 (5 x 3 + 4).
- Divisor 4: The closest multiple to the condition is 19 (4 x 4 + 3).
- Divisor 3: The closest multiple to the condition is 19 (3 x 6 + 1).
- Divisor 2: The closest multiple to the condition is 19 (2 x 9 + 1).
Since the number 19 satisfies all the conditions, it is the smallest number that leaves remainders from 1 to 9 when divided by the corresponding divisors.