Finding the Least Number Greater Than 1000 with Given Remainders

The Problem

We are given a number x that, when divided by N, leaves a remainder of 8. Additionally, one-third of x, when divided by N, leaves a remainder of 31. We need to find the least such number greater than 1000.

Setting Up the Problem

We denote the number as x and the divisor as N. The conditions provided can be expressed as:

x ≡ 8 (mod N) 1/3 ≡ 31 (mod N)

The second condition can be interpreted as follows: for some integer k, the equation 1/3 ≡ 31 (mod N) can be rewritten as 1/3 31 - kN. Multiplying through by 3 to eliminate the fraction, we get 1 93 - 3kN, rearranged to 3kN 92. This implies that N must be a divisor of 92.

The Divisors of 92

The divisors of 92 are: 1, 2, 4, 23, 46, and 92. Given thatx ≡ 8 (mod N), N must also satisfy N > 8. Therefore, the possible values for N are 23, 46, and 92.

Checking Each Value of N

N 23

In this case, the general form is x 23k - 8. Setting x ≤ 1000, we solve for k: 23k - 8 ≤ 1000 23k ≤ 1008 k ≤ 43.8 Therefore, the smallest integer value for k is 44. Calculating x: x 23·44 - 8 1016 - 8 1016

N 46

The general form is x 46k - 8. Setting x ≤ 1000, we solve for k: 46k - 8 ≤ 1000 46k ≤ 1008 k ≤ 21.87 Therefore, the smallest integer value for k is 22. Calculating x: x 46·22 - 8 1012 - 8 1020

N 92

The general form is x 92k - 8. Setting x ≤ 1000, we solve for k: 92k - 8 ≤ 1000 92k ≤ 1008 k ≤ 10.85 Therefore, the smallest integer value for k is 11. Calculating x: x 92·11 - 8 1012 - 8 1020

Conclusion

The least number x ≥ 1000 that satisfies both conditions occurs at:

For N 23, x 1016 For N 46, x 1020 For N 92, x 1020

Thus, the least such number greater than 1000 is: 1016.