N is a four-digit positive integer. Dividing N by 9 , the remainder is 5. Dividing N by 7, the remainder is 3. Dividing N by 5, the remainder is 4. What is the smallest possible value of N?
n is a four-digit positive integer.
Dividing n by 9, the remainder is 5.
Dividing n by 7, the remainder is 3.
Dividing n by 5, the remainder is 4.
What is the smallest possible value of n?
n≡5(mod9)n≡3(mod7)n≡4(mod5)Let m=9∗7∗5=315
Because 9 and 7 and 5 are relatively prim (gcd(9,7,5)=1),
we can go on.
n=5∗7∗5∗17∗5(mod9)+3∗9∗5∗19∗5(mod7)+4∗9∗7∗19∗7(mod5)+315k|k∈Zn=175∗(135(mod9))+135∗(145(mod7))+252∗(163(mod5))+315k135(mod9)|35≡−1(mod9)≡1−1(mod9)135(mod9)≡−1(mod9)145(mod7)|45≡3(mod7)≡13(mod7)≡3ϕ(7)−1(mod7)≡36−1(mod7)≡35(mod7)≡243(mod7)145(mod7)≡5(mod7)163(mod5)|63≡3(mod5)≡13(mod5)≡3ϕ(5)−1(mod5)≡34−1(mod5)≡33(mod5)≡27(mod5)163(mod5)≡2(mod5)
n=175∗(135(mod9))+135∗(145(mod7))+252∗(163(mod5))+315kn=175∗(−1)+135∗5+252∗2+315kn=−175+675+504+315kn=1004+315k|1004≡59(mod315)n=59+315knmin.=59