Given:
A sequence of numbers leaves the remainders 1, 2, and 7 when divided by 2, 3, and 8 respectively.
Formula Used:
Chinese Remainder Theorem for solving simultaneous congruences:
x≡a1modn1
x≡a2modn2
x≡a3modn3
Calculation:
We need to find the smallest positive integer x such that:
x≡1mod2
x≡2mod3
x≡7mod8
Let's solve the first two congruences:
x≡1mod2⇒x=2k+1 for some integer k.
Substitute in the second congruence:
2k+1≡2mod3
2k≡1mod3
k≡2mod3 (since the multiplicative inverse of 2 mod 3 is 2)
k=3m+2 for some integer m.
Substitute back:
x=2(3m+2)+1
x=6m+5
Now, consider the third congruence:
6m+5≡7mod8
6m≡2mod8
3m≡1mod4 (since 6 ≡ 3 mod 8)
m≡3mod4 (since the multiplicative inverse of 3 mod 4 is 3)
m=4n+3 for some integer n.
Substitute back:
x=6(4n+3)+5
x=24n+18+5
x=24n+23
Thus, the sequence is of the form x=24n+23.
We need the second term in this sequence:
For n=1:
x=24(1)+23
x=47
The second term in the sequence is 47.