Show that an integer n can be written as a sum of 5’s and/or 7’s with no regard to order for any n>=24.?

Take n ≧ 24, where n - an integer
Then n=24,25,26,...
i.e. n=24a+b, where a,b -- an integer, a≧1, b≧0
n=(5+7)2a +(5x3-7x2)b
n=5(2a+3b)+7(2a-2b)
∴ n=5p+7q, where p=2a+3b, q=2a-2b

In particular：
i) When a=b, q=2a-2b=0 => n=5p ；i.e. n is written as a sum of 5’s
ii) If n=23,
then n= 5(0)+23 = 5(1)+18 = 5(2)+13 = 5(3)+8 =5(4)+3
But none of 23,18,13, 8, 3 can be written as a multiple of 7 (i.e. 7b)
∴ 23 cannot be written as 5a+7b

Conclusion : n can be written as a sum of 5’s and/or, where n≧24


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Note :
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(a) " n written as a sum of 5’s and/or 7’s " 的意思是 : -
把 n 寫成 :
i) " a sum of 5’s and 7’s " : n = 5p +7q , where p,q - an integer , p≠0 , q≠0 ; 或
ii) " a sum of 5’s or 7’s " : n = 5p , or n = 7q

(b) 這類數並不需要什麼特別公式 , 主要是學 手法 (mathematical skill) ;
試多幾題就會「上手」!

24 = 2(5) + 2(7) , 34 = 2(5) + 24 , 44 = 2(5) + 34 , ...
25 = 5(5) , 35 = 2(5) + 25 , 45 = 2(5) + 35 , ...
26 = 5 + 3(7) , 36 = 2(5) + 26 , 46 = 2(5) + 36 , ...
27 = 4(5) + 7 , 37 = 2(5) + 27 , 47 = 2(5) + 37 , ...
28 = 4(7) , 38 = 2(5) + 28 , 48 = 2(5) + 38 , ...
29 = 3(5) + 2(7) , 39 = 2(5) + 29 , 49 = 2(5) + 39 , ...
30 = 6(5) , 40 = 2(5) + 30 , 50 = 2(5) + 40 , ...
31 = 2(5) + 3(7) , 41 = 2(5) + 31 , 51 = 2(5) + 41 , ...
32 = 5(5) + 7 , 42 = 2(5) + 32 , 52 = 2(5) + 42 , ...
33 = 5 + 4(7) , 43 = 2(5) + 33 , 53 = 2(5) + 43 , ...