☆犬☆ 發問時間： 科學數學 · 4 年前

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.?

2 個解答

• SC147
Lv 6
4 年前
最佳解答

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

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Note :

~~~~

(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」!

• 4 年前

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 , ...