# Proof that every number ≥ $8$ can be represented by a sum of fives and threes.

Can you check if my proof is right?

Theorem. $\forall x\geq8, x$ can be represented by $5a + 3b$ where $a,b \in \mathbb{N}$.

Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\ x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\ x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$

Inductive step:

$n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\ b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\ c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\ \\ S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$

Basis stays true, because $8,9,10 \in S$

Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.

If $x \in a_n$ then $x+1 \in b_x$,

If $x \in b_x$ then $x+1 \in c_x$,

If $x \in c_x$ then $x+1 \in a_x$.

• This seems to me a good formulation, even though it is not the usual one. (Maybe it is better!) It requires only the proofs of the last three statements, for which you can get a good hint by looking at how you change $m$ and $n$ in the sum $x=3m+5n$ when you transition from $x=a_1$ to $b_1$, from $b_1$ to $c_1$, and from $c_1$ to $a_2$. Mar 8 '15 at 20:16
• This is the well-known Frobenius Theorem/Chicken McNuggets Theorem/Coin problem/etc
– smci
Mar 9 '15 at 10:26
• do you need to use induction? Mar 9 '15 at 12:31

Proof by induction.
For the base case $n=8$ we have $8=5+3$.
Suppose that the statement holds for $k$ where $k\gt 8$. We show that it holds for $k+1$.

There are two cases.

1) $k$ has a $5$ as a summand in its representation.

2) $k$ has no $5$ as a summand in its representation.

For case 1, we delete "that $5$" in the sum representation of $k$ and replace it by two "$3$"s ! This proves the statement for $k+1$.

For case 2, since $k\gt 8$, then $k$ has at least three "$3$"s in its sum representation. We remove these three $3$'s and replace them by two fives! We obtain a sum representation for $k+1$. This completes the proof.

• Very nice -- best proof in this thread, in my opinion. Mar 8 '15 at 19:57
• Very pretty, well done for using one base case only. Mar 8 '15 at 20:30
• It'd be really neat if there was an animation of this proof! Mar 9 '15 at 6:12
• You also need to prove that the sum representation of every number has at least three 3s or at least one 5 to ensure at least one of the cases always applies. This is not difficult. 8 is 3 + 5, case 1, which becomes 3 + 3 + 3 = 9. By case 2, 9 becomes 5 + 5 = 10. From then on, the cycle of replacing always removes one 5 and adds two 3s, removes one 5 and adds two 3s, then removes three 3s and adds two 5s. Thus, all n ≥ 8 congruent to 0 (mod 3) contain three or more 3s; all n ≥ 8 congruent to 1 (mod 3) contain two 5s; and all n ≥ 8 congruent to 2 (mod 3) contain one 5. Mar 9 '15 at 8:03
• @ClementCherlin You don't need any complicated argument to see that every number $\ge8$ with a sum representation has at least three $3$'s or one $5$. If both are false then it contains no $5$'s and at most two $3$'s; thus it is at most $6$. Mar 9 '15 at 11:12

You are almost done, actually you can prove it even without induction:

If $\forall x \ge 8$ and you must prove that $x \in \mathbb{N} \land x = 5a + 3b$

Let $x_1$ be 8 and consider your base cases:

$x_1 = 8 = 3 ⋅ 1 + 5 ⋅ 1$

$x_2 = 9 = 3⋅3 + 5 ⋅ 0$

$x_3 = 10 = 3⋅0 +5⋅2$

This is true for the first 3 $n$ (1, 2 and 3) of $x_n$.

Now, consider the following:

$x_4 = 11 = x_1 + 3$

$x_5 = 12 = x_2 + 3$

$x_6 = 13 = x_3 + 3$

$x_7 = 14 = x_4 + 3$

$x_8 = 15 = x_5 + 3$

$...$

$x_n = x_{n-3} + 3 \,\,\,\,\,\,\forall n \gt 3$

As you can see from the pattern, it's obvious that you are summing only multiples of 3 basing on your previous results (which were also sums of multiples of threes and multiples of fives).

And you may also note that as soon as you sum 5 threes ($3+3+3+3+3 = 15 = 3⋅5$) you can safely replace them with 3 fives ($5 + 5 + 5 = 15 = 3⋅5$)

The previous equation leads to a recurrence relation and you can say with certainty that your statement is true $\forall x \ge 8$

• I upvoted this. I like the idea that you only focus on 3 !
– mick
Dec 28 '17 at 21:36

I would avoid induction and use the elementary Euclidean division algorithm (Eda).

Let $n\geq8$ be an integer. Then, by the Eda, there exist integer $q$ and $r$ such that $r\in\{0,1,2\}$ and $$n=3q+r.$$

• If $r=0$, we are done since $n=3q$.

• If $r=1$, then $q\geq3$ (because $n\geq8$). Hence $n=3(q-3)+10$.

• If $r=2$, then $q\geq2$ (because $n\geq8$). Hence $n=3(q-1)+5$.

• This is very nice and deserves more up-votes. The inductive proof that has 100+ votes is also nice, but there is something to be said for directness. Mar 16 '15 at 16:48
• I upvoted this. To adress Silvio's comment ; I notice that the induction and this proof are very similar ... ironically an argument both ways imho : a reason to upvote this or a reason not to.
– mick
Dec 28 '17 at 21:34

Write $x=8n+k$ for $0\leq k<8$ and $n\geq 1$. Because $8n=(3+5)n$, the problem is clear if $k=0,3,5$. Let's consider the remaining 5 cases: \begin{align*} k=1 & \implies 8n+1=5(n-1)+3(n+2),\\ k=2 & \implies 8n+2=5(n+1)+3(n-1),\\ k=4 & \implies 8n+4=5(n-1)+3(n+3),\\ k=6 & \implies 8n+6=5n+3(n+2),\\ k=7 & \implies 8n+7=5(n+2)+3(n-1). \end{align*} Note that $k=4$ is based on $k=1$ (just add one more $3$) and, similarly, $k=6$ and $k=7$ follow from $k=1$ and $k=2$, respectively. So in fact, we have really only considered 2 cases.

This is some mixture of Bezout's Lemma and Frobenius coin problem. Frobenius coin problem claims that if $a_1$ and $a_2$ are coprime numbers then every number bigger or equal to $(a_1-1)(a_2-1)$ can be written in the form $a_1x + a_2y$ for some non-negative integers $x,y$.

Write $a_1=3$ and $a_2=5$ in you case and you've got the answer. Anyway a inductive proof wouldn't be much harder.

• Of course, using the Frobenius coin problem / lemma doesn't avoid the inductive proof, it just hides it in the proof of the Frobenius coin lemma. Mar 8 '15 at 22:31
• The $(a_1-1)(a_2-1)$ result is due to J.J. Sylvester.
– MJD
Mar 9 '15 at 12:50

1.if the integer can be written as $3s$, then set $a=0$

2.if the integer can be written as $3s+1$, set $a=2$, $10+3b=3s+1$,b=s-3

3.if the integer can be written as $3s+2$, set $a=1$, $b=s-1$

I am sure you have got the idea.

Your base case work is fine.

For the induction, you can assume (1) $k\ge 10$ and (2) all cases from $n=8$ to $n=k$ hold.

Then it is simply a matter of noting that we have $k-2=5a+3b$ and so we can say that $k+1 = 5a+3(b+1)$ as required.

• This is the best answer to the original question. Mar 9 '15 at 17:58

Really late to the party, but here is a very short and sweet approach:

Claim: For each integer $n\geq 8$, let $S(n)$ be the statement that $n$ is expressible as a sum of $3$'s and $5$'s.

Base step: Since $8=3+5, 9=3+3+3, 10=5+5, S(8), S(9)$, and $S(10)$ are true.

Inductive step ($S(k)\to S(k+3)$): Let $k\geq 8$ and assume that $S(k)$ is true, where for some non-negative integers, $a,b,k = a\cdot 3+b\cdot 5$. Then $$k+3=(a+1)\cdot 3+b\cdot 5,$$ a sum of $3$'s and $5$'s, so $S(k+3)$ is true.

By mathematical induction, three cases of induction actually, for all $n\geq 8, S(n)$ is true. $\blacksquare$

Any number is of the form $5k+1$, $5k+2$, $5k+3$, $5k+4$ and $5k+5$.

If $n=5k+3$ then nothing to prove.

$n=5k+1 \implies n=5(k-1)+2\cdot3$

$n=5k+2 \implies n=5(k-2)+4\cdot3$

$n=5k+4\implies n=5(k-4)+8\cdot3$

$n=5k+5\implies n=5(k-2)+5\cdot 3$

Hence for all $k\ge5$ the result holds.

Hence for all $n\ge 25$ the result hold.

What is left is to check the remaining cases and that is also true.

Hence the proposition is true.

Let $P(m)$ be the proposition that $k$ can be represented by a sum of fives and threes for all $m-2 \le k \le m$. $P(10)$ is true, since $8=5+3$, $9=3+3+3$, and $10=5+5$. This is the base case. And if $P(m)$ is true, then so is $P(m+1)$, since (applying the inductive hypothesis at the *): $$m+1=(m-2)+3 =^*(3a+5b)+3=3(a+1)+5b.$$ By induction on $m$, then, $P(m)$ is true for all $m\ge 10$; hence $k$ can be represented by a sum of fives and threes for all $k \ge 8$.

Starting with 8, let's investigate the pattern. 5*1 + 3*1 = 8, 5*0 + 3*3 = 9, 5*2 + 3*0 = 10, you can keep writing this out for it to be easier to see. In general you can increase every number by 1 simply by increasing the previous integer by 6 increase the 3*2 and decreasing by 5 (5*1) hence increasing the integer by 1 (6-5). In the case that your previous value of 5 was 0 and you can't decrease, you will increase the 5 twice (5*2=10) and decrease the 3 3 times (3*3=9) the difference is 1. Hence, every number can be generated this way.

• Welcome to MSE. Please edit and use MathJax to properly format math expressions. Jul 27 '19 at 14:55
• @LeeDavidChungLin sorry, I apologize. I need to learn how to use it. Is my proof correct? Jul 27 '19 at 15:08

Lets call $T$ every multiple of three: $T$ satisfies the condition. Numbers are $T, T-1,$ or $T-2$.

for $n > 3\cdot 2+1$:

$$n=T \implies n=3k$$ $$n=T-1 \implies n= 3\cdot 2 - 1 + 3(k-2) = 5 + 3(k-2)$$ $$n=T-2 \implies n= 3\cdot 4 - 2 + 3(k-4) = 10 + 3(k-2)$$

I'm attempting to answer the question intuitively:

Every third number is a multiple of 3, and can definitely be written as a sum of 3s and 5s: $3x$ + $5$ x $0$.

Now we need to deal with numbers $y$ for which $y mod 3 = 1$ and $y mod 3 = 2$

For $y mod 3 = 2$: We know that $5 mod 3 = 2$, so 5 x 1 + 3$x$ should give us all numbers $y$ where $y mod 3 =2$ For $y mod 3 = 1$: We know that $5$ x $2 mod 3 = 1$, so 5 x 2 + 3$x$ should give us all numbers $y$ where $y mod 3 =1$

The smallest case in which we can use this proof is 5 x 2 + 3 x 0 = 10

We can show it is possible for 8 and 9 as 5+3 =8 and 3 x 3 =9

Hence, we can show that is possible for all numbers greater than or equal to 8.

This is also proves that the maximum multiplier we would need for 5 is 2.