# Doing addition and multiplication between the numbers of the set $S$ to find $100$ and $75$.

Consider the set $$S=\{0.7,1.4,3.5,7,14,17.5,35,52.5,70\}.$$
Now play with these numbers by doing addition and multiplication and we have to get $$100$$ and $$75$$. Subtraction and division are not legal. Like: $$14 \times 7 =98$$, but I cannot find $$2$$ to add with $$98$$ to get $$100$$. Similarly I cannot get $$75$$ also only by multiplication between these numbers and addition between them.

I don't believe this is possible. Rewrite the question, multiplying everything by $$10$$, and we get $$S = \{7, 14, 35, 70, 140, 175, 350, 525, 700\}$$. Now we're trying to add and multiply to get $$1000$$ and $$750$$. Every integer in $$S$$ is divisible by $$7$$. So adding or multiplying any two of them results in another multiple of $$7$$. Neither $$1000$$ nor $$750$$ are divisible by $$7$$, so they cannot be obtained in this way.

EDIT: As fleablood mentions below, my answer is only valid once we show that $$7$$ divides $$m$$ if and only if $$7$$ divides $$10^k\cdot m$$, and that if $$7$$ divides both $$10m$$ and $$10n$$, then $$7$$ divides $$10^j\cdot m + 10^k\cdot n$$. Then powers of $$10$$ can be safely ignored when they arise in combining multiplications and additions.

The first statement is immediate from the fact that $$10^k$$ factorises into primes $$(2\cdot 5)^k$$, meaning that $$10^k\cdot m$$ is divisible by $$7$$ when and only when $$m$$ is.

Now we prove the second statement. Suppose $$7$$ divides both $$10m$$ and $$10n$$. Then $$10^j\cdot m + 10^k\cdot n = 10^{j-1}(7r) + 10^{k-1}(7s)$$ for some integers $$r$$ and $$s$$. Rewriting this as $$7\cdot(10^{j-1}r + 10^{k-1}s)$$, it becomes clear that the sum was divisible by $$7$$.

But fleablood's full answer is quite a bit better than mine. Working with all the elements multiplied by $$10$$ introduces unneeded complexity.

• Slight quibble. multiplying $10a*10b$ will not give us $10(a*b)$ so the results of what we play around with in the new $S$ don't correspond to the results if we played around in the old $S$. But what we get will be a power of $10$ times are result (we'll need to show that explicitly) so we can't just show $1000$ and $750$ are impossible but $100*10^k$ and $75*10^k$ are all impossible. – fleablood Aug 5 '19 at 17:48
• You are correct @fleablood. I'll edit the answer to address this. – marcelgoh Aug 5 '19 at 17:52
• It's actually more serious than I thought. $a*b = ab$ and $c + ab = c+ab$. If we multiply everything by $10$ we get $10a*10b = 100ab$ and $10c + 100ab = 10(c + 10ab)$. The corelation gets weaker. Now if $7\not \mid 10a,10b,10c$ we can prove $7\not \mid 10(c+10ab)$ but... it may not be so smooth. – fleablood Aug 5 '19 at 17:57
• Hmmm, so I guess the answer is still flawed. In any case I think your solution is better. – marcelgoh Aug 5 '19 at 18:02
• Your answer is still good. A thing we just need to state an inductive steps $7|10^k m \iff 7|m$ and if $7|10k$ and $7|10j$ then $7|10^l j + 10^m k$. so the powers of $10$ won't matter. – fleablood Aug 5 '19 at 18:29

All these numbers are an integer multiples of $$0.7$$. So any sum or product will be an integer multiple $$0.7^k$$ for some positive integer $$k$$.

$$75$$ and $$100$$ are not such numbers.

If $$75= m*(.7)^k$$ then $$75*10^k = m*7^k$$ but $$7$$ is prime and divides neither $$75$$ or $$10$$ so it can not divide $$75*10^k$$.

Similarly if $$100 = n*(.7)^k$$ then we'd have $$100*10^k=n*7^k$$ but $$7|100*10^k$$ is impossible.