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.
Please help me to solve this.


2 Answers 2


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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – fleablood
    Aug 5, 2019 at 17:48
  • $\begingroup$ You are correct @fleablood. I'll edit the answer to address this. $\endgroup$
    – marcelgoh
    Aug 5, 2019 at 17:52
  • $\begingroup$ 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. $\endgroup$
    – fleablood
    Aug 5, 2019 at 17:57
  • $\begingroup$ Hmmm, so I guess the answer is still flawed. In any case I think your solution is better. $\endgroup$
    – marcelgoh
    Aug 5, 2019 at 18:02
  • 1
    $\begingroup$ 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. $\endgroup$
    – fleablood
    Aug 5, 2019 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.


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