All your last digit are belong to us

Logic Level 1

$\large{\text{1 \$\square\$ 2 \$\square\$ 3 \$\square\$ \$\ldots\$ \$\square\$ 100}}$

Each square in the above expression is replaced by either a $+$ or a $\times$ . For each digit $0,1,2,\ldots,9$ , can it appear as the last digit of the resulting expression?

As an explicit example, replacing everything with $+$ gives the result $1+2+\ldots+100 = 5050$ , so $0$ can be the last digit; the question is whether all ten digits can appear by substituting the squares appropriately (of course different substitution for different digits).

Yes No

3 solutions

Chew-Seong Cheong
Nov 5, 2015

Consider $S_0 = 1 \times 2 \times 3 \times ... \times 9 + 10 + 11 \times 12 \times 13 \times ... \times 100 = 9! + 10 + \dfrac{100!}{10!} = ...0+10+...0 = ...\color{#3D99F6}{0}$ . We note that $9!$ ends with $0$ , $10$ ends with $0$ and $\frac{100!}{10!}$ also ends with $0$ , therefore, $S_0$ ends with $0$ .

Similarly,

$\begin{array} {} \small S_1 = 1 \times 2 \times 3 \times ... \times 10 + 11 + 12 \times 13 \times 14 \times ... \times 100 = 10! + 11 + \dfrac{100!}{11!} = ...0+11+...0 = ...\color{blue}{1} & \small \text{ends with } \color{blue}{1} \\ \small S_2 = 1 \times 2 \times 3 \times ... \times 11 + 12 + 13 \times 14 \times 15 \times ... \times 100 = 11! + 12 + \dfrac{100!}{12!} = ...0+12+...0 = ...\color{blue}{2} & \small \text{ends with } \color{blue}{2} \\ \small S_3 = 1 \times 2 \times 3 \times ... \times 12 + 13 + 14 \times 15 \times 16 \times ... \times 100 = 12! + 13 + \dfrac{100!}{13!}= ...0+13+...0 = ...\color{blue}{3} & \small \text{ends with } \color{blue}{3} \\ ... & ... \\ \small S_9 = 1 \times 2 \times 3 \times ... \times 18 + 19 + 20 \times 21 \times 22 \times ... \times 100 = 18! + 19 + \dfrac{100!}{19!} = ...0+19+...0 = ...\color{blue}{9} & \small \text{ends with } \color{blue}{9} \end{array} $

Therefore, the answer is $\boxed{\text{Yes}}$ .

Moderator note:

Oh, that's a nice simple construction!

Ivan Koswara
Nov 5, 2015

The answer is yes.

The trick is that $\ldots 1 \square \ldots 2$ gives a last digit of $3$ if it's a $+$ and a last digit of $2$ if it's a $\times$ . Thus we can decrease the last digit by one by simply changing such $+$ to $\times$ .

Our entire expression will consist entirely of $+$ 's, except for some pairs of $\ldots 1 \square \ldots 2$ that are filled with $\times$ instead. For example,

$1 + 2 + 3 + \ldots + 100 = 5050$
$1 \times 2 + 3 + \ldots + 100 = 5049$
$1 \times 2 + 3 + \ldots + 10 + 11 \times 12 + 13 + \ldots + 100 = 5158$
$1 \times 2 + 3 + \ldots + 10 + 11 \times 12 + 13 + \ldots + 20 + 21 \times 22 + 23 + \ldots + 100 = 5577$

The last digit decreases by one for each $\times$ we insert. Since there are enough pairs ( $(1,2), (11,12), (21,22), \ldots, (91,92)$ ), we can obtain all last digits.

Moderator note:

Nice observation of how to reduce the units digit by 1 each time.

Lu Chee Ket
Dec 9, 2015

Just like having many coins that we can get all specific figures wanted, a change of $+$ into $\times$ produces 1, 5 or 9 considered not near for each pick:

3   2   -1
5   6   1
7   12  5
9   20  11
11  30  19
13  42  29
15  56  41
17  72  55
...

187 8742    8555
189 8930    8741
191 9120    8929
193 9312    9119
195 9506    9311
197 9702    9505
199 9900    9701

Non side by side with effective many 1, 5 and 9 are there to sufficiently able to make wanted digit of 1 to 9 from 0, and it becomes obvious and certain particularly when only one resultant digit is wanted at once.

Answer: $\boxed{Yes}$

All your last digit are belong to us

3 solutions

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