Instantly share code, notes, and snippets.

# requinix/OQrmDNv.md Created Oct 28, 2015

Why does {26& divided by %2# equal #?

The image shows some long division where {26& ÷ %2# = # with no remainder. Put another way, # × %2# = {26&.

The problem is to find the numeric values of the four symbols used: # % { &. Each is a single digit and presumably not one of 2, 6, or 0 (which were written normally), nor is a digit reused with multiple symbols.

The most obvious symbol to start with is #. # × # = & with maybe something carried into the ten's place. Of the 10 digits, 7 are available to try:

• 1 × 1 = 1 (#=1, &=1) - no: it reuses the 1 for two symbols
• 3 × 3 = 9 (#=3, &=9) - possible
• 4 × 4 = 16 (#=4, &=6) - no: 6 is not available
• 5 × 5 = 25 (#=5, &=5) - no: it reuses the 5 for two symbols
• 7 × 7 = 49 (#=7, &=9) - possible
• 8 × 8 = 64 (#=8, &=4) - possible
• 9 × 9 = 81 (#=9, &=1) - possible

So # is one of [3, 7, 8, 9] and & is one of [1, 4, 9].

Now let's try to look at # × 2# = 6& with maybe something carried into the hundreds place. Consider the four values for # and &:

• 3 × 23 = 69 (#=3, &=9)
• 7 × 27 = 189 (#=7, &=9)
• 8 × 28 = 224 (#=8, &=4)
• 9 × 29 = 261 (#=9, &=1)

Only two of those have a 6 in the ten's place: (#=3, &=9) and (#=9, &=1).

Now try # × %2# = {26&. The new % symbol only means a little bit more effort.

For (#=3, &=9) we have 3 × 23 = 69 with no carry. That means # × % = {2. There are now five remaining possible values for %:

• 3 × 1 = 3 (%=1)
• 3 × 4 = 12 (%=4)
• 3 × 5 = 15 (%=5)
• 3 × 7 = 21 (%=7)
• 3 × 8 = 24 (%=8)

There is only one that has a 2 in the one's place, leaving us with one possible solution: (#=3, &=9, %=4, {=1)

The other possibility for # and & is where (#=9, &=1), and 9 × 29 = 261. This time there is a carry of 2 so that means # × % + 2 = {2, or you can safely subtract the 2 to get a simpler # × % = {0. There are five different possible values this time:

• 9 × 3 = 27 (%=3)
• 9 × 4 = 36 (%=4)
• 9 × 5 = 45 (%=5)
• 9 × 7 = 63 (%=7)
• 9 × 8 = 72 (%=8)

None of these have a 0 in the one's place.

So we're only left with the one solution we found before: (#=3, &=9, %=4, {=1). Trying these values out in the original equation, # × %2# = {26& becomes 3 × 423 = 1269 which is correct.