3×13=39; 156_4=39; 45-6=39; 12+27=39
In 1978—that is, let me do the math—40 years ago—education was the theme of the Children's Stamps. Alongside a child selling stamps, the stamps designed by Babs van Wely featured a reading, a writing, and a calculating child. I really wanted to talk about that calculating child. Because it turns out: arithmetic in primary school is doing very poorly.
I didn't come up with that myself; no, that is the result of a so-called PPON, a Periodic Survey of Educational Level. An apparently alarming result, otherwise today's de Volkskrant wouldn't have opened its front page with it.
Addition, subtraction, and especially multiplication and division—hordes of primary school children struggle with them. What can be done about it? No, even the PPON team doesn't have an immediate solution. Would it perhaps be an idea to dedicate a series of stamps to arithmetic?! As far as I'm concerned, they don't necessarily have to be children's stamps, as long as they are stamps that encourage arithmetic. There are plenty of possibilities! Just calculate how many sums can be invented that add up to 39. Or do you think that stamps aren't nearly 'cool' or 'awesome' enough to reach today's primary school youth?!
This is an edited version of a previously published article on Postzegelblog. 
Is it possible to use the digits 1 through 9 once, and use addition, subtraction, multiplication, and division to arrive at 39?
'@ His
Of course, and in a particularly simple way. Undoubtedly there are many other possibilities, but the numbers below are also almost in order.
8×7=56; 56-9=47; 47-6=41; 41-5=36; 36-4=32; 32+(3×2)=38; 38+1=39.
The account is good !
Well done, Patrick.
Handing out stamps in class........a good idea! We might even end up with a few regular collectors.
Another elegant solution to reach 39:
1+2-3+4+5+6+7+8+9=39
Is zero allowed too? The 3 and the 9 are already in the result, so I won't use them.
1 × 2 × 4 × 5 + 6 – 7 + 8 × 0 = 39. If I do have to use that:
(1 + 2 + 3 + 4 + 5) × (8 – 6) + 7 × 0 + 9 = 39.
(68 + 5 – 23) × 4 – 90 – 71= 39
The zero makes it much easier. Without zero:
Hans Kremer's solution (above) can be done with or without zero in different ways:
0 -1 – 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 39.