by Stephen Hogg
Published Friday January 05 2024
Agent J discovered that S.P.A.M.’s IQ booster drug comprises five elements with atomic numbers Z under 92. Some information had been coded as follows: Z1[4;6], Z2[1;4], Z3[4;6], Z4[0;6] and Z5[7;2], where Z5 is lowest and Z3 is highest. Code key: [Remainder after dividing Z by 8; Total number of factors of Z (including 1 and Z)]. The drug’s name is the elements’ chemical symbols concatenated in encoded list order.
J subsequently discovered the Z3 and Z5 values. Now MI6 had just fifteen possible sets of atomic numbers to consider. Finally, J sent a sixth element’s prime Z value (a catalyst in the drug’s production, not part of it). She’d heard it was below Z1 and above Z2, without knowing these. MI6 boffins were now sure of the drug’s name.
Give the catalyst’s atomic number.
Sunday Times Teasers Discussion Group 
The order of the five elements is pretty much given but confirmed when considering the number of different factors in each atomic number. There remains some choice of values, which is confirmed given the single separating prime value for the catalyst.
I’m afraid I guessed that three particular 2-letter elements in a certain order would likely figure in the name.
There is an orderly route to the answer. As Turner suggests, there is a humorous twist if you work out the name of the drug. However it is not necessary to do so.
I found 17 atomic numbers satisfying the the required [remainder, factors].
I am not sure what is meant by “just fifteen possible sets[!] of atomic numbers”.
I would not know how to exclude 2 atomic numbers from my set of 17 ?
” Now MI6 had just fifteen possible sets of atomic numbers to consider.” means that if you know Z3 and Z5 there are 15 possible permutations of (Z1, Z2, Z4). And so it places constraints on Z3 and Z5.
Thanks John,
I will now try to choose Z3 and Z5 in such away that there remain exactly 15 permutations of Z1, Z2, Z4
Simple but fun.
I was initially inclined to choose the smallest possible Z5 and the largest possible Z3 out of my list of 17 atomic numbers.
I inspected Brian’s hidden answer to find that my trial with max Z3 is apparently wrong.
Why I could not find out as yet.
Choosing the smallest possible Z5 and the largest possible Z3 out of your list of 17 atomic numbers leaves more than fifteen possible sets of atomic numbers to consider.
Thanks, John
I shoud have noticed that, having easily checked 15 sets with Brian’s Z3,Z5.