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Beelzebub 1358 solution

The last Beelzebub puzzle: grid contains (Richard) WHITELEGG, (Michael) MACDONALD-COOPER, COLIN (Gumbrell) and PAUL (Henderson), the four setters over the years.

Across

1              WHITE; whit e

5              LEGGING; L egging

11           HAIR-EEL; air in heel

12           AINEE; hidden

13           INTERROGATEE; anag.

16           EMPLOY; e + p l o’ in my

17           GEAR UP; ear in gup

19           CAPITANI; a/c (rev.) + in a tip (rev.)

22           ROUNDURE; ro(d) + R in undue

24           PHAEIC; hae(t) in pic

25           COOPER; O in coper

29           AUSTRALIA DAY: a + anag.

30           UREIC; i.e. in cru (all rev.)

31           VILLAIN; villa in

32           LARCHEN; arch in len(t)

33           ADULT; adult(erate)

Down

1              WHITECAP; anag. + pace (rev.)

2              HANGMAN; Han G-man

3              TRELLISED; ill in desert (all rev.)

4              EERY; E(ll)ery Q.

5              LEROY; hidden

6              ELOIGN; Elo + anag.

7              GAG REIN; E in Gag(a)rin

8              INTERRUPT; err up in I t t

9              NEED; ne(Ar)ed

10           GEEK; G(R)eek

14           APPRAISER; app raiser

15           MACDONALD; do in anag. In mad

18           PEER GYNT; anag. In pt

20           TRIARCH; a R in t rich

21           PREVAIL; re v in pail

23           OCTAVE; o + T in cave

25           COLIN; l in coin

26           PAUL; paul(dron)

27           AURA; (L)aura

28           ZILA; z I l A

 

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This is the website of Paul Henderson, who sets crosswords for The Independent (London) under the pseudonyms Phi, for the Daily Telegraph (London) under the pseudonym Kcit, and anonymously for The Times (London) amongst many other outlets. For a more detailed biography see the About Me page.

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