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Find all of these common words within this Boggle grid, each containing the letter E.
Each letter after the first must be a horizontal, vertical, or diagonal neighbor of the one before it.
No individual letter cube may be used more than once in a word.
B - - - E
B E - -
B E - -
C - - E
C - - E
C - - - - - E -
D E - -
D E - -
F - - E
F - - E -
F - - E
G E - - -
H E - -
H E - -
L - E -
L - - E
M - - E
I thought about throwing you in the deep end on this one, but even just the last part of the puzzle is challenging enough, so I’ll lead you up to that point:
We saw from the Prime Balance puzzle (POTW #380) that if we want a number of distinct prime number blocks equally spaced around a circle to have a centre of gravity at the centre of the circle, the least number of blocks is six. This is because if you want distinct rational weights, the number of weights cannot be a prime or a prime power. If we add in a further stipulation that the number of blocks must be odd, the lowest number that is odd but neither a prime nor a prime power is 15.
Using the fact that 15 is 3 x 5 we can set up a system of equations as we did before:
A = V
B = W+T
C = X+U
D = Y
E = Z+T
F = V+U
G = W
H = X+T
I = Y+U
J = Z
K = V+T
L = W+U
M = X
N = Y+T
O = Z+U
Whatever we choose for T to Z the system of blocks will be balanced around the centre of the circle. We still have to ensure that the numbers A to O are all prime and all different. What is the least possible total weight of the 15 blocks?
Without any electronic assistance, find the value of the fraction:
This right-angled triangle is shown with its inscribed circle (incircle). The distances a and b make up the distance from each end of the base to the tangent point of the incircle. c is the height of the triangle, perpendicular to the base.
If I tell you that a, b and c are all whole numbers, and all different, what is the minimum sum of a, b and c?
Three circles of radius 70, 20 and 50 are mutually tangent.
A fourth circle is introduced, which passes through the centre of the 50 circle, and which is also tangent to both the 20 circle and the 70 circle at the point that they are tangent to one another.
What is the radius of this fourth circle?
The answer to each clue is a 3-, 6-, or 9-letter word. If you split these into 3-letter chunks, you will find 20 different ‘triples’ each appearing twice.
For example if one answer was ‘bellow’ then ‘bel’ and ‘low’ would appear somewhere else too, for instance in ‘disbelief’ and ‘lowest’, but then ‘dis’, ‘ief’ and ‘est’ would have to appear somewhere else too, etc, until all triples have appeared exactly twice.
Age
Be present at
Christian religious leader
Condensed water droplets
Endurable
Finish
Imperfections
'In the name of Allah'
Mold
Opinion, recommendation
Outcast
Part of a circle
Part of speech
Part of triple jump
Set upon
Shiite religious leader
Softcover
Suitable for purpose
Tropical fruit
Underneath the ...
Vanilla ..., rapper
You can change one word into another of the same length by a series of steps whereby at each step you change one of the letters and rearrange to form a valid word. This differs from usual word ladders where you aren’t permitted to rearrange the letters.
LEAD and GOLD both contain L and D so would only require two steps to change between them, so just one intermediate word, for instance: LEAD – GLAD – GOLD.
Can you change CARBON into SILVER in only five steps (four intermediate words)?
Can you explain this curious sequence, and why it doesn’t continue after 90?
1
8
2
100
1000000000000000000000000000000000 (33 zeroes)
9
11
19
90
Four lines are drawn from the midpoints of four sides of a square to a point in the interior, dividing the square into four regions. The areas of three of those regions are given. If the bottom left point is at (0,0), what are the coordinates of the interior point?
What is the area of this quadrilateral, drawn in a quarter circle?
Watching the Oscars this year I noticed that one of the nominees for Best Picture was a remake of one of the very first winners almost a century ago. In fact, in four of the last five years, a Best Picture nominee has been a remake of a film to previously be nominated for Best Picture. I trawled the lists and found several more examples for you to try and discover below. Where a film has three or more words in the title, I’ve given the initials; where it has only a single or two-word title, I’ve given an anagram.
Winner 1930, nominated 2022: AQOTWF
Winner 1961, nominated 2021: WSS
Nominated 1933 and 2019: TIME NOT WELL
Nominated 1937 and 2018: ASIB
Nominated 1935 and 2012: SMELLIER BASES
Nominated 1952 and 2001: GERONIMO ULU
Nominated 1943 and 1978: HCW
Nominated 1936 and 1968: RAJ
Nominated 1934 and 1963: ACE PATROL
Winner 1935, nominated 1962: MOTB
Enter the letters E, N, I, G, M and A into the circles at the end of the starfish tentacles, so that:
Each starfish will have each letter once each.
Each straight line (indicated by an arrow) will have each letter once each.
Each triangle (indicated by a bold line) will be surrounded by each letter once each.
A school year is split into three equal classes. Every student either is either a boy or a girl, with the total number of boys and girls being equal.
Class A has twice as many girls as class B, and four times as many boys as class C.
Class C has two more girls than class B has boys.
How many students are there altogether?
One circle is tangent to the inside of another circle as shown. Two diameters of the larger circle pass behind the smaller circle, and the exposed sections of the lines measure 40, 60, 70 and 55 respectively.
What is the area of the shaded region?
I’m writing this before publishing last week’s puzzle, so it’s possible you used a dissection to establish the area already, in which case you have possibly already solved this one too.
Still using the same shape from the last puzzle; we previously discovered that the area was exactly 1. This being the case it ought to be possible to dissect the shape into a number of pieces and reassemble those pieces to form a unit square. For instance, below is a method using five parts, but what is the fewest number of pieces with which you can achieve this?
I have a shape as below: a concave pentagon. Two of the points are the ends of the base of a unit semicircle. Two other points lie on the semicircle; at those points the angles are right angles. Four of the five edges are equal in length, and the fifth length (being the base of the semicircle) is equal to 2.
What is the area of the pentagon?
What is the largest prime number that CANNOT be expressed as the sum of three positive composite odd numbers?
A, B, C and D are four points on a circle. Lines AC and BD cross at E.
AB=10, BC=10, BE=4.
What is the length of ED?
The answer to each clue is a 3-, 6-, or 9-letter word. If you split these into 3-letter chunks, you will find 20 different ‘triples’ each appearing twice.
For example if one answer was ‘bellow’ then ‘bel’ and ‘low’ would appear somewhere else too, for instance in ‘disbelief’ and ‘lowest’, but then ‘dis’, ‘ief’ and ‘est’ would have to appear somewhere else too, etc, until all triples have appeared exactly twice.
100% probability
A group of related objects
A mongrel dog
Bunk or flower?
Chicken leg used in percussion?
Conducted or escorted
Corruptly paid money
Cut short
Dictionary of synonyms
Fourfold
Lots
More repulsive
Old name for Sulphur
Recite or chant in monotone
Reddish brown apple colour
Slang for a police officer
Small plate for a kitten's milk
The definite article
Whimsical
Xerox machine