Another Sequence - Teaser 8

What's the next letter in this series?

   H, H, L, B, B, C, N, O, F, N

Hint

1834 — 1907

Answer

   S

Explanation

These are the first letters of the English names of the first few elements (based on their atomic number): Hydrogen, Helium, Lithium, Beryllium, Boron, Carbon, Nitrogen, Oxygen, Fluorine, Neon ... and, of course, Sodium.  Dmitri Mendeleev, born on 27 January 1834, published his Periodic Table of Elements in 1860.

How Many Presents from my True Love?

On the first day of Christmas my true love gave to me a partridge in a pear tree.
On the second day of Christmas my true love gave to me two turtle doves and a partridge in a pear tree.
...
On the twelfth day of Christmas my true love gave to me twelve drummers drumming, eleven pipers piping, ten lords-a-leaping, nine ladies dancing, eight maids-a-milking, seven swans-a-swimming, six geese-a-laying, five golden rings, four calling birds, three french hens, two turtle doves and a partridge in a pear tree.

So how many presents did I receive all together?

This can be solved using the differences method, which is nicely explained by Ken Ward.

In the table below, n represents the day of Christmas (starting from 0, the day before I start receiving presents), p(n) represents the number of presents I receive on that day, P(n) represents the total number of presents received so far, δ¹ is the first difference (the difference between successive terms of P(n) — which is p(n) of course), δ² is the second difference (that between successive terms of δ¹) and δ³ is the third difference.

n	0	1	2	3	4	5	6	...
p(n)	0	1	3	6	10	15	21	...
P(n)	0	1	4	10	20	35	46	...
δ1		1	3	6	10	15	21	...
δ2			2	3	4	5	6	...
δ3				1	1	1	1	...

As the δ³ values are all the same, the equation we're after is a cubic polynomial.  That is, it's of the form:

[1]P(n) = an³ + bn² + cn + d

Now we can do some substitution.  We can see immediately that

P(0) = 0a + 0b + 0c + d = 0 ⇒ d = 0

And now, from the next three terms, we can create 3 simultaneous equations:

[2]P(1) = a + b + c = 1
[3]P(2) = 8a + 4b + 2c = 4
[4]P(3) = 27a + 9b + 3c = 10

We can eliminate c, first from [2] and [3] and then from [2] and [4] to produce:

[5]6a + 2b = 2
[6]24a + 6b = 7

From which we can deduce that (24 – 18a) = 1 ⇒ 6a = 1, or a = ¹⁄₆.

Substituting that back in [5], we find that 1 + 2b = 2 ⇒ 2b = 1, or b = ¹⁄₂.

Finally we can use [2] to determine that ¹⁄₆ + ¹⁄₂ + c = 1, or that c = ¹⁄₃.

Now we can substitute these values back in [1] and tidy up a bit:

P(n)	= 1⁄6n3 + 1⁄2n2 + 1⁄3n
	= 1⁄6n(n2 + 3n + 2)
	= 1⁄6n(n + 1)(n + 2)

So, by the 12^th day of Christmas I will have received ¹⁄₆.12(12 + 1)(12 + 2) or 2.13.14 = 364 presents. Not bad going.

Teaser 5

What is the next digit in this sequence?

   1, 1, 1, 1, 5, 1, 7, 2, 3, 5

Hint

There's a corresponding sequence: 12, 6, 4, 3, 12, 2, 12, 3, 4, 6

Answer

   11

Explanation

Start with 1/12 (one-twelfth), and then keep adding 1/12. You end up with the sequence 1/12, 2/12, 3/12, 4/12, …. Next reduce each term to its simplest form. This results in: 1/12, 1/6, 1/4, 1/3, 5/12, 1/2, 7/12, …. The numerators from these fractions give you the original sequence.

Teaser 4

What is the next letter in this sequence?

O, T, T, F, F, S, ...

Hint

Try initials

Answer

S

Explanation

One, Two, Three, Four, Five, Six, Seven. Easy really!

Teaser 3

What is the next number in this sequence?

4, 2, 3, 4, 6, 2, 4, ...

Hint

What is the next digit in this sequence?

4, 2, 3, 4, 5, 2, 4, ...

Answer

8

Explanation

What has 4 letters, is has 2, the has 3, etc.