#math trivia for #September28: #271 is a permutation of digits 1, 2, 7. Which other day-numbers (1-365) are a permutation of these digits?

— Burt Kaliski Jr. (@modulomathy) September 29, 2013

There are six possible permutations of any three distinct digits; in this case, the six possibilities are:

- 127
- 172
- 217
- 271
- 712
- 721

However, only the first four are day-numbers, so the answer is 127, 172, 217.

]]>#math trivia for #September21: #264 is the sum of a square and a cube, and also of two squares and a cube. What are they?

— Burt Kaliski Jr. (@modulomathy) September 22, 2013

Answer: The square and cube are 256 = 16^{2} and 8 = 2^{3}.

The two squares and cube are 4 = 2^{2}, 196 = 14^{2}, and 64 = 4^{3}; or 100 = 10^{2}, 100, and 64. (Or, trivially adapting the first answer, 0, 256 and 8.)

(For simplicity, I’ve assumed the cubes are positive; it would be interesting to see if allowing negative cubes — and thus potentially much larger squares — would yield further solutions.)

]]>#math trivia for #June11: #163 has three different digits where two (1,3) divide into the other (6). What’s the next number like this?

— Burt Kaliski Jr. (@modulomathy) June 12, 2012

The next number with three different digits where two divide into the other is 182.

Several numbers in between satisfy the divisibility requirement, but the digits are not distinct: 166, 171, 177, 181.

Because the “other” digit must have two distinct, smaller divisors, the only possible values it can have are 4, 6, 8 and 9.

]]>#math trivia for #September20: #263 has no digits in common with its base-N representation for which N < 10?

— Burt Kaliski Jr. (@modulomathy) September 21, 2013

The base-N representations of 263 for N = 2, …, 9 are:

- 100000111 base 2
- 100202 base 3
- 10013 base 4
- 2023 base 5
- 1115 base 6
- 524 base 7
- 407 base 8
- 322 base 9

The representations that don’t include at least one of the digits in 263 are N = 2, 6 and 8. In addition, if we count unary representation (which is sometimes considered base-1), then the representation for N = 1 doesn’t include one of the digits either. (The unary representation of 263 consists of 263 1’s.)

]]>#math trivia for #September19: #262 is a palindrome that’s twice a palindrome. Which other day-numbers (1-365) have this property?

— Burt Kaliski Jr. (@modulomathy) September 20, 2013

We’re restricted to even palindromes, so our set of candidates (same forward and reverse) is:

- 2, 4, 6, 8
- 22, 44, 66, 88
- 202, 212, 222, 232, 242, 252, 262, 272, 282, 292

Among these, the only ones that are not twice a palindrome are 212, 232, 252, 272 and 292; the carry breaks the symmetry. The others are all twice a palindrome.

]]>#math trivia for #September18: #261 = 29*9; September is the 9th month. Is day 29*M always in month M? What about other multiples of M?

— Burt Kaliski Jr. (@modulomathy) September 19, 2013

Yes, day 29*M is always in month M:

- Day 29: January 29
- Day 58: February 27
- Day 87: March 28
- Day 116: April 26
- Day 145: May 25
- Day 174: June 23
- Day 203: July 22
- Day 232: August 20
- Day 261: September 18
- Day 290: October 17
- Day 319: November 15
- Day 348: December 14

In a leap year, the dates from March onward would be one day later, but would still remain in the month.

What about other multiples of M? We need a multiplier K such that K*M falls in December; if that happens then most of the others should fall into place (though we’ll have to check February because it is shorter than the others).

In a non-leap year, the first day of December is Day 335, which gives a minimum value for K of 28, because 28*12 = 336. The last day of December is Day 365, which gives a maximum value for K of 30.

We know that 29 works. What about 28 and 30?

To check 28, subtract one day per month off each of the dates above. This will work, because the day part of each of the dates is greater than the month part.

To check 30, add one day per month. This will be a problem for February: Day 30*2 = 60 in a non-leap year falls on March 1.

In a leap year, both 28 and 30 would work. Day 30*2, the only questionable case, would fall on February 29.

]]>#math trivia for #September16: #259 has two squares side by side. Do any day-numbers have two squares overlapping?

— Burt Kaliski Jr. (@modulomathy) September 17, 2013

The side-by-side squares are of course 25 and 9. The problem doesn’t state exactly how much overlapping is allowed, so many forms are possible. Presumably, every digit is involved in at least one of the two squares. At least one digit must be involved in only one square, otherwise the solution is trivial (i.e., every square would overlap itself).

- abc where these portions are squares
- a and abc: 100, 121, 144, 169, 196
- ab and abc: 169, 256, 361
- ab and bc: none
- abc and b: 144, 196
- abc and bc: 100
- abc and c: 100, 121, 144, 169, 289, 324, 361

- bc where these portions are squares:
- b and bc: 16, 49
- bc and c: 49, 64, 81

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#math trivia for #September17: #260 is number of license plates with letter then number. How many if both digits can be letter or number?

— Burt Kaliski Jr. (@modulomathy) September 18, 2013

The count of 260 comes from the product of the number of letters, A-Z, which is 26, and the number a single-digit numbers, 0-9, which is 10. There are 260 license plates of this two-digit form. (The first digit is the letter, the second is the number.)

If both digits can be letter or number, then there are 36 possibilities for each digit. The number of license plates of this form is 36*36 = 1296.

Wikipedia gives an overview of the letter / number combinations allowed on different countries’ license plates, whose forms generally involve many more digits than the simple examples analyzed here.

]]>#math trivia for #September15: #258 consists of three digits in an arithmetic progression. Which other numbers in the 200s do?

— Burt Kaliski Jr. (@modulomathy) September 16, 2013

The digits 2, 5 and 8 form an arithmetic progression because each successive pair has the same difference: 5-2 = 8-5 = 3. The answers consist of the numbers with digits

- 2 a b

where a-2 = b-a. Alternatively, we have a = 2+d and b = 2+2d where d is the difference between consecutive digits. It is clear that d must be between -1 and 3 so that a and b will be between 0 and 9. The answers are thus:

- 210
- 222
- 234
- 246

#math trivia for #September14: #257 is a prime of the form 2^(2^n)+1. What is n? Which other day-numbers (1-365) have this form?— Burt Kaliski Jr. (@modulomathy) September 15, 2013

Answer:

- n = 3 gives 2^(2^n) + 1 = 2^(2^3) + 1 = 2^8 + 1 = 257.
- Other day-numbers of this form are

3 = 2^(2^0) + 1, for n = 0

5 = 2^(2^1) + 1, for n = 1

17 = 2^(2^2) + 1, for n = 2

These are all **Fermat numbers**, named for the mathematician who studied numbers of this form. They’re all also **Fermat primes**, as they both have the form and are prime numbers.

It was once conjectured that all Fermat numbers are primes, but at this point the only Fermat numbers known to be prime are the four mentioned here and the Fermat number for n = 4, which is 65537. The larger ones that have been studied are all composite — but because they grow so quickly as a function of n, it will take a long time with current methods to determine the primality of even larger ones, and nothing has yet been proved about whether other primes do or don’t exist.

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