Fewest Firsts

As we just noted, in 2026, all seven days of the week appear as the first of the month at least once. But you know, I decided that I don’t like that at all. Instead, I want as few days of the week as possible to appear as the first of the month in a given year.

To accomplish this, I have been granted the authority to change the number of days in each of that year’s 12 months, provided that there are still 365 or 366 days in the year and each month has at least 28 days and at most 31 days.

What are the fewest days of the week that can appear as the first of the month in such a calendar year? (And for fun, rather than for credit: How many such calendars can you design with this property?)

Let $N^*$ be the minimal number of days of the week that can appear as the first of the month of any calendar year. First from the Classic problem we see that the furthest away that two consecutive months can start is 3 days of the week apart, so this means that the smallest that we could possibly hope for is to have a calendar with only three possible days of the week for firsts of the month. That is, $N^* \gt 2,$ or equivalently that $N^* \geq 3.$

Therefore, all we have to do is provide some calendar year where $N=3$ is attained. For instance, take the following setup where March, June, and September have 29 days and all other months have 31 days. This would give $3 \cdot 29 + 9 \cdot 31 = 87 + 279 = 366$ days in the year. Similarly, if January 1 occurs on $x \in \mathcal{D},$ then February 1 occurs on $x+3,$ March 1 occurs on $x+6,$ April 1 occurs on $x+7 \equiv x,$ May 1 occurs on $x+3,$ June 1 occurs on $x+6,$ July 1 occurs on $x+7\equiv x,$ August 1 occurs on $x+3,$ September 1 occurs on $x+6,$ October 1 occurs on $x+7 \equiv x,$ November 1 occurs on $x + 3$ and finally, December 1 occurs on $x+6.$ Putting these altogther gives a subset $\mathcal{N} = \{ x, x + 3, x + 6 \} \subsetneq \mathcal{D},$ with $N= |\mathcal{N}| = 3.$ Therefore, we have that $N^* \leq N = 3$ and hence the fewest days of the week that can appear as the first of the month is $N^*=3.$

Can every day be the first?

In 2026, every day of the week is the first day of the month at least once:

• Monday is June 1.
• Tuesday is September 1 and December 1.
• Wednesday is April 1 and July 1.
• Thursday is January 1 and October 1.
• Friday is May 1.
• Saturday is August 1.
• Sunday is February 1, March 1, and November 1.

Is 2026 special in this regard? If so, when is the next year when one of the days of the week is not represented among the firsts of the month? Otherwise, if 2026 is not special in this regard, then why not?

First let's note that if a month has $28$ days, looking at you February, then the next month will start on the same day of the week that it starts. If a month has $29$ days, again looking less frequently at you February, then then next month will start on the next day of the week. If a month has $30$ days then the next month will start two days after it, and finally for all of those $31$ day months, the following month will start three days of the week after it.

Let's further represent the days of the week as numerals, $\mathcal{D} = \mathbb{Z} / 7\mathbb{Z}.$ If January 1 occurs on some $x \in \mathcal{D}$ then February 1 occurs on $x+3.$ If the year is not a leap year then we have March 1 also on $x+3,$ followed by April 1 on $x+6,$ May 1 on $x+8 \equiv x+1,$ June 1 on $x+4$, July 1 on $x+6,$ August 1 on $x+9 \equiv x+2$, September 1 on $x+5,$ October 1 on $x+7 \equiv x,$ November 1 on $x+3$ and finally December 1 on $x+5.$ Gropuing these all together we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6\} = \mathcal{D}.$

If on the other hand, this is a leap year, then we have January 1 and February 1 on $x$ and $x+3$ as before, but then March 1 on $x+4,$ April 1 on $x+7 \equiv x,$ May 1 on $x+2,$ June 1 on $x+5,$ July 1 on $x+7 \equiv x$, August 1 on $x+3,$ September 1 on $x+6,$ October 1 on $x+8 \equiv x+1,$ November 1 on $x+4,$ and finally December 1 on $x+6.$ Again, grouping these all togehter, we see that we have all the days of the week covered $\{ x, x+1, x+2, x+3, x+4, x+5, x+6 \} = \mathcal{D}.$

Therefore, there is nothing special about 2026. Each and every year under the Gregorian system has this property where each day of the week is represented among the firsts of the month.