Can Every Day Be First

Fiddler

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?

Solution

Given an non-leap year,

1. Let January 1^st be $n\pmod 7$.
2. February 1^st is $n+31\pmod 7 \equiv n+3 \pmod 7$.
3. March 1^st is $n+3+28\pmod 7 \equiv n+3 \pmod 7$.
4. April 1^st is $n+3+31\pmod 7 \equiv n+6 \pmod 7$.
5. May 1^st is $n+6+30\pmod 7 \equiv n+1 \pmod 7$.
6. June 1^st is $n+1+31\pmod 7 \equiv n+4 \pmod 7$.
7. July 1^st is $n+4+30\pmod 7 \equiv n+6 \pmod 7$.
8. August 1^st is $n+6+31\pmod 7 \equiv n+2 \pmod 7$.
9. September 1^st is $n+2+31\pmod 7 \equiv n+5 \pmod 7$.
10. October 1^st is $n+5+30\pmod 7 \equiv n \pmod 7$.
11. November 1^st is $n+31\pmod 7 \equiv n+3 \pmod 7$.
12. December 1^st is $n+3+30\pmod 7 \equiv n+5 \pmod 7$.

All 7 residues exist for any non-leap year.

Given an leap year,

1. Let January 1^st be $m\pmod 7$.
2. February 1^st is $n+31\pmod 7 \equiv m+3 \pmod 7$.
3. March 1^st is $m+3+29\pmod 7 \equiv m+4 \pmod 7$.
4. April 1^st is $m+4+31\pmod 7 \equiv m \pmod 7$.
5. May 1^st is $m+30\pmod 7 \equiv m+2 \pmod 7$.
6. June 1^st is $m+2+31\pmod 7 \equiv m+5 \pmod 7$.
7. July 1^st is $m+5+30\pmod 7 \equiv m \pmod 7$.
8. August 1^st is $m+31\pmod 7 \equiv m+3 \pmod 7$.
9. September 1^st is $m+3+31\pmod 7 \equiv m+6 \pmod 7$.
10. October 1^st is $m+6+30\pmod 7 \equiv m+1 \pmod 7$.
11. November 1^st is $m+1+31\pmod 7 \equiv m+4 \pmod 7$.
12. December 1^st is $m+4+30\pmod 7 \equiv m+6 \pmod 7$.

All 7 residues exist for any leap year as well.

Answer

2026 is not unique. Every day of the week is the first day of the month at least once for any year.

Extra Credit

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?)

Solution

Consider the number of days in a month :

1. $28 \equiv 0 \pmod 7$
2. $29 \equiv 1 \pmod 7$
3. $30 \equiv 2 \pmod 7$
4. $31 \equiv 3 \pmod 7$

13 Months

$28 \times 12 = 336$ Adding a 13^th month of $29$ or $30$ days would bring the total to $365$ or $366$, which would yield only two days of the week that would appear as the first of the month. This yields 26 solutions if 13 months is possible.

12 Months

$\dfrac{365}{12} \approx 30.42$. However, consider $\dfrac{366}{3} = 122$ and $\dfrac{364}{4} = 91$.

4 Month Cycle of 122 days

$122 = $

1. $29+31+31+31 \equiv \boxed{1+3+3+3}$
   • $29+31+31 \equiv 1+3+3 \equiv 0$ is a cycle of 3 months that returns to the day of the week at the beginning of the cycle.
   • $31+31+29+31+31+29+31+31+29+31+31+31 \equiv (3+3+1) + (3+3+1) + (3+3+1) + (3+3+3)$ would require $\boxed{4}$ days of the week to be the first of the month.
   
   
2. $30+30+31+31 \equiv \boxed{2+2+3+3}$
   • $30+30+31 \equiv 2+2+3 \equiv 0$ is a cycle of 3 months that returns to the day of the week at the beginning of the cycle.
   • $31+30+30+31+30+30+31+30+30+31+31+31 \equiv (3+2+2) + (3+2+2) + (3+2+2) + (3+3+3)$ would require $\boxed{5}$ days of the week to be the first of the month.

3 Month Cycle of 121 days

$121 = $

1. $29+31+31 \equiv \boxed{1+3+3}$
   • One solution that would$\boxed{4}$ days of the week to be the first of the month is the same as the first solution above.
   • Another solution is $31+31+29+31+31+29+31+31+29+31+31+30 \equiv (3+3+1) + (3+3+1) + (3+3+1) + (3+3+2)$ that would require $\boxed{4}$ days of the week to be the first of the month. This solution is a 365-day year.
   
   
2. $30+30+31 \equiv \boxed{2+2+3}$
   • $30+30+31 \equiv 2+2+3 \equiv 0$ is a cycle of 3 months that returns to the day of the week at the beginning of the cycle.
   • $31+30+30+31+30+30+31+30+30+31+30+31 \equiv (3+2+2) + (3+2+2) + (3+2+2) + (3+2+3)$ would require $\boxed{4}$ days of the week to be the first of the month.

Answer

I couldn't find a solution better than 4 firsts of the month.

Rohan Lewis

2026.03.02