### More Mods

What is the units digit for the number 123^(456) ?

### Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

### Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

# Days and Dates

##### Stage: 4 Challenge Level:

Well done to Chuyi from Loughborough High School, Charkrit from Traill International School, and an anonymous solver, who all explained that as there are seven days in a week, the sequence of Wednesdays will go up in sevens.

Evie from St Anne's School sent us some good thinking about when her birthday will fall:

My birthday is $52$ days away. You divide $52$ by the number of days in a week, so you can find out how many weeks there are. $52 \div 7= 7$r$3$. So my birthday is in 7 weeks and 3 days.
If today is Sunday, we count on 3 days. Therefore, my birthday will be on a Wednesday.

Every year, your birthday shifts forwards one day. This is because $365 \div 7 = 52$r$1$. In $52$ weeks, it will still be a Wednesday, but there is a remainder. So in 2010, my birthday will be on a Thursday.

Of course, a leap year has 366 days so the remainder would be two instead of one, which is why your birthday shifts forward two days every leap year!

Abinhav from Bangkok Patana School sent us a very clear explanation about remainders when we divide by seven:

In $2, 9, 16$ and $23$ days from now, it will be a Wednesday. What other numbers of days from now will be Wednesdays?
Answer: $30, 37, 44, 51$ (keep adding $7$ each time)

Can you generalise what you have noticed?
If the number of days from now is termed $n$, and the remainder upon dividing by $7$ is $r$, then when $n\div 7$ yields $r = 2$, the day will be Wednesday.
From this we can see that if today is a Monday, then the day $n$ days from now will be moved forward $r$ days, starting from Monday. So for example in $24$ days from now, it will be a Thursday since $r$ is $3$. So the day moves forward $3$ days.

Abinhav went on to explain what happens to remainders when you add:

Remainder of numbers ADDED together divided by $7$. If $r_{1}$ is the remainder from the first division, and $r_{2}$ is the remainder from the second division, then the remainder from the total is always $r_{1} + r_{2}$. This occurs since when we add, we simply take the number we had origianlly and add on a supplement. So if originally the number was $9$, giving $r_{1} = 2$; and the second number was $15$, giving $r_{2} = 1$, then $9 + 15 = 24$. This can also be written as $7 + r_{1} + 14 + r_{2}$. So from this we get that $24 = 21 + r_{1} + r_{2}$. Since the number gained from adding multiples of $7$ will always be a multiple of $7$, the final remainder is always $r_{1} + r_{2}$.

Finally, an explanation of what happens to remainders when you multiply:

Remainder when two numbers are multiplied and then divided by $7$. $15 \times 26 = 390$. When $15$ is divided by $7$, remainder is $1$. When $26$ is divided by $7$, r$= 5$. When $390$ (their product) is divided by $7$, the remainder is $5$.
Hence when we multiply two numbers; upon division the final remainder is also the product of their original remainders. This occurs since when we multiply, we take any number and add those many more lots of that particular number. Say the remainder from the first number is $r_{1}$ and the remainder from the second number is $r_{2}$.
So for $15 \times 26$, we can rewrite this as $$(14 + r_{1})(21 + r_{2})$$ From this we get
$$294 + 14r_{2} + 21r_{1} + r_{1}r_{2}$$
Since the first three terms will always be multiples of 7, the final remainder is always $r_{1}r_{2}$ (in this case, 5).

Of course, in the case where $r_{1}r_{2}$ is greater than 7, the remainder will simply be what would be left if you divided $r_{1}r_{2}$ by 7.