Problems that I was asked to write up.
(Let me know if you’d like others written up.)
Page 67
a is not divisible by 2 or 3 then a 2  1(mod 24)
8d – Prove that if the integer
So assume a is an integer tat is not divisible by 2 or 3.
2
24
|
(a
1)

So we want to show

By our “quiz theorem”, since the canonical prime factorization of
2

show that
 
3 | a 1
24  2 3 3, we need only
8 | a 1 and
2
in order to show this. By the difference of squares law,

a2 1  a 1a 1 This is the product of the two integers on either side of a . Since
a is not a multiple of 3, and in any set of 3 consecutive integers, one is a multiple of 3 (and
we know the middle one isn’t), either (a 1) or (a 1) must be a multiple of 3.

For 8, note that since 2 does not divide a , we could use part a o=f this
problem. (To prove
part a, note that since a is odd, both (a 1) and (a 1) must be even. And, noting that


every other even number is a multiple of 4, their product contains 3 twos, so it is divisible


by 8. Done.



Page 73
2d. Triangular numbers are those natural numbers t of the form
some n .


Recall that we showed (by induction) that
t
t  1 2  3 ... n for

n(n 1)
.
2
In such a case, in section 4.3 we note the units digit of a product of two numbers can be
found my multiplying the unitsdigit of the product of the units digits of the original
numbers. So the units digit of
t is the units digit of the product of the units digits of n and
n(n 1)
n 1 divided by 2. Just examine the 10 cases and check the results of
mod10 :
2


0 1
45
1 2 mod10  1, 2  3 mod10  3, 3 4 mod10
mod10  0,
 6,
mod10  0,
2
2
2
2
2
8  9
56
67
78
90
mod10  5,
mod10  1,
mod10  8,
mod10  6,
mod10  0
2
2
2
2
2
Done.

27a. The formula tells us that:
a10  1 0  2  0  3 7  4  2  5  3 6  2  7  5  8  6  9  9mod11
 (21mod11 8 15 mod1112 mod11 35 mod11 48 mod11 81mod11) mod11
 (10  8  4 1 2  4  4) mod11
 33mod11  0
So the check digit is correct. (Try part b for fun. The check digit is not correct.)
