Solutions to the Quiz # 5 for Dr. Z.’s Number Theory Course for Oct. 24, 2013

1. (5 points) Using divisiblity tests, determine for each of the following integers (written in the usual, base 10, way) whether it is divisible by (i) 7 (ii)13 . Explain! a. 490735357707 b. 169130403130 Sol. to 1a: We break it up into blocks of three:

490|735|357|707

For (i)(divisibility by 7) we need to find 490 − 735 + 357 − 707 (mod 7) .

Most people first found

490 − 735 + 357 − 707 = −595

and then took it (mod 7) getting zero (and some people complained that they can’t use a calculator!, these are simple addition/subtraction problems!). But people who do not like to calculate could have done (mod 7) for each individual number and realized that they are all 0! Ans. to 1a(i): 490735357707 is divisible by 7 For 1a(ii), it is the same way, but now we take it (mod 13). First way: Again we get −495. Since 495 (mod 13) = 10, it is not divisible by by 13. Second way: 490 (mod 13) = 9, 735 (mod 13) = 7, 357 (mod 13) = 6, 707 (mod 13) = 5 So

490 − 735 + 357 − 707 (mod 13) = 9 − 7 + 6 − 5 = 3 .

So it is not divisible by 13. Sol. to b: We break it up into blocks of three:

169|130|403|130

The alternating sum is

(169 + 403) − (130 + 130) = 312 .

Since 312 is not divisible by 7, 169130403130 is not divisible by 7, but since 312 is divisible by 13, 169130403130 is divisible by 13.

Ans. to 1b(i): not divisible by 7. Ans. to 1b(ii): divisible by 13.

2. (5 points) What day of the week is

(i) Oct. 24, 2014 (ii) Oct. 24, 2015 (iii) Oct. 24, 2016 (iv) Oct. 24, 2017

(v) Oct. 24, 2018 Sol. of 2: Every non-leap year, it advances by 1 day (since 365 (mod 7) = 1). In a leap year (after Feb. 29), it advances by 2 days. So, since today is Thursday, (i) Friday (ii) Sat. (iii) Monday (since 2016 is a leap year, and Oct. is after Feb. 29)) (iv) Tues. (v) Wed.