1.
Solution
Any multiple of 7 would give us 0 mod 7, for example 220*7=1540. Just add 4, 1544.
2. Every season a baseball team plays 162 games. Suppose that Cal Ripken plays in his 2500th consecutive game on the 45th game of the year. In which game did his streak begin?
Solution 1
Each 162 games backwards puts us at game #45. 2499 mod 162=69 so we count 69 games backwards from #45: the first 44 and last 25 games of the previous season. So the streak began on game 138. (Note: games 138-162 constitute 25 games.)
Solution 2
Let's model the season of 162 games on the congruence classes 0,1,...161 mod 162. Suppose it starts on x mod 162, then 2499 games later it is at 44 mod 162. Thus x+2499=44 mod 162, whose solution is x=137, thus game 138.
3. Show that 1231=6 mod 7. What is 1231^1000 mod 7?
Solution
1231=7*175+6, so 1231 =6=-1 mod 7, thus 1231^1000=(-1)^1000=1 mod 7.
4. Is 12345678987654321 divisible by 9?
Solution
Cast out nines: eliminate
2,3,4,
8,1,
9,
5,6,7
so it is 0 mod 9, so it is divisible by 9.
5 ab=a+b mod 9 while ba=b+a mod 9, so ab=ba mod 9, or 9 divides ab-ba.
6. Let n be any positive integer whose sum of digits is 15. What is n mod 3, and 2^n mod 7?
Solution
Since any power of 10 is 1 mod 3, n is congruent to 15 mod 3, or 0 mod 3. Note that 2^3=8=1 mod 7, thus any power of 2^3 will be 1 mod 7.
1. Find an integer x between 0 and 22 such that 3*14-73=x mod 23.
Solution
42-73=-31=-8=15 mod 23.
2. It takes Mars 687 Earth days to rotate around the sun once. On Earth January 1, 2009 we make a Martian calendar, and call this day Martian Day 1 of Year 1. What day on the Martian calendar is April 1, 2011?
Solution
The number of earth days is 365+365+31+28+31+1=821=1*687+234, so we get year 2, day 234 of the Martian calendar.
3. If x=0 mod 9 and y=4 mod 9, can x*y have a last digit other than 6 or 0?
Solution
Yes. Let x=18, y=4 then x*y=72 ends in a 2.
4. Let a1 a2... ak be the digits of an integer written in base 3. Complete the following theorem, and explain why it is true:
a1 a2... ak is divisible by 3 if, and only if, ak ...
Solution
a1 a2... ak in base 3 is divisible by 3 if, and only if, the last digit is 0. This is completely the same as an integer base 10 is divisible by 10 if, and only if, the last digit is a 0.
5. Without explicitly doing any multiplications, explain why 2345454*99889=234285053606 is wrong.
Solution
Cast out nines: 2345454 is 0 mod 9, so the answer must be 0 mod 9, but 2+3+4+2+8+5+0+5+3+6+6+0+6=8 mod 9.
6. Show that it is impossible to find integers x and y such that 20*x+36*y=2.
Solution
Since 4 divides 20 and 4 divides 36, 4 must divide any integer combination of 20 and 36. This says 4 must divide 2, which is impossible.
7. Explain why casting out nines works for subtraction.
Solution
Let x and y be two integers, and consider x-y. We know that x mod 9= sum of the digits of x. We know that y mod 9= sum of the digits of y. We know that x-y mod 9= sum of the digits of x-y. Since x-y mod 9=x mod 9 -y mod 9 casting out nines works for subtraction.
8. An unusual foreign country has coins worth 15 cents and 17 cents. Suppose that person A and person B each have many coins of each type, and A wants to pay B 5 cents. How may this be accomplished?
Solution
We try to write 15*x+17*y=1, which is possible since GCD(15,17)=1. Then we multiply by 5. -9*15+8*17=1, so if A pays 8 of the 17 cent coins, and gets back 9 of the 15 cent coins, A has paid 1 cent. To pay 5 cents do this 5 times.