M11 Divisibility 1 (Using Algebra to Prove) - PgDivByN1

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Divisibility 1Problem

1 A conjecture from arithmetic

You might know almost at a glance that 1011 − 1 is evenly di-visible by 9, but how could you tell quickly what, if anything,711 − 1 is evenly divisible by? First, look for a pattern.

1. Show that 1011 − 1 and 107 − 1 are both divisible by 9.Give a solid explanation why 1 less than any power of 10must be divisible by 9.

2. Compute 63 − 1 and 67 − 1 and find a small (one-digit) Can you show why that factordivides all numbers of the form6n − 1 (where n is a wholenumber)?

factor that both have in common. Are 62 − 1 and 65 − 1also divisible by that factor?

3. Calculate 3n−1 for several values of n. Explain why theseare all divisible by 2.

4. Calculate 112 − 1, 113 − 1, 114 − 1, and 115 − 1 and findthe largest obvious common factor. Unless you use fancy Four of the factors of 1137 − 1 are

2591, 136151713, 36855109, and2615418118891695851, but youwere not expected to find them!

software, you probably can’t conveniently calculate 1137,but you might be able to explain why 1137 − 1 ends witha zero. That will go a long way toward helping you showthat 11n − 1 is always divisible by.... Explain.

5. Numbers of the form 10n − 1 had 9 as a common factor;numbers like 3n − 1 had 2 as a common factor; you founda common factor in 6n − 1; and you also found a commonfactor in 11n − 1. What’s the pattern here? State a con-jecture of the form “If you subtract 1 from any power ofsome number a, the result is divisible by....”

Problems with a Point: January 16, 2001 c© EDC 2000

Divisibility 1: Problem 2

2 A proof from algebra

The rest of this problem set is devoted to proving that the pat-tern you’ve seen for 3n − 1, 6n − 1, 10n − 1, and 11n − 1 holdstrue for all numbers of the form an − 1.

1. (a) Multiply (x − 1)(x + 1).(b) Multiply (x − 1)(x2 + x + 1).(c) Multiply (x − 1)(x3 + x2 + x + 1).

2. (a) Multiply x(x5 + x4 + x3 + x2 + x + 1).(b) Multiply −1 × (x5 + x4 + x3 + x2 + x + 1).(c) Multiply (x − 1)(x5 + x4 + x3 + x2 + x + 1).

3. (a) Multiply x(x7 + x6 + x5 + x4 + x3 + x2 + x + 1).(b) Multiply (−1)(x7 + x6 + x5 + x4 + x3 + x2 + x + 1).(c) Multiply (x − 1)(x7 + x6 + x5 + x4 + x3 + x2 + x + 1).

4. If you have worked carefully, you will see a pattern.Explain why products like

(x − 1)(x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1)

always have only two terms, and say what those two termsmust be.

5. On the basis of your pattern, what are x3−1x−1

and x5−1x−1

? Why must x �= 1 in this problem?

Show the computation that proves you are correct.

6. And now, the piece de resistance. Piece de resistance is fromFrench, roughly meaning the“great accomplishment.”

(a) Explain why an − 1 can always be factored (if n is aninteger and n > 1), and give an illustrative example.

(b) Relate this back to problem 5 in section 1.


Divisibility 1: Problem 3

3 Epilogue: An application to arith-

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This result also explains why a whole number whose digits add For example, the sum of thedigits in the number 142857 is1 + 4 + 2 + 8 + 5 + 7 = 27.27 is divisible by 9, so 142857must be divisible by 9.(142857 = 9 · 3 · 13 · 407)

up to a multiple of 9 must, itself, be a multiple of 9.

1. If 10n − 1 is divisible by 9, then 10n must not be divisibleby 9. What is the remainder when you divide 10n by 9?Explain how you know that’s true.

2. If b is a whole number less than 9, then b × 10n is notdivisible by 9. What is the remainder when you divideb × 10n by 9? Explain how you know.

3. If b and c are both whole numbers and each of them isless than 5, show what the remainder is when you divideb×105+c×103 by 9. Show that you get the same remainderif you divide c × 107 + b × 106 by 9.

4. If b + c = 9, what remainder do you get when you divideb × 105 + c × 103 by 9? Explain why.

5. Extend your explanation to work for five-digit numbers.Without actually dividing, predict the remainder when26453 is divided by 9. Tell whether 1350243 is divisibleby 9 or not.

The process called “casting outnines,” in which one compares the“digital sum” in a calculation tothat of the result, was long usedas a quick check for errors inarithmetic. For example, to checkthe calculation 361 × 53 = 19123,we change each of the numbers361, 53, and 19123 to the sum ofits digits to get 10, 8, and 16,and then reduce those the sameway to get 1, 8, and 7. When theoriginal calculation is performedwith these reduced numbers, itlooks like 1 × 8 = 7, which is notcorrect, so the original calculationis also not correct.


Divisibility 1: Hints 1

Hints


Hint to problem 1. Look at numbers of the form 10n.Hint to problem 2. Show why the units digits of 63,67, 62 and 65 are the same.Hint to problem 3. Are powers of 3 odd or even?Hint to problem 4. What is the units digit of anypower of 11?


Hint to problem 5. If ab

= c then cb = a. What correspond to a, b, and c inthis problem?


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Hint to problem 1. 10n is 1 more than 10n − 1, amultiple of 9, so...Hint to problem 4.

b × 105 + c × 103 = b(9m + 1) + c(9n + 1)= 9(mb + nc) + (b + c)= . . .

Hint to problem 5. Dividing 26453 is like dividingeach of 20000, 6000, 400, 50, and 3 by 9. The remaindersof the divisions, taken separately, are 2, 6, 4, 5, and 3.


Divisibility 1: Answers 1

Answers


1. Numbers of the form 10n are written as a 1 followed byn zeros (for example, 107 = 10, 000, 000). So numbers ofthe form 10n − 1 consist of n 9s (for example, 107 − 1 =9, 999, 999). These are all obviously divisible by 9.

2. 63−1 = 215, 67−1 = 279935, 62−1 = 35 and 65−1 = 7775,all divisible by 5.

3. All powers of 3 must be odd. Subtracting 1 makes themeven. So 3n − 1 is always divisible by 2.

4. 112 − 1 = 120, 113 − 1 = 1330, 114 − 1 = 14640, and115 − 1 = 161050. These are all multiples of 10. In fact,the units digit of all powers of 11 must be 1, so 11n − 1 isalways divisible by 10.

5. So far, it seems that numbers of the form an−1 are divisibleby a−1. “If you subtract 1 from any power of some numbera, the result is divisible by a − 1.”


1.

(x − 1)(x + 1) = (x2 + x) + (−x − 1) = x2 − 1(x − 1)(x2 + x + 1) = (x3 + x2 + x) + (−x2 − x − 1) = x3 − 1

(x − 1)(x3 + x2 + x + 1) = (x4 + x3 + x2 + x) + (−x3 − x2 − x − 1) = x4 − 1

Here’s another way of looking at it:

(x + 1)(x − 1) = (x2 − x) + (x − 1) = x2 − 1(x2 + x + 1)(x − 1) = (x3 − x2) + (x2 − x) + (x − 1) = x3 − 1

(x3 + x2 + x + 1)(x − 1) = (x4 − x3) + (x3 − x2) + (x2 − x) + (x − 1) = x4 − 1



2. x6 + x5 + x4 + x3 + x2 + x− x5 − x4 − x3 − x2 − x − 1

x6 − 1

3. x8 + x7 + x6 + x5 + x4 + x3 + x2 + x− x7 − x6 − x5 − x4 − x3 − x2 − x − 1

x8 − 1

4. Multiplying (x6 + x5 + x4 + x3 + x2 + x + 1) by x raisesthe exponent of each term, leaving a new highest power ofx (in this case, x7), and no 1. Multiplying the same seriesby −1 leaves the exponents unchanged, but all negativecoefficients. Adding the results, all but the highest andlowest exponent terms cancel each other out. Productslike (xn + . . .+x3 +x2 +x+1)(x−1), therefore, have onlytwo terms, xn+1 and −1.

5. As long as x �= 1, x3−1x−1

= x2 + x + 1 because (x2 + x +1)(x − 1) = x3 − 1.Also, x5−1

x−1= x4 +x3 +x2 +x+1. Of course, if x = 1, these

expressions involve division by 0, which makes no sense.

6. Problem 4 in section 2 showed that an−1a−1

= an−1 + an−2 +. . . + a2 + a + 1. An example is (104 − 1) ÷ (10 − 1) =9999 ÷ 9 = 1111.


metic

This result also explains why a whole number whose digits addup to a multiple of 9 must, itself, be a multiple of 9.

1. When you divide 10n by 9, the remainder is 1, because 10n

is 1 more than 10n − 1, a multiple of 9.

2. If b < 9, then dividing b + 9ab by 9 leaves a remainderof b. This is because 10n is 1 plus a multiple of 9, sob × 10n = b(1 + 9a) = b + 9ab.



3. If b < 5 and c < 5 and both are whole numbers, then

b×105+c×103 = b(9m+1)+c(9n+1) = 9(mb+nc)+(b+c)

Because b + c < 9, dividing [9(mb + nc) + (b + c)] by 9leaves a remainder of b + c. The exponents are irrelevantin this calculation, and so whether you divide 403000 or34000000 by 9, in either case the remainder is 7.

4. If b + c = 9, what remainder do you get when you divideb × 105 + c × 103 by 9? ANSWER:There is no remainder.

b × 105 + c × 103 = b(9m + 1) + c(9n + 1)= 9(mb + nc) + (b + c)= 9(mb + nc) + 9= 9(mb + nc + 1)

So, there is no remainder because 9(mb + nc + 1) is amultiple of 9.

5. Dividing 26453 leaves a remainder of 2. (20 → 2 + 0 = 2).The number 1350243 is divisible by 9.


Divisibility 1: Solutions 1

Solutions


1. Numbers of the form 10n are written as a 1 followed byn zeros (for example, 107 = 10, 000, 000). So numbers ofthe form 10n − 1 consist of n 9s (for example, 107 − 1 =9, 999, 999). Just looking at these tells us that they are alldivisible by 9 and that the quotient after dividing by 9 isa number consisting of n 1s.

2. 63 − 1 = 215, 67 − 1 = 279935, 62 − 1 = 35 and 65 − 1 =7775, all clearly divisible by 5. In fact, all numbers of theform 6n − 1 are divisible by 5. We use the multiplicationalgorithm to show why. The product of 6 and any numberthat ends with 6 is another number that ends with 6.

abcde6×6

mnpqr6

some number ending with 6

another number ending with 6

Now, because 6 ends in 6, 62 also ends in 6, and so 63 endsin 6, and so on for all higher powers of 6.

Because the units digit of 6n is 6 for all n, the units digitof 6n − 1 is 5, which makes 6n − 1 a multiple of 5 for all n.

3. All powers of 3 must be odd. Subtracting 1 makes themeven. So 3n − 1 is always divisible by 2.

4. The units digit of all powers of 11 must be 1. Subtracting1 from any of them produces numbers that end in 0—multiples of 10. So 11n − 1 is always divisible by 10. Whymust all powers of 11 end with 1? We already know thatthe first few powers do. So, pick one of these—say, 11n—and show that if it ends in 1 (which it does), then 11n+1

must also end in 1. 11n·11 = 11n·(10+1) = 11n·10+11n· =a number ending in 0 (a multiple of 10) plus a numberending in 1. Alternatively, base your argument on themultiplication algorithm again.

5. So far, it seems that numbers of the form an−1 are divisibleby a−1. “If you subtract 1 from any power of some numbera, the result is divisible by a − 1.”




1.

(x − 1)(x + 1) = (x2 + x) + (−x − 1) = x2 − 1(x − 1)(x2 + x + 1) = (x3 + x2 + x) + (−x2 − x − 1) = x3 − 1

(x − 1)(x3 + x2 + x + 1) = (x4 + x3 + x2 + x) + (−x3 − x2 − x − 1) = x4 − 1

Here’s another way of looking at it:

(x + 1)(x − 1) = (x2 − x) + (x − 1) = x2 − 1(x2 + x + 1)(x − 1) = (x3 − x2) + (x2 − x) + (x − 1) = x3 − 1

(x3 + x2 + x + 1)(x − 1) = (x4 − x3) + (x3 − x2) + (x2 − x) + (x − 1) = x4 − 1

2. x6 + x5 + x4 + x3 + x2 + x− x5 − x4 − x3 − x2 − x − 1

x6 − 1

3. x8 + x7 + x6 + x5 + x4 + x3 + x2 + x− x7 − x6 − x5 − x4 − x3 − x2 − x − 1

x8 − 1

4. Multiplying (x6 + x5 + x4 + x3 + x2 + x + 1) by x raisesthe exponent of each term, leaving a new highest power ofx (in this case, x7), and no 1. Multiplying the same seriesby −1 leaves the exponents unchanged, but all negativecoefficients. Adding the results, all but the highest andlowest exponent terms cancel each other out. Productslike (xn + . . .+x3 +x2 +x+1)(x−1), therefore, have onlytwo terms, xn+1 and −1.

5. As long as x �= 1, x3−1x−1

= x2 + x + 1 and x5−1x−1

= x4 + x3 +x2 + x + 1. Of course, if x = 1, these expressions involvedivision by 0, which makes no sense.

6. Problem 4 in section 2 showed that an−1a−1

= an−1 + an−2 +. . . + a2 + a + 1. An example is (104 − 1) ÷ (10 − 1) =9999 ÷ 9 = 1111.




metic

This result also explains why a whole number whose digits addup to a multiple of 9 must, itself, be a multiple of 9.

1. If 10n − 1 is divisible by 9, then 10n must not be divisibleby 9. What is the remainder when you divide 10n by 9?SOLUTION: 10n is 1 more than 10n − 1, a multiple of 9.So, dividing 10n by 9 leaves a remainder of 1.

2. If b is a whole number less than 9, then b × 10n is notdivisible by 9. What is the remainder when you divideb×10n by 9? SOLUTION: Because 10n is 1 plus a multipleof 9, b × 10n = b(1 + 9a) = b + 9ab. Dividing b + 9ab by 9leaves a remainder of b if b < 9.

3. If b < 5 and c < 5 and both are whole numbers, then

b×105+c×103 = b(9m+1)+c(9n+1) = 9(mb+nc)+(b+c)

Because b + c < 9, dividing [9(mb + nc) + (b + c)] by 9leaves a remainder of b + c. The exponents are irrelevantin this calculation, and so whether you divide 403000 or34000000 by 9, in either case the remainder is 7.

4. If b + c = 9, what remainder do you get when you divideb × 105 + c × 103 by 9? SOLUTION:

b × 105 + c × 103 = b(9m + 1) + c(9n + 1)= 9(mb + nc) + (b + c)= 9(mb + nc) + 9= 9(mb + nc + 1)

So, there is no remainder because 9(mb + nc + 1) is amultiple of 9.

5. Dividing 26453 is like dividing each of 20000, 6000, 400,50, and 3 by 9. The remainders of the divisions, takenseparately, are 2, 6, 4, 5, and 3. Because their total, 20, isgreater than 9, still more 9s can be divided out. But aftertaking out those two more nines, there is still a remainderof 2. (20 → 2 + 0 = 2). Reasoning the same way aboutthe number 1350243, we add 1+3+5+0+2+4+3 = 18.Exactly two more nines can be divided out, so 1350243 isdivisible by 9.


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M11 Divisibility 1 (Using Algebra to Prove) - PgDivByN1