5.1 Is This a Multiple Of?

Sums of consecutive numbers

Kavya is fiddling with a simple but charming idea, expressing numbers as sums of consecutive whole numbers:

9 = 4+5,\qquad 14 = 2+3+4+5,\qquad 18 = 5+6+7,\qquad 21 = 10+11 = 6+7+8 = 1+2+3+4+5+6.

Look at 21, it can be built in three separate ways! That single observation opens up a whole cluster of questions worth pondering.

Math Talk Puzzle these out alongside Kavya:

  • Can every natural number be expressed as a sum of consecutive numbers?
  • Which numbers allow more than one such expression?
  • Every odd number 2k+1 equals k + (k+1), a run of two. So does every even number also have a consecutive-sum form?
  • Allowing negatives, can you write 0 as a sum of consecutive numbers? (Try (-2)+(-1)+0+1+2.)

A famous snag lurks here. The powers of two, 1, 2, 4, 8, 16, \dots, are precisely the numbers that cannot be split into a sum of two or more consecutive numbers. Every other number can. That is a delightful claim to put to the test yourself.

Placing + and - between four numbers

Pick four numbers in a row, say 4, 5, 6, 7. Drop a + or a - sign in front of each of the last three. How many expressions can you form? Because each of the three signs is freely + or -, there are 2\times2\times2 = 8 outcomes:

\begin{aligned} &4+5+6+7 = 22, &\quad &4+5+6-7 = 8,\\ &4+5-6+7 = 10, &\quad &4+5-6-7 = -4,\\ &4-5+6+7 = 12, &\quad &4-5+6-7 = -2,\\ &4-5-6+7 = 0, &\quad &4-5-6-7 = -14. \end{aligned}

Try This Scan the eight outcomes: 22, 8, 10, -4, 12, -2, 0, -14. Every one is even! Now redo this with another run of four, say 6,7,8,9. Are they still all even? (Yes.) Then try four non-consecutive numbers, does the parity still stay uniform?

This is no accident, and we need not slog through every case. Here are two ways to prove it.

Why all eight share one parity Argument 1 (flip a single sign). Take any expression such as a+b-c-d. Reverse the sign of b to get a-b-c-d. The change is (a+b-c-d)-(a-b-c-d) = 2b, an even quantity. Reversing any one sign always shifts the total by twice that number, an even shift. Since each of the eight expressions can be reached from any other by reversing some signs, they all carry the same parity.

Argument 2 (parity laws). Recall \text{odd}\pm\text{odd}=\text{even}, \text{even}\pm\text{even}=\text{even}, \text{odd}\pm\text{even}=\text{odd}. In each line a+b and a-b have identical parity. So a\pm b, then a\pm b\pm c, then a\pm b\pm c\pm d, the parity never hinges on the signs we picked.

Breaking even

We can read parity straight off the page without computing. An expression turns out even exactly when an even number of its odd parts are added or subtracted (the odd parts cancel two at a time).

Worked Example

Without computing, which of these are even? \;58+26,\quad 914-330,\quad 6\times271\times5,\quad 650-413,\quad 703+128,\quad 217\times305,\quad 482-377.

  • 58+26: even + even = even.
  • 914-330: even - even = even.
  • 6\times271\times5: the factor 6 makes it even.
  • 650-413: even - odd = odd.
  • 703+128: odd + even = odd.
  • 217\times305: odd \times odd = odd.
  • 482-377: even - odd = odd.

The same logic carries over to algebraic expressions. The expression 4m+2q is always even, for two reasons: both 4m and 2q are even, and also 4m+2q = 2(2m+q) carries a factor of 2. But x^2+5 is not always even: if x=4 then x^2+5 = 21 (odd), yet if x=3 then x^2+5 = 14 (even), because x^2 inherits the parity of x.

Try This Which of these give an even number for every integer value? \;2a+2b,\; 3g+5h,\; 4m+2n,\; 2u-4v,\; 13k-5k,\; 6m-3n,\; x^2+5,\; b^2+1,\; 4k\times3j. Jot one example and one non-example wherever it clarifies.

Pairs to make fours

Add two even numbers. Is the result always a multiple of 4? Try 8+20=28 (yes) and 6+14=20 (yes) but 10+12=22 (no!). So sometimes. To pin down exactly when, sort even numbers by their remainder on division by 4: every even number is either 4p (remainder 0) or 4p+2 (remainder 2).

Case Algebra Multiple of 4?
both multiples of 4 4p+4q = 4(p+q) Yes
neither a multiple of 4 (4p+2)+(4q+2) = 4(p+q+1) Yes (the two 2’s pair up into a 4)
one of each 4p+(4q+2) = 4(p+q)+2 No (remainder 2)

This echoes parity exactly: two “remainder 2” numbers combine into a “remainder 0” number, just as odd + odd = even.

Always, Sometimes, or Never?

A claim about factors and multiples becomes most useful once we know whether it is always true, only sometimes true, or never true.

Worked Example

Weigh each statement.

  1. If 8 divides two numbers, it divides their sum, always true. Write them as 8a and 8b; then 8a+8b = 8(a+b). The same goes for the difference: 8a-8b = 8(a-b).
  2. If a number is divisible by 8, then any two numbers adding to it are each divisible by 8, sometimes true. 96 = 64+32 (both multiples of 8) but also 96 = 70+26 (neither is).
  3. If a number A is divisible by 7, then every multiple of A is divisible by 7, always true. If A=7j then A\times m = 7(jm).
  4. If a number is divisible by 12, it is divisible by every factor of 12, always true. 12m = 2\times6\times m = 3\times4\times m, so 2,3,4,6 each divide it.
  5. If a number is divisible by 7, it is divisible by every multiple of 7, sometimes true. 56 = 7\times8 is divisible by 14 and 28, but not by 21.

Two general principles drop out of this:

Divisibility facts to keep - If A is divisible by k, then every multiple of A is divisible by k. - If A is divisible by k, then A is divisible by every factor of k. - If a divides M and a divides N, then a divides M+N and M-N. - If A is divisible by k and by m, then A is divisible by the LCM of k and m (since A’s prime factorisation must already contain that of \operatorname{LCM}(k,m)).

So “divisible by both 9 and 4” forces “divisible by \operatorname{LCM}(9,4)=36” (always true), while “divisible by both 6 and 4” only forces “divisible by \operatorname{LCM}(6,4)=12”, not necessarily 24.

One more, a trap worth meeting: “adding an odd number to an even number yields a multiple of 6” is never true. Odd plus even is odd, but every multiple of 6 is even. In symbols, 2n+(2m+1)=6j would force 2(n+m) = 6j-1, i.e. an even number equal to an odd number, impossible.

What remains?

What does the family of numbers that leave remainder 3 on division by 5 look like? Multiples of 5 take the form 5k, so “3 more than a multiple of 5” is 5k+3:

k=0,1,2,3,4 \;\Rightarrow\; 5k+3 = 3,\,8,\,13,\,18,\,23,\dots

The very same family can also be described as “2 short of a multiple of 5”, i.e. 5k-2 (with k\ge1): 3,8,13,18,23,\dots, an identical list. Expressions that look quite different can name the very same set of numbers.

Figure it Out: Multiples & Parity

Practice

  1. The sum of four consecutive numbers is 50. What are the numbers?
  2. Suppose p is the greatest of five consecutive numbers. Describe the other four in terms of p.
  3. Always / sometimes / never true? Justify with algebra and (counter)examples.   (i) The sum of two even numbers is a multiple of 3.   (ii) If a number is not divisible by 18, then it is also not divisible by 9.   (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.   (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.   (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
  4. Find numbers that leave remainder 1 when divided by 3 and remainder 1 when divided by 4. Write an algebraic expression for all such numbers.
  5. Marble riddle: grouped in 4’s, three are left over; grouped by 3, two remain; grouped by 5, four remain; grouped by 7, none left over; fewer than 200. How many marbles?
  6. Imran lists numbers that leave remainder 4 when divided by 7. He claims: “Add any three of them and you get a multiple of 7.” Is he right? If not, what is the remainder?
  7. 853 leaves remainder 6 on division by 7, and 2406 leaves remainder 5. Without calculating, find the remainder of (i) 2406+853 and (ii) 2406-853 on division by 7.
  8. Find a number that leaves remainder 1 (÷3), remainder 2 (÷4) and remainder 3 (÷5). What is the smallest such number, and why?
  1. 11, 12, 13, 14 (their sum is 50). (Check: 11+12+13+14 = 50.)
  2. p-1,\ p-2,\ p-3,\ p-4.
    1. Sometimes, 2+4=6 (multiple of 3) but 2+6=8 (not).   (ii) Sometimes, 27 is not divisible by 18 yet is divisible by 9; 30 is divisible by neither.   (iii) Sometimes, 9 and 11 aren’t divisible by 6 and their sum 20 isn’t either, but 8+10 = 18 is.   (iv) Always, 6x+9y = 3(2x+3y).   (v) Sometimes, 18+9 = 27 is a multiple of 9, but 12+9 = 21 is not.
  4. A number that is “1 more than a common multiple of 3 and 4” has the form \operatorname{LCM}(3,4)\cdot n+1 = 12n+1: the numbers are 1, 13, 25, 37, 49, \dots
  5. “3 short of a multiple of 4, 1 short of a multiple of 3, 1 short of a multiple of 5” all say the number is 1 short of a common multiple of 3, 4, 5, i.e. of the form 60k-1: that gives 59, 119, 179, \dots Of these, 119 is divisible by 7 (119 = 7\times17). So 119 marbles.
  6. Write them as 7a+4,\ 7b+4,\ 7c+4. Their sum is 7(a+b+c)+12 = 7(a+b+c+1)+5, so the remainder is 5, not 0. The claim is false.
  7. Let 2406 = 7p+5 and 853 = 7q+6. (i) 2406+853 = 7(p+q)+11 = 7(p+q+1)+4 → remainder 4. (ii) 2406-853 = 7(p-q)-1 = 7(p-q-1)+6 → remainder 6.
  8. Each remainder is exactly 2 less than its divisor, so the number is 2 less than a common multiple of 3, 4, 5. The smallest such common multiple is \operatorname{LCM}(3,4,5)=60, so the smallest number is \mathbf{58}.