Divisibility Conjectures by Students

Math 131

Spring 05

Divisibility Conjectures

The conjectures are arranged in an order that highlights mathematical relationships (rather than in numerical order).

Divisibility by 4

K: If the number ends in zero (unless the number is a multiple of 20) it will not go into the number evenly. It will instead be a decimal number. (counterexample)

K: If the number ends in 4 or 8 and in the tens place the number is a multiple of 2, then it will work (conjecture)

S: My conjecture of the multiples of 4 patterns are that if the tens place is even 0,4,and 8 are its multiples but if the tens place is odd then 2 and 6 are its multiples.

M: 1. If the number ends in zero, unless the number is a multiple of 20, the number will go into the number evenly without a remainder. The answer will be in decimal form.

M: 2. If the number ends in 24,33,48, or 64, the answer will come out even without a remainder.

T: My conjecture is about the number 4. We found that no matter how big the number is, that in order to tell if it's divisible by 4 or not all you have to do is look at the last two digits. If the number in the tens place is an odd number and the number in the ones place is a 2 and 6, then it is divisible by 4. Also, if the number in the tens place is an even number and the number in the ones place is either a 0, 4 or 8, then it is also divisible by 4.

Divisibility by 8

K: If n is a multiple of 4, then 10n is a multiple of 8. If 10n is a multiple of 8, then 10n+8 is also a multiple of 8.

Example: n=8; 10(8)=80. 80+8=88.

K: The digit in the ones place of a multiple of 8 will always be 8, 6, 4, 2, or 0, in that order. As the number in the ones place decreases by 2, the digit in the tens place increases by one.

Example: 8, 16, 24, 32, 40…

K: When the number in the tens place is multiple of four, it will be followed by a zero. The next multiple of 8 after that will be the same multiple of four followed by an 8.

Example: 40, 48…80, 88…120, 128.

Divisibility by 9

A: The 9s are a part of every other table in some way
9x3 is just 3x9 so if you know the 3s table you'll know part of the 9s
And you could always work backwards for the big ones
9x10 is 90, is minus 9 from 90 and u get 81, that’s 9x9
Minus 9 from 81 and u get 72
There’s a pattern for the whole thing
18, 27, 36.... The 2nd digit is always one less
And the 1st digit is always 1 more
It’s a very easy pattern
just count 1,2,3,4,5,6,7,8,9.... Then count 9-1, so 8,7,6,5,4,3,2,1,0 and match them up.

M: I believe that if digits in N add up to a multiple of 9, then N is divisible by 9. On the multiplication table that goes up to 20, each number in the 9's column adds up to 9, except for 99 which adds up to 18.

Divisibility by 3

M: In class, Z and I worked on the 3's. We found some patterns and made a conjecture

-Each number goes up by 3

-Every tenth number ends in zero (therefore each tenth number is a multiple of 10)

-The one's place digit go in a pattern on 3,6,9,2,5,8,1,4,7,0 and then begin again.

-If digits in N add up to a multiple of 3, than N is divisible by 3

Divisibility by 6

M: When looking at the 6's to see if the conjecture [for divisibility by 3] would work there as well, I noticed that all of the digits of N in the 6 column add up to multiples of 3 as well. Therefore, I thought that maybe when N adds up to a multiple of 3 it could be divisible by six as well, however, that is not the case because I found a counter example. The digits in 15 add up to 6, which is a multiple of 3 and 6, however, 15 is not divisble by 6.

Divisibility by 12

Divisibility by 11

K: Until 11x10, the answer to 11xn will be nn.

K: The digits in the answer to 11x any two digit number (x) will increase in this pattern: the digit in the tens place will increase by 1, starting with 1. The digit in the ones place will also increase by 1, but starting with 0, so that the digit in the tens place is always one more than the digit in the ones place.

-M: Each number increases by 11

-The ending digit goes in a pattern of 1,2,3,4,5,6,7,8,9,0 and then begins again

-When you add the digits together, the numbers go up by 2's: 1+1=2, 2+2=4, 3+3=6 etc.

-When you multiply a single digit by 11, you simply double the number

-When you multiply a number in the teens by 11, you get a 1, one more than the ending digit, and the ending digit:

12*11= 132

16*11= 176

-When you multiply a number in the 20's by all, you get a 2, two more than the ending digit, and the ending digit:

21*11= 231

27*11= 297

J (also see more extended version): The first digit and last digit of the multiplier is the first and last digit of the product. Example 52 × 11 =5_2. Upon further examination, I also discovered that if you add the digits of the multiplier, you would get the missing number in between. 5 + 2 =7, therefore 52 × 11 =572. This works for two digit multipliers. When the sum of your two digits exceed 9 however, it is then carried to the next place just like normal addition, example 88× 11 = 8_8. When you do 8+8=16, 6 will go into the blank spot, and the one is carried to the 8 in the hundreds place giving you 88× 11 = 968. When we move to three digit multipliers however, it works a little differently. Example 112 × 11 = 1_ _ 2. What you do is add the second and the third digits of the multiplier together 1+2=3, placing 3 in the second missing dash 112 × 11 = 1_ 32. Next add the first and second digits of the multiplier together 1+1=2. Therefore, 112 × 11 = 1232. Four digit numbers work the same way as three digit numbers. Example 6954 × 11 = 6_ _ _4. Working with the last two digits first, 5+4=9, thus 6954 × 11 = 6_ _94. Then the next two digits to the left, 9+5= 14, placing the 4 in the next dash and regrouping the 1 to the numbers in the next place, 6954 × 11 = 6_ 494. Next work with the last two digits to the left, 6+9+1(regrouped)=16. place the 6 in the last dash and regroup the 1 with the 6 in the 60 thousands place. Therefore 6954 × 11 = 76494.

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