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A set of rules and examples to help determine the divisibility of numbers by certain digits and numbers. It covers divisibility by 2, 3, 4, 5, 6, 7, 8, 9, and 11. Examples of numbers and their divisibility checks.
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Let ai denote the digits of a number, e.g. 542 is represented by a 3 a 2 a 1 where a 3 = 5, a 2 = 4, and a 1 = 2. The number anan− 1 · · · a 2 a 1 is divisible by if is divisible by or 2 a 1 2 3 an + an− 1 + · · · + a 2 + a 1 3 4 a 2 a 1 4 = 00 5 a 1 5 6 anan− 1 · · · a 2 a 1 2 and 3 7 anan− 1 · · · a 2 − 2 a 1 7 8 a 3 a 2 a 1 8 = 000 9 an + an− 1 + · · · + a 2 + a 1 9 10 a 1 = 0 11 (a 1 + a 3 + a 5 + · · ·) − (a 2 + a 4 + a 6 + · · ·) 11 12 anan− 1 · · · a 2 a 1 3 and 4 13 anan− 1 · · · a 2 − 9 a 1 13
Example 0.
2: Since the last digit of 23578 , which is 8 , is divisible by 2 , then 23578 is divisible by 2. The last digit of 2365 is 5. Since 5 is not divisible by 2 , 2365 is not divisible by 2.
3: Let us see if 231 is divisible by 3. Summing up the digits of 231 we get 2 + 3 + 1 = 6. Since 6 is divisible by 3 then 231 is divisible by 3. Let us check if 23574 is divisible by 3. Summing up the digits of 23574 gives us 2 + 3 + 5 + 7 + 4 = 21. Since 21 is divisible by 3 , then 23574 is divisible by 3. But suppose that you could not remember that 21 is divisible by 3. Then you could sum up the digits of 21 to get 2 + 1 = 3. Clearly 3 is divisible by 3. So 21 is divisible by 3 and therefore 23574 is divisible by 3. Is 27934576592477693217 divisible by 3? Let us sum up the digits again 2+7+9+3+4+5+7+6+5+9+2+4+7+7+6+9+3+2+1+7 = 105. If we are not sure if 105 is divisible by three we can sum up its digits.
So 1 + 0 + 5 = 6. Now we know 6 is divisible by 3 , so 105 is divisible by 3 and therefore 27934576592477693217 is divisible by 3. The number 16 is not divisible by 3 since 1 + 6 = 7 is not divisible by
Let us try 23761. It’s digits sum to 2 + 3 + 7 + 6 + 1 = 19 and 19 is not divisible by 3 since 1 + 9 = 10 is not divisible by 3.
4: The number 116 is divisible by 4 since the last two digits of 116 is 16 and 16 is divisible by 4. How about the number 1352? The last two digits of 1352 is 52. Now 4 will divide 52 ( 52 ÷ 4 = 13). So 4 divides 1352. Will 4 divide 125487100? Yes since the last two digits are 00.
5 It is easy to see that 5 will divide 25 , 40 , 14520 , 785425 , and 789875412545 , since each one ends in a 0 or a 5.
6: To check if 132 is divisible by 6 requires two steps. First is 132 divisible by 2? Yes since the last digit is divisible by 2. Next we sum up its digits: 1 + 3 + 2 = 6. Since 6 is divisible by 3 , 132 is divisible by 3. Therefore 132 is divisible by 6. Let us try 1764. Clearly 1764 is divisible by 2 and it is divisible by 3 since 1 + 7 + 6 + 4 = 18 = 9. So 1764 is divisible by 6. The number 1371 is not divisible by 6 since 1371 is not divisible by 2. Note that it is divisible by 3 , 1+3+7+1 = 12, but that does not matter since it is not divisible by 2. The number 1678 is not divisible by 6 since it is not divisible by 3. This can be seen since 1+6+7+8 = 22. It does not matter that it is divisible by 2. It has to be divisible by both. We can see that 3211 is not divisible by 6 since it is not divisible by either 2 or 3.
7: Let us start off with some easy examples. First 14 is divisible by 7 since 1 − 2(4) = 1 − 8 = − 7 and − 7 is divisible by 7. Next 21 is divisible by 7 since 2 − 2(1) = 2 − 2 = 0 and 0 is divisible by 7. Is 924 divisible by 7? We apply the rule to get 92 − 2(4) = 92 − 8 = 84. If we are not sure whether 7 divides 84 we can apply the rule again.
To check to see if 11 divides 2607 we make the calculation: (2 + 0) − (6 + 7) = 2 − 13 = − 11. Since 11 divides − 11 it divides 2607. Now try 38236. So (3 + 2 + 6) − (8 + 3) = 11 − 11 = 0. So 11 divides
Finally does 11 divide 3042149? Well since 11 divides (3 + 4 + 1 + 9) − (0 + 2 + 4) = 17 − 6 = 11 it divides 3042149.
12: Is 132 divisible by 12? If it is it has to be divisible by both 3 and 4. Since 1 + 3 + 2 = 6 it is divisible by 3 and since 32 is divisible by 4 , 132 is divisible by 12. What about 264? Again we test for divisibility by 3 and 4. Since 2 + 6 + 4 = 12 is divisible by 3 and 64 is divisible by 4 , 264 is divisible by
Can we divide 1800 by 12? First we add the digits of 1800 to get 1 + 8 + 0 + 0 = 9. So 1800 is divisible by 3. The last two digits of 1800 are 00 , so 1800 is divisible by 4. Therefore 1800 is divisible by 12.
13: Is 65 divisible by 13? By the rule we compute
6 − 9(5) = 6 − 45 = − 39.
Since 39 is divisible by 13 , 65 is divisible by 13. Lets try 377. 37 − 9(7) = 37 − 63 = − 26 So 377 is divisible by 13 , since 26 is divisible by 13. Is 2041 divisible by 13?
204 − 9(1) = 195
19 − 9(5) = 19 − 45 = − 26 So 2041 is divisible by 13.