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# Divisibility/Multiples/Factors: Tips and hints

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Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
chetan2u wrote:
Hi,
taking the method used for 7, we can check for some more. lets see till 30..
7 :- Subtract 2 times the last digit from remaining number. Repeat the step if required. If the result is divisible by 7, the original number is also divisible by 7
13 :- Add 4 times the last digit to the remaining number. Repeat the step if required. If the result is divisible by 13, the original number is also divisible by 13
15 :- ends with 5 or 0 and sum of the digit is div by 3
16 :- last 4 digits should be div by 16. wil not be required as it itself becomes a long calculation
17 :- Subtract 5 times the last digit from remaining number. Repeat the step if required. If the result is divisible by 17, the original number is also divisible by 17
18 :- should be even number and the sum of digits div by 9..
19 Add 2 times the last digit to the remaining number. Repeat the step if required. If the result is divisible by 19, the original number is also divisible by 19
23 Add 7 times the last digit to the remaining number. Repeat the step if required. If the result is divisible by 23, the original number is also divisible by 23
29 Add 3 times the last digit to the remaining number. Repeat the step if required. If the result is divisible by 29, the original number is also divisible by 29

Although you will have to remember the numeric value to be added/subtracted in 13,17,19,23,29,31...etc..
One pattern can be kept in mind .. anything ending in 1 and 3 will have addition and 7 ,9 will have subtraction...

Num ending w 9 have addition, you have written subtraction in the last statement
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Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
Please can someone help with this -

The rule is that any n consecutive integers are divisible by n!- so how is (x-1)x(x+1) divisible by 24 (if x is odd) and not 6 (3! since they are 3 consecutive integers?)

thanks
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Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
tickledpink001 wrote:
Please can someone help with this -

The rule is that any n consecutive integers are divisible by n!- so how is (x-1)x(x+1) divisible by 24 (if x is odd) and not 6 (3! since they are 3 consecutive integers?)

thanks

If x is odd, then (x - 1) and (x + 1) are consecutive even integers, making one of them a multiple of 4. Thus, (x - 1)(x + 1) must be divisible by 2*4 = 8. Additionally, since (x - 1), x, and (x + 1) are three consecutive integers, one of them must be a multiple of 3. Therefore, (x - 1)x(x + 1) must be divisible by 8*3 = 24 for any odd value of x.

Hope it helps.
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Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
Bunuel wrote:
tickledpink001 wrote:
Please can someone help with this -

The rule is that any n consecutive integers are divisible by n!- so how is (x-1)x(x+1) divisible by 24 (if x is odd) and not 6 (3! since they are 3 consecutive integers?)

thanks

If x is odd, then (x - 1) and (x + 1) are consecutive even integers, making one of them a multiple of 4. Thus, (x - 1)(x + 1) must be divisible by 2*4 = 8. Additionally, since (x - 1), x, and (x + 1) are three consecutive integers, one of them must be a multiple of 3. Therefore, (x - 1)x(x + 1) must be divisible by 8*3 = 24 for any odd value of x.

Hope it helps.

Thanks Bunuel.

I'm still a bit unclear - I understand both rules and how (x - 1)x(x + 1) must be divisible by 8*3 = 24 for any odd value of x, but what I'm trying to understand is this-

Given we know that (x - 1)x(x + 1) are 3 consecutive integers, shouldn't the product of these terms should first and foremost be divisible by 3! only? and subsequently therefore any other multiple of 3! ?
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Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
tickledpink001 wrote:
Bunuel wrote:
tickledpink001 wrote:
Please can someone help with this -

The rule is that any n consecutive integers are divisible by n!- so how is (x-1)x(x+1) divisible by 24 (if x is odd) and not 6 (3! since they are 3 consecutive integers?)

thanks

If x is odd, then (x - 1) and (x + 1) are consecutive even integers, making one of them a multiple of 4. Thus, (x - 1)(x + 1) must be divisible by 2*4 = 8. Additionally, since (x - 1), x, and (x + 1) are three consecutive integers, one of them must be a multiple of 3. Therefore, (x - 1)x(x + 1) must be divisible by 8*3 = 24 for any odd value of x.

Hope it helps.

Thanks Bunuel.

I'm still a bit unclear - I understand both rules and how (x - 1)x(x + 1) must be divisible by 8*3 = 24 for any odd value of x, but what I'm trying to understand is this-

Given we know that (x - 1)x(x + 1) are 3 consecutive integers, shouldn't the product of these terms should first and foremost be divisible by 3! only? and subsequently therefore any other multiple of 3! ?

Yes, the product of three consecutive integers, like (x - 1)x(x + 1), is always divisible by 3! = 6. However, when the middle integer x is odd, the product becomes divisible by 24, a multiple of 6. Not sure what is the source of the confusion here.
Re: Divisibility/Multiples/Factors: Tips and hints [#permalink]
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