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# whats the best way to approach divisibility questions?

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whats the best way to approach divisibility questions? [#permalink]

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25 Jan 2006, 15:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I saw this question in an earlier post and always have problems with these....
Thanks!
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.
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26 Jan 2006, 00:30
At frist split up the given stem

xy*xy*y*y

1) says that x is divisible by 7 - but what about y ? It could be 0, or 1, each yielding a different result. We still don't know so statement 1 is insufficient.

2) xy is divisible by 49, so do we know anything about the respective values? No, xy can be 0 or 49 ... it's not suffcient

1+2) Aren't sufficient as well, since the case that y is 0 can't be excluded

This is one of the easy divisibility questions, will try to find a difficult one to explain the method
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26 Jan 2006, 05:01
Don't know if the following question goes into the direction you want to, but give it try:

A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11 ?

(1) 3
(2) 21
(3) 59
(4) 68

Formal answer as provided in the document available in this forum:

When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation

d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7
Therefore, d1r2 + r1 = 8*7 + 3 = 59
VP
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26 Jan 2006, 11:52

Can you pls point me to the document that you mentioned.

Don't know if the following question goes into the direction you want to, but give it try:

A certain number when successfully divided by 8 and 11 leaves remainders of 3 and 7 respectively. What will be remainder when the number is divided by the product of 8 and 11 ?

(1) 3
(2) 21
(3) 59
(4) 68

Formal answer as provided in the document available in this forum:

When a number is successfully divided by two divisors d1 and d2 and two remainders r1 and r2 are obtained, the remainder that will be obtained by the product of d1 and d2 is given by the relation

d1r2 + r1.

Where d1 and d2 are in ascending order respectively and r1 and r2 are their respective remainders when they divide the number.

In this case, the d1 = 8 and d2 = 11. And r1 = 3 and r2 = 7
Therefore, d1r2 + r1 = 8*7 + 3 = 59
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26 Jan 2006, 14:10
This is something I havent seen before, thank you. Are there other formulas for figuring out if something is divisible by a certain number. I think my problem is that i don't know the best way to approach the questions.

At frist split up the given stem

xy*xy*y*y

1) says that x is divisible by 7 - but what about y ? It could be 0, or 1, each yielding a different result. We still don't know so statement 1 is insufficient.

2) xy is divisible by 49, so do we know anything about the respective values? No, xy can be 0 or 49 ... it's not suffcient

1+2) Aren't sufficient as well, since the case that y is 0 can't be excluded

This is one of the easy divisibility questions, will try to find a difficult one to explain the method
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heres another example of a problem i have trouble with.... [#permalink]

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26 Jan 2006, 21:10
just not sure the best way to approach these....any help is appreciated as I have a test coming up soon..
Thanks!
IF X is the integer, is X/3 an integer?

1).72/X is the integer

2).81/X is the integer
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26 Jan 2006, 21:54
1) 72/x = 3*3*8/x. -> not sufficient. x could be 8, which is not a multiple of 3.

2) 81/x = 3^4/x -> x must be a multiple of 3.

Ans B
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Re: whats the best way to approach divisibility questions? [#permalink]

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26 Jan 2006, 21:55
ayl989 wrote:
I saw this question in an earlier post and always have problems with these....
Thanks!
Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.

Is x^2*y^4 divisible by 49?

(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.

1) If x is an integer divisible by 7, it must be a multiple of 7. x can take values 7, 14, 21, 28... etc. Notice that x will be of the form 7*1, 7*2, 7*3.....7*n

So x^2 = 49n, x^2*y^4 = 49n(y^4). This will always be divisible by 49 since the numerator and denominator cancel out leaving you with just y^4. Y can be a fraction, an integer, a whole number, but it doesn't matter.

2) x*y = 49 -> This will result in (x,y) = (1,49), (49,1), (7,7) and no other combinations. All 3 combinations will be able to cancel out the 49.

Ans is therefore D
VP
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26 Jan 2006, 22:30
ywilfred, what about x = 1 ? I think answer should be E.

ywilfred wrote:
1) 72/x = 3*3*8/x. -> not sufficient. x could be 8, which is not a multiple of 3.

2) 81/x = 3^4/x -> x must be a multiple of 3.

Ans B
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26 Jan 2006, 22:42
lhotseface wrote:
ywilfred, what about x = 1 ? I think answer should be E.

ywilfred wrote:
1) 72/x = 3*3*8/x. -> not sufficient. x could be 8, which is not a multiple of 3.

2) 81/x = 3^4/x -> x must be a multiple of 3.

Ans B

Ah yes.. i forgot all about '1'. You're right about E then.
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27 Jan 2006, 02:43
How is 1 divisible by 7?
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27 Jan 2006, 02:58
How is 1 divisible by 7?

he meant y
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27 Jan 2006, 15:40
thanks for the answers to the questions. but I was looking more for a method on how to approach these types of problems. what do you look to try for first? what order/method is used?

I just seem to get stuck and was wondering if anyone could provide pointers and not just find the answer.
Thanks again.
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29 Jan 2006, 18:49
I'm not sure what you meant. But what I did previously was work a lot on these type of questions and then fine-tune my own method. There is no hard and fast way to solve such problems (probably the same thing with RC) but you just have to expose yourself to more of such questions and soon you'll feel much more comfortable facing them.
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29 Jan 2006, 18:50
ayl989 wrote:
anyone?

If you need help with picking numbers, try reading the sticky about picking numbers. The sticky is in this math forum.
29 Jan 2006, 18:50
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