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When positive integer n is divided by 3, the remainder is 2; [#permalink]
 18 Nov 2007, 09:06
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When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n – 2 is divisible by 5
2) t is divisible 3
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Re: DS - remainder [#permalink]
18 Nov 2007, 12:33
gluon wrote: When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n – 2 is divisible by 5
2) t is divisible 3
i guess, an example/plugging numbers would be better:
1. n could be 2 or 17 or 32 etc. insufficient.
2. t could be 3, 18, 33, or so on. so also insuff.
togather we can conclude on the reminder of "nt" when it is divided by 15.
if nt is 2x3, r is 6.
if nt is 17x18, r is 6.
so C.
Note: when calculating reminder never do as under:
(17x18)/15 = (17x6)/5. so reminder is 2. its wrong because 17x18 is already divided by 3. to get the proper reminder of nt, 2 should be multiplied by 3. so the reminder is 6.
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Re: DS - remainder [#permalink]
24 Nov 2007, 16:19
Himalayan wrote: gluon wrote: When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n – 2 is divisible by 5
2) t is divisible 3 i guess, an example/plugging numbers would be better: 1. n could be 2 or 17 or 32 etc. insufficient. 2. t could be 3, 18, 33, or so on. so also insuff. togather we can conclude on the reminder of "nt" when it is divided by 15. if nt is 2x3, r is 6. if nt is 17x18, r is 6. so C. Note: when calculating reminder never do as under:
(17x18)/15 = (17x6)/5. so reminder is 2. its wrong because 17x18 is already divided by 3. to get the proper reminder of nt, 2 should be multiplied by 3. so the reminder is 6.17*9=153
153/15= 10 and r=3
7*9=63
63/15= 4 and r=3
Is that right?
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Re: DS - remainder [#permalink]
24 Nov 2007, 18:19
Himalayan wrote: gluon wrote: When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n – 2 is divisible by 5
2) t is divisible 3 i guess, an example/plugging numbers would be better: 1. n could be 2 or 17 or 32 etc. insufficient. 2. t could be 3, 18, 33, or so on. so also insuff. togather we can conclude on the reminder of "nt" when it is divided by 15. if nt is 2x3, r is 6. if nt is 17x18, r is 6. so C. Note: when calculating reminder never do as under:
(17x18)/15 = (17x6)/5. so reminder is 2. its wrong because 17x18 is already divided by 3. to get the proper reminder of nt, 2 should be multiplied by 3. so the reminder is 6.same approach...but was taking a lot of time..guess need to plug in numbers
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hold on a second ... in the stem, we are told that when n is divided by 3, the remainder is 2
statement 1 tells us that n-2 is divisible by 5, i.e. n-2 = 5, 10, 15, 20, 25...
therefore, n=7,12,17,22,27 ...
but if we take these values of n and divide by 3, i dont get a remainder of 2
can someone point out where i went wrong in my logic ?
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Re: DS - remainder [#permalink]
25 Nov 2007, 01:35
GMAT TIGER wrote: Deckard wrote: Himalayan wrote: gluon wrote: When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?
1) n – 2 is divisible by 5
2) t is divisible 3 i guess, an example/plugging numbers would be better: 1. n could be 2 or 17 or 32 etc. insufficient. 2. t could be 3, 18, 33, or so on. so also insuff. togather we can conclude on the reminder of "nt" when it is divided by 15. if nt is 2x3, r is 6. if nt is 17x18, r is 6. so C. Note: when calculating reminder never do as under:
(17x18)/15 = (17x6)/5. so reminder is 2. its wrong because 17x18 is already divided by 3. to get the proper reminder of nt, 2 should be multiplied by 3. so the reminder is 6.17*9=153 153/15= 10 and r=3 7*9=63 63/15= 4 and r=3 Is that right? t cannot be 9. Got it. I was too concentrated on the statements and I forgot to consider the text. Thanx.
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Re: DS - remainder
[#permalink]
25 Nov 2007, 01:35
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