When positive integer n is divided by 3, the remainder is 2;

[s.ashwin.rao]

Great...Thanks Nelson.

[tim]

:)

[jp.jprasanna]

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 by 3.

Solution:
n = 3a + 2, t = 5b + 3
Statement 1:
n = 5c + 2 and n = 3a + 2 => n = 15d + 2
nt = (15d + 2)*(5b + 3) = 15d(5b + 3) + 10b + 6
Remainder = 10b + 6; it will have multiple values

Not Sufficient

Statement 2:
t = 3x and t = 5b + 3 => t = 15y + 3
nt = ( 3a + 2)*(15y + 3)
Same as above

Not sufficient

Statement 1 and 2
n = 5c + 2 and n = 3a + 2 => n = 15d + 2
t = 3x and t = 5b + 3 => t = 15y + 3
nt = (15d + 2) * (15y + 3) = 15d(15y + 3) + 2*15y + 2*3 = 15z + 6

Remainder = 6

Sufficient
Ans C

Dear Ron / Jamie - Could you please let me knw how to reach n = 15d + 2 & t = 15y + 3 from the given statements? Algebra?

Statement 1:
n = 5c + 2 and n = 3a + 2 => n = 15d + 2

Statement 2:
t = 3x and t = 5b + 3 => t = 15y + 3

[tim]

if n = 5c + 2 and n = 3a + 2, that means n is 2 more than a multiple of 5 and 2 more than a multiple of 3. therefore it must be 2 more than a multiple of their LCM, which is 15..

if t = 3x and t = 5b + 3, it is a multiple of 3 that is 3 more than a multiple of 5. well, a multiple of 3 is also 3 more than a multiple of 3, so we have the same situation as we have with statement 1: we are 3 more than a multiple of 3 and 3 more than a multiple of 5, therefore 3 more than a multiple of 15..

[jp.jprasanna]

if n = 5c + 2 and n = 3a + 2, that means n is 2 more than a multiple of 5 and 2 more than a multiple of 3. therefore it must be 2 more than a multiple of their LCM, which is 15..

if t = 3x and t = 5b + 3, it is a multiple of 3 that is 3 more than a multiple of 5. well, a multiple of 3 is also 3 more than a multiple of 3, so we have the same situation as we have with statement 1: we are 3 more than a multiple of 3 and 3 more than a multiple of 5, therefore 3 more than a multiple of 15..

Hi tim - sorry for asking this again....

if n = 2A +1 and n = 3B +2

then nos overlapping would be 5,8,11,......but im still not sure how to express n that satisfy both the condition in algebraic form with out listing out the nos.... i.e after listing then out i was able to figure out n = 4C+1 ... but is there a quicker way? (without listing the nos out)

[kg.tests]

Hello,

Here 's how i approached this problem. But I do not know if I'am right. Could someone please tell me if my reasoning is right.

From the question stem it is given that

when n is divided by 3 the remainder is 2

so possible values of n are

n = 2, 5, 8, 11, 14, 17, 20, 23,..... ----- (1)

when t is divided by 5 the remainder is 3

so t can have values

t = 3, 8, 13, 18, 23, 28, 33, ...... -------(2)



from stmt 1

n -2 is divisible by 5

so possible values of n

n = 3, 8, 13, 18, 23... (3)

from 1 and 3 the common values are

n = 8, 23, ...... (4)

when we multiply the values of n in (4) with the values of t in (2)

we get different remainders when divided by 15.

so clearly it is NS.

stmt 2

t is divisible by 3

so t can have values

t = 3, 6, 9, 12, 15, 18, 21, ...... (5)

taking common values of t from (2) and (5)

t = 3, 18, ....... (6)

multiplying values of t in (6) with values of n in (1 ) and dividing by 15 we get different values for the remainder

So NS.

Combining (1) and (2)

n = 8, 23, ..... (4)
t = 3, 18 .... (6)

dividing nt/15 we get a common remainder ..

sufficient...

Ans C

[tim]

Hi tim - sorry for asking this again....

if n = 2A +1 and n = 3B +2

then nos overlapping would be 5,8,11,......but im still not sure how to express n that satisfy both the condition in algebraic form with out listing out the nos.... i.e after listing then out i was able to figure out n = 4C+1 ... but is there a quicker way? (without listing the nos out)

not sure what you mean by "asking this again", because i can't tell that you've asked about this case before. at any rate, you've got numbers that are not consistent with your equations. care to try again? the bottom line is that if you aren't comfortable with the theory, listing numbers is probably your safest bet..

[tim]

kg.tests, your approach looks valid..

[kg.tests]

thanks Tim. :)

[tim]

:)

[qqixiaofan]

This solution is great

I used the following approach for this one.

Statement 1: (n-2) is divisible by 5. But (n-2) also must be divisible by 3 since n has a remainder of 2 when n is divided by 3. Thus, (n-2) is divisible by 15 => when n is divided by 15, the remainder is 2.

Statement 2: t is divisible by 3 => (t-3) is aslo divisible by 3. But (t-3) is divisible by 5 since t has a remainder of 3 when t is divided by 5. So (t-3) is divisible by 15 and t has a remainder of 3 when t is divided by 15.

Combining Statement 1 and Statement 2: when nt is divided by 15, the remainder is 2X3=6.

[RonPurewal]

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