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by N:Dure » Mon Jan 03, 2011 5:22 am
If n is a positive integer, which of the following has a remainder of 3 when divided by 4, 5 and 6?

a) 12n+3
b) 24n+3
c) 80n+3
d) 90n+2
e) 120n+3

My approach was to see which of the quotients of n is divisible by the 3 no.s and has a remainder of 3. The OA takes the LCM of 4,5,6 and does amazing things with it... is my approach correct?


37) The positive integers m and n leave remainders of 2 and 3 respectively when divided by 6. m > n. What's the remainder when m-n is divided by 6?

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38) The remainder when m+n is divided by 12 is 8, and when m-n is divided by 12 is 6. If m >n, then what's the remainder when mn is divided by 6?

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by Anurag@Gurome » Mon Jan 03, 2011 5:37 am
N:Dure wrote:If n is a positive integer, which of the following has a remainder of 3 when divided by 4, 5 and 6?

a) 12n+3
b) 24n+3
c) 80n+3
d) 90n+2
e) 120n+3
If an integer, x leaves a remainder of 3, when divided by 4, 5, and 6, then x must be 3 more than a integer, I which is divisible by 4, 5 and 6. Now I is nothing but the multiple of LCM of 4, 5, and 6. Therefore x must be 3 greater than the multiple of LCM of 4, 5, and 6. Hence x = (60m + 3), where m is a non-negative integer. Only option E is of this form.

The correct answer is E.

Note: What you have done, i.e. "to see which of the quotients of n is divisible by the 3 no.s and has a remainder of 3" is same I explained above. The quotient of n divisible by all the three numbers 4, 5 and 6 is nothing but the multiple of the three numbers.
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by Anurag@Gurome » Mon Jan 03, 2011 5:53 am
N:Dure wrote:The positive integers m and n leave remainders of 2 and 3 respectively when divided by 6. m > n. What's the remainder when m-n is divided by 6?

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Tricky Approach: Pick numbers m and n accordingly. For example m = 14 and n = 9. Thus, (m - n) = 5

Methodical Approach:

When divided by 6, m leaves a remainder of 2.
Thus m is of the form (6a + 2), where a is any non-negative integer.

When divided by 6, n leaves a remainder of 3.
Thus n is of the form (6b + 3), where b is any non-negative integer.

Therefore, (m - n) = (6a + 2) - (6b + 3) = 6(a - b) - 1 = 6(a - b - 1) + 5
Thus, (m - n) leaves a remainder of 5 when divided by 6.

The correct answer is E.
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by Anurag@Gurome » Mon Jan 03, 2011 6:08 am
N:Dure wrote:The remainder when m+n is divided by 12 is 8, and when m-n is divided by 12 is 6. If m >n, then what's the remainder when mn is divided by 6?

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Tricky Approach: Pick numbers m and n accordingly. For example (m + n) = 8 and (m - n) = 6. Thus, m = 7 and n = 1. Therefore, mn = 7 which leaves a remainder of 1 when divided by 6.

Methodical Approach:

(m + n) is of the form (12a + 8), where a is any non-negative integer.
(m - n) is of the form (12b + 6), where b is any non-negative integer.

Therefore, m = [(12a + 8) + (12b + 6)]/2 = (6a + 6b + 7)
Therefore, n = [(12a + 8) - (12b + 6)]/2 = (6a - 6b + 1)

Therefore, mn = (6a + 6b + 7)(6a - 6b + 1) = (36a² - 36ab + 6a +36ab - 36b² + 6b + 42a - 42b + 7) = (36a² - 36b² + 48a -36b + 6 + 1) = Multiple of 6 + 1

Hence mn leaves a remainder of 1 when divided by 6.

The correct answer is A.
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