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 Help on 3 math question on HCF , LCM
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By: Kuro (offline)  Saturday, March 01 2014 @ 07:55 PM CST (Read 989 times)  
Kuro

1)If 504k is a perfect square, find the least possible integer value of k

2)Find a number which when divided by 3 gives a remainder of 1;when divided by 4 gives a remainder of 2;when divided by 5 gives a remainder of 3;and when divided by 6 gives a remainder of 4?

3)The whole number 239 has the following property: When divided by 2,3,4,5 and 6 it leaves a remainder of 1,2,3,4 and 5 respectively. How many 3-digit whole numbers ,up to 300 will have this property, excluding 239?

(There's no answer key)

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By: jo sarah (offline)  Sunday, March 02 2014 @ 07:30 PM CST  
jo sarah

Hi,

1)If 504k is a perfect square, find the least possible integer value of k

Solution:
504 = 3^2 x 7 x 2^3
504k = 3^2 x 7 x 2^3 x k...............[Note: perfect squares must have even powers]
so, smallest 504k = 3^2 x 7^2 x 2^4 = (3^2 x 7 x 2^3) x (7 x 2)
which means, k = 7 x 2 = 14 Big Grin


2)Find a number which when divided by 3 gives a remainder of 1;when divided by 4 gives a remainder of 2;
when divided by 5 gives a remainder of 3;and when divided by 6 gives a remainder of 4?

Solution:
Let the no. be n, and a, b, c, d be integers such that
n = 3a + 1 = 4b + 2 = 5c + 3 = 6d + 4
Then, 3a + 1 = 4b + 2; i.e., a = (4b + 1)/3
and, 3a + 1 = 5c + 3; i.e., a = (5c + 2)/3
and, 3a + 1 = 6d + 4; i.e., a = (6d + 3)/3 = 2d + 1......[1]

From above, 4b + 1 = 5c + 2 = 6d + 3
when, 4b + 1 = 5c + 2; i.e., b = (5c + 1)/4..............[2]
and, 4b + 1 = 6d + 3; i.e., b = (6d + 2)/4 = (3d + 1)/2

again, from above, 5c + 1 = 6d + 2
leading to, c = (6d + 1)/5...............................[3]
i.e., c is a multiple of 5, which when 1 is subtracted from it will be a multiple of 6.
thus, 6d must be a multiple of 6 ending with 4 in ones place (can't end with 9 because multiples of 6 are even)

So, now with that fact,
if 6d = 24, d = 4; from [3], c = 5; a = 9, from [1]; b = 13/2! (no! because not integer)
next, if 6d = 54, d = 9; a = 19; c = 11; b = 14; giving n = 58 (for all cases of a, b, c, d)
A number satisfying the conditions is 58. Laughing Out Loud

[if there be more nos., try out 14 x 6, 19 x 6, etc., for 6d...have fun and enjoy!]


3)The whole number 239 has the following property: When divided by 2,3,4,5 and 6 it leaves a remainder of
1,2,3,4 and 5 respectively. How many 3-digit whole numbers ,up to 300 will have this property, excluding 239?

[pardon me, I am out of time: try analysing this in similar manner as no. 2. thanks.]

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