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Hard LCM and HCF problems

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Hard LCM and HCF problems

by aagar2003 » Sun Jun 20, 2010 6:24 pm
1. Find the least five digit number which on divided by 12, 18, 21, and 28 leaves the same remainder.
A. 11019
B. 10059
C. 10119
D. 12089
E. 10289

2. Find the greatest number of four digits which is exactly divisible by each of 12, 18, 40 and 45.
A. 9550
B. 9720
C. 8500
D. 8650
E. 9972

3. The sum and the differences of the L.C.M. and the H.C.F. of two numbers are 312 and 264, respectively. Find the numbers if their sum is 168.
A. 60,108
B. 108,84
C. 92,84
D. 72,84
E. 96,72


4. Find the L.C.M. of 6/7, 12/13, 7/15.
A. 84
B. 1/84
C. 13
D. 1/13
E. None of Above

5. Find the greatest number that will divide 63, 45 and 69 so as to leave the same remainder.
A. 6
B. 9
C. 8
D. 10
E. 3

Answers:
1 C
2 B
3 E
4 A
5 A

Can somebody please explain in detail their working?
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by hardik.jadeja » Sun Jun 20, 2010 8:31 pm
[email protected] wrote:1. Find the least five digit number which on divided by 12, 18, 21, and 28 leaves the same remainder.
A. 11019
B. 10059
C. 10119
D. 12089
E. 10289
Are you sure the answer of the first question is C?

When we divide 10119 by 12 or 18, remainder is 3.
When we divide 10119 by 21, remainder is 18.
When we divide 10119 by 28, remainder is 11.

The answer should be 10081. When 10081 is divided by 12,18,21 or 28, remainder is 1.

This is how I came to this figure.

LCM of 12,18,21,28 is 252. When 252 is multiplied by 40, it forms smallest possible five digit number 10080 that is divisible by 252.

Since 10080 is divisible by all 12,18,21 and 28. 10081 must be the answer as it leave the remainder 1 when divided by 12,18,21 or 28.
Last edited by hardik.jadeja on Sun Jun 20, 2010 9:14 pm, edited 1 time in total.
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by hardik.jadeja » Sun Jun 20, 2010 8:40 pm
[email protected] wrote: 2. Find the greatest number of four digits which is exactly divisible by each of 12, 18, 40 and 45.
A. 9550
B. 9720
C. 8500
D. 8650
E. 9972
Solution for second question:

LCM for 12, 18, 40 and 45 is 360.

The least five digit number divisible by 360 is 10080 (360*28). So the greatest possible four digit number divisible by 360 has to be 360*27 = 9720.

Pick B.
Last edited by hardik.jadeja on Sun Jun 20, 2010 9:15 pm, edited 1 time in total.
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by hardik.jadeja » Sun Jun 20, 2010 8:50 pm
[email protected] wrote: 4. Find the L.C.M. of 6/7, 12/13, 7/15.
A. 84
B. 1/84
C. 13
D. 1/13
E. None of Above
L.C.M. of 6/7, 12/13, 7/15 = LCM of 6,12 and 7 / HCF of 7,13,15 = 84/1 = 84.

Pick A.
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by hardik.jadeja » Sun Jun 20, 2010 8:56 pm
[email protected] wrote: 5. Find the greatest number that will divide 63, 45 and 69 so as to leave the same remainder.
A. 6
B. 9
C. 8
D. 10
E. 3
Start with the highest number 10. When we divide 63, 45 or 69 with 10, the remainder is going to be different. We dont need to calculate anything here.

Next highest number is 9. 63 and 45 are divisible by 9 but 69 is not. Again no calculation required.

Next highest number is 8. When we divide 63, 45 or 69 with 8, the remainder is going to be different. You can tell this without calculating anything. But if you want, you can try with couple of numbers.

Next number is 6. Bingo. When we divide 63, 45 or 69 with 6, the remainder is 3.

Pick A.
Last edited by hardik.jadeja on Sun Jun 20, 2010 9:23 pm, edited 1 time in total.
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by hardik.jadeja » Sun Jun 20, 2010 9:12 pm
[email protected] wrote: 3. The sum and the differences of the L.C.M. and the H.C.F. of two numbers are 312 and 264, respectively. Find the numbers if their sum is 168.
A. 60,108
B. 108,84
C. 92,84
D. 72,84
E. 96,72
Lets say the LCM of the two numbers is X and HCF is Y.

So X + Y = 312
and X - Y = 264

If we add these two equations, we get X(LCM) = 288 and Y(HCF) = 24

LCM of two numbers * HCF of two numbers = Multiplication of two numbers.

So, XY = 6912 (Multiplication of two numbers)

Now we know that multiplication and sum of the numbers we need is 6912 and 168, respectively.

Lets check the options now

A. 60,108 (60 + 108 = 168)
B. 108,84 (108 + 84 does not equal to 168)
C. 92,84 (92 + 84 does not equal to 168)
D. 72,84 (72 + 84 does not equal to 168)
E. 96,72 (96 + 72 = 168)

Only A and E left. Now lets check the multiplication.

A. 60,108 -> 60*108 = 6480
E. 96,72 -> 96*72 = 6912 .... Bingo

Pick E.
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by aagar2003 » Sun Jun 20, 2010 11:20 pm
hardik.jadeja wrote:
[email protected] wrote:1. Find the least five digit number which on divided by 12, 18, 21, and 28 leaves the same remainder.
Are you sure the answer of the first question is C?
I agree that answer C seems incorrect. However, I hadve no clue how to answer the question.
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by gmatmachoman » Sun Jun 20, 2010 11:52 pm
hardik.jadeja wrote:
[email protected] wrote: 3. The sum and the differences of the L.C.M. and the H.C.F. of two numbers are 312 and 264, respectively. Find the numbers if their sum is 168.
A. 60,108
B. 108,84
C. 92,84
D. 72,84
E. 96,72
Lets say the LCM of the two numbers is X and HCF is Y.

So X + Y = 312
and X - Y = 264

If we add these two equations, we get X(LCM) = 288 and Y(HCF) = 24

LCM of two numbers * HCF of two numbers = Multiplication of two numbers.

So, XY = 6912 (Multiplication of two numbers)

Now we know that multiplication and sum of the numbers we need is 6912 and 168, respectively.

Lets check the options now

A. 60,108 (60 + 108 = 168)
B. 108,84 (108 + 84 does not equal to 168)
C. 92,84 (92 + 84 does not equal to 168)
D. 72,84 (72 + 84 does not equal to 168)
E. 96,72 (96 + 72 = 168)

Only A and E left. Now lets check the multiplication.

A. 60,108 -> 60*108 = 6480
E. 96,72 -> 96*72 = 6912 .... Bingo

Pick E.
One more way of doing it:

After finding the HCF as 24, look for the options where both the numbers as divisibe by 24.

Directly u can see only E has numbers that are both divisible by 24. pick E
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