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Re: How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
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Hi apeksha104

We have to consider a & b prime as the HCF is given 12. Now if we take a & b as 16 & 2 as you have indicated, the HCF will no longer be 12. (It will be 24 as 2 is another common factor among 2 & 16)

HCF is the highest common factor. So for the numbers having 12 as their HCF, it is necessary to have prime nos.

Ask if you have further doubts.

apeksha104 wrote:
Why is it that there should be just prime numbers for a and b? Why not instances like 14+4 = 18 or 16+2 = 18
I didn't get the line, "Since the highest common factor is 12, a and b should be prime"
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How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
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KarunMendiratta wrote:
Carcass wrote:
How many pairs of natural numbers whose HCF is 12 add up to 216?

6
3
9
18
17


Since HCF is \(12\)
Let the numbers be \(12a\) and \(12b\)

\(12a + 12b = 216\)
\(12 (a + b) = 216\)
\((a + b) = \frac{216}{12} = 18\)

Remember the HCF is \(12\) so \(a\) and \(b\) could only be prime numbers.

We can have three such pairs:

\(1 + 17 = 18\)
\(5 + 13 = 18\)
\(7 + 11 = 18\)

Hence, option B

How is 1 a prime number?

Thanks, for pointing out
Not primes but co-primes, made the changes
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How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
3
1
KarunMendiratta wrote:
Carcass wrote:
How many pairs of natural numbers whose HCF is 12 add up to 216?

6
3
9
18
17


Since HCF is \(12\)
Let the numbers be \(12a\) and \(12b\)

\(12a + 12b = 216\)
\(12 (a + b) = 216\)
\((a + b) = \frac{216}{12} = 18\)

Remember the HCF is \(12\) so \(a\) and \(b\) could only be prime numbers.

We can have three such pairs:

\(1 + 17 = 18\)
\(5 + 13 = 18\)
\(7 + 11 = 18\)

Hence, option B

I guess its not abut prime number. From the available combinations we have to choose the ones which won't affect the HCF.

Since a + B = 18, so available combinations are
1+17 (No common factor so acceptable)
2+16 (Common Factor of 2)
3+15 (Common Factor of 3)
4+14 (Common Factor of 2)
5+13 (No common factor so acceptable)
6+12 (Common Factor of 6)
7+11 (No common factor so acceptable)
8+10 (Common Factor of 2)

All the aforementioned common factors will change the HCF in \(12a + 12b = 216\)

Therefore Option B.
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Re: How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
I do not get the co-prime importance here. Could someone explain how we choose 3 pairs?
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Re: How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
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scuva wrote:
I do not get the co-prime importance here. Could someone explain how we choose 3 pairs?



Hey, I think that I understand why. It took me some time to understand.
Look on this pair:
24 | 192
24 + 192 = 216

But if you check, the question asks about HCF (HIGHEST common).
It you look on 24 and 192,
24:24=1 and 192:24=8 so the hcf here is not 12 but 24.

Now check:
12 | 204 (HCF = 12) - Correct
24 | 192 (HCF = 24)
36 | 180 (HCF = 36)
48 | 168 (HCF = 24)
60 | 156 (HCF = 12) - Correct
72 | 144 (HCF = 24)
84 | 132 (HCF = 12) - Correct
96 | 120 (HCF = 24) - Correct
108 | 108 (HCF = 108)

Divide each pair by 12, and you will see why it should be co-prime.
60 | 156 ===> 5 | 13
But
24 | 192 ===> 2 | 16 ===> Here you also have 2 so the HCF is 12x2=24.
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How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
Natural numbers = positive integers
12*prime 1 = one value
12* prime 2 = second value
There could also be non-primes as the multiplier as long as those two multiplier values did not also share any common factors.
Add them together to get 216

12*1=12
12*2=24
12*3=36
12*4=48 (not prime)
12*5=60
12*6=72 (not prime)
12*7=84
12*8=96 (not prime)
12*9=108 (not prime)
12*10=120 (not prime)
12*11=132
12*12=144 (not prime)
12*13=156
12*14=168 (not prime)
12*15=180 (not prime)
12*16=192 (not prime)
12*17=204
cannot go higher, because 12*19=228, which is greater than 216

Then I looked at the above for any values that add up to 216 and which did not also share any common factors in what I used to multiply them.
Pair 1: 12 multiplied by 1&17
Pair 2: 12 multiplied by 5&13
Pair 3: 12 multiplied by 7&11

You can see a pattern here in that the multipliers that match together add up to 18. This makes sense, since we're looking at 12 * (multiplier 1 + multiplier 2) = 216, so the two multipliers with equal 18.

(Other pairs shared common factors. For example, 2&16 multiplied by 12 look like they might work, but we see the multipliers share a common factor of 2, so we know they could not be the solution, since then the values of 2*12 and 16*12 would share a common factor of 2*12=24. Using this logic, we can quickly see all the evens are basically out, since their match is also always even. And then we can see 15&3 likewise share a common factor of 3. And finally, 9 with itself, 9*12 + 9*12, well 9 shares a common factor of 9 with itself. It turns out only the prime number options work out!)

Answer choice = C (3)
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Re: How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
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Re: How many pairs of natural numbers whose HCF is 12 add up to 216? 6 3 [#permalink]
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