The sum of two numbers is 84 and their HCF is 12 . How many such pairs are there?
Answers
Step-by-step explanation:
Given , HCF = 12
∴ Numbers = 12p and 12q , where p and q are prime to each other.
∴ 12p + 12q = 84
⇒ 12 (p + q) = 84
⇒ p + q =
84
= 7
12
∴ Possible pairs of numbers satisfying this condition = (1 , 6), (2 , 5) and (3 , 4).
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Answer:
both the numbers have the highest common factor 12 it clearly means that the both the numbers are the multiples if 12
So let the numbers be 12x and 12y
As given their sum is 84
ATQ,12x+12y=84
12 (x+y)=84
x+y=7
So the number of pairs with 7 as their sum are (1,6),(2,5),(3,4)
That is three pairs
So the numbers are (12,72),(24,60), (36,48)
Hope it helpss
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Ajay Narayanan
, Another engineer among the masses...
Answered 4 years ago
since both numbers are divisible by 12, let them be 12x12x and 12y12y, where x and y are co-prime. Assuming natural numbers, xx and yy lie between 11 and 8383. since
12(x+y)=8412(x+y)=84
x+y=84/12=7x+y=84/12=7
so the possible pairs are 12∗12∗(1,6),(2,5)(3,4)(1,6),(2,5)(3,4);
or (12,72),(24,60),(36,48)(12,72),(24,60),(36,48)
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