If the HCF of the two numbers be 13 and lcm 273 then the sum of the numbers will be?
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Let the two numbers be a and b.
Since the H.C.F. is 13, we can write
a=13xa=13x
b=13yb=13y
for any two positive integers x and y.
Given L.C.M. = 273. We can write
13×x×y=27313×x×y=273
⟹xy=273/13⟹xy=273/13
xy=21xy=21
Given a>13a>13and b>13b>13,
13x>13⟹x>113x>13⟹x>1
13y>13⟹y>113y>13⟹y>1
Now, xy=21xy=21
We need to find x and y.
So, factors of 21 are 1,3,7 and 21.
(x,y) can't be (1,21) or (21,1) since x and y should be greater than 1.
Therefore, (x,y) = (3,7) or (7,3)
Thus,
a = 13×3 = 39
b = 13×7 = 91
Thus, (a,b) = (39,91) or (91,39)
Sum of a and b = (a+b) = 39+91 = 130
Since the H.C.F. is 13, we can write
a=13xa=13x
b=13yb=13y
for any two positive integers x and y.
Given L.C.M. = 273. We can write
13×x×y=27313×x×y=273
⟹xy=273/13⟹xy=273/13
xy=21xy=21
Given a>13a>13and b>13b>13,
13x>13⟹x>113x>13⟹x>1
13y>13⟹y>113y>13⟹y>1
Now, xy=21xy=21
We need to find x and y.
So, factors of 21 are 1,3,7 and 21.
(x,y) can't be (1,21) or (21,1) since x and y should be greater than 1.
Therefore, (x,y) = (3,7) or (7,3)
Thus,
a = 13×3 = 39
b = 13×7 = 91
Thus, (a,b) = (39,91) or (91,39)
Sum of a and b = (a+b) = 39+91 = 130
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