If H is the HCF of 4052 and 12376 and H=4052×A+12576×B then find the

1.	If H is the HCF of 4052 and 12376 and H=4052×A+12576×B then find the value of (H+A+B)
Answer» {tex}12576 = 4052 \\times 3 + 420{/tex}{tex}4052 = 420 \\times 9 + 272{/tex}{tex}420 = 272 \\times 1 + 148{/tex}{tex}272 = 148 \\times 1 + 124{/tex}{tex}148 = 124 \\times 1 + 24{/tex}{tex}124 = 24 \\times 5 + 4{/tex}{tex}24 = 4 \\times 6 + 0{/tex}HCF of 12576 and 4052 is \'4\'.

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