1.If the sum of two numbers is is 1215 and their hcf is 81. Find the number of such pair.
2. Find the hcf of 65 and 117 and express it in the form of 65m+117
1>
It is given that the sum of two numbers is 1215 and their H.C.F is 81
Since the H.C.F of two numbers is 81, the two numbers are 81x and 81y.
Now, 81x + 81y = 1215 [Sum of two numbers is 1215]
81 (x + y) = 1215
(x + y) = 15
So,
For x = 7, y = 8; in this case the numbers are 7 × 81 = 567 and 8 × 81 = 648
x = 1, y = 14; in this case the numbers are 1 × 81 = 81 and 14 × 81 = 1134
x = 2, y = 13; in this case the numbers are 2 × 81 = 162 and 13 × 81 = 1053
x = 4, y = 11; in this case the numbers are 4 × 81 = 324 and 11 × 81 = 891
So the number of such pairs are 4
2>
Here is the answer to your question.
Among 65 and 117; 117 > 65
Since 117 > 65, we apply the division lemma to 117 and 65 to obtain
117 = 65 × 1 + 52 … Step 1
Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain
65 = 52 × 1 + 13 … Step 2
Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain
52 = 4 × 13 + 0 … Step 3
In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers
The H.C.F. of 65 and 117 is 13
From Step 2:
13 = 65 – 52 × 1 … Step 4
From Step 1:
52 = 117 – 65 × 1
Thus, from Step 4, it is obtained
13 = 65 – (117 – 65 × 1) × 1
⇒13 = 65 × 2 – 117
⇒13 = 65 × 2 + 117 × (–1)
In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1