Question

1)The sum of two numbers is 216 and their HCF is 27. Find the numbers.
2)Find the largest number which divides 62, 132 and 237 to leave the same remainder in
each case.



solve this both questions​

Answer 1

Answer:

$\huge\underline\mathfrak\blue{Solution1}$

$\impies$ Let the require numbers be 27 a and 27 b

$\impies$27 a + 27 b = 216

$\impies$ a + b = 8

$\impies$ Now co-primes with sum 8 are (1 , 7 ) and ( 3 , 5 )

$\impies$ The require numbers (27 x 1, 27 x 7) and ( 27 x 3, 27 x 5)

$\impies$ (27, 189 ) and ( 81 , 135 )

$\huge\underline\mathfrak\blue{Solution2}$

$\impies$ As the given numbers are 62, 132 and 237.

$\impies$ As we have to find largest number which divides 62,132 and 237 leaving same remainder in each case.

$\impies$ let us assume that remainder left is m

$\impies$ So, numbers which are completely divisible are

$\impies$ 62-m , 132-m, 237-m

$\impies$ Now subtract the pairs

$\impies$ 132-m -62+m = 70

$\impies$ 237-m-132+m = 105

$\impies$ Now find the HCF of 70 and 105

$\impies$ 105 = 70 \times 1 + 35 \\ 70 = 35 \times 2 + 0

$\impies$ So, HCF is 35

$\impies$ So, the largest number which divides 62,132 and 237 and leaves same remainder is 35

$\impies$ Proof:

$\impies$ 62 = 35 \times 1 + 27 \\ \\ 132 = 35 \times 3 + 27 \\ \\ 237 = 35 \times 6 + 27

Step-by-step explanation:

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