Question

find the largest number which when divides by 98 120 and 153 leaves 2 remainder​

Answer 1

Answer:

Answer:35

Solution:

As the given numbers are 62, 132 and 237.

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

let us assume that remainder left is m

So, numbers which are completely divisible are

62-m , 132-m, 237-m

Now subtract the pairs

Step-by-step explanation:

Now find the HCF of 70 and 105

\begin{lgathered}105 = 70 \times 1 + 35 \\ 70 = 35 \times 2 + 0\end{lgathered}

105=70×1+35

70=35×2+0

So, HCF is 35

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

Proof:

\begin{lgathered}62 = 35 \times 1 + 27 \\ \\ 132 = 35 \times 3 + 27 \\ \\ 237 = 35 \times 6 + 27\end{lgathered}

62=35×1+27

132=35×3+27

237=35×6+27              

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