Question

what is the least natural number which when divided by 7 and 8 leaves a remainder of 2 in each case​

Answer 1

Answer:

504 is the least number exactly divisible by 6,7,8,9 and 12. Hence, least number which when divided by 6,7,8,9 and 12 leaves remainder 2 in each case=504+2=506.

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Answer 2

Answer:

The least natural number when divided by 7 and 8 leaves a remainder of 2 in each case is 58.

Step-by-step explanation:

To solve this problem, we should know about the formula of dividends.

• The dividend formula is provided by$\text{Dividend} = (\text{Divisor}\times \text{Quotient})+\text{Remainder}$
• The dividend is referred to as the number that is being divided.
• The divisor is referred to as the number that is dividing the dividend.
• The quotient is referred to as the resultant number of divisions.
• The remainder is referred to as the number that is left after dividing the dividend by the divisor.

Step 1: Calculate LCM of 7 and 8.

The LCM of 7 and 8 is

$\begin{aligned}y&=\text{LCM} (7,8)\\&=7\times2\times2\times 2\\&=56\end{aligned}$

LCM is 56. This gives the divisor.

Step 2: Find the least natural number.

Given the number when divided by 7 and 8, it gives a remainder of 2. Let's take the number or dividend as x. Then, x is written as,

$x=(56\times k)+2$

Here, k is a natural number (quotient) from 1, 2, 3...etc.

Therefore,

for $k = 1 \Rightarrow x = 58.$

This 58, when divided by both 7 and 8, gives the remainder 2. Therefore, 58 is the answer.

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