Question

What is the least number which when divided by 7,3
8 and 11 always gives 6 as remainder?
a.624
b. 622
c. 616
d. 654​

Answer 1

Answer:

no option is correct.please mark it as the brainliest

Answer 2

Answer:

The answer to the given question is none of the above.

Step-by-step explanation:

The options are given, among those which option is correct, the number that when divided by 7,3

8 and 11 always give 6 as the remainder?

let's divide every number given.

let's divide option a by all the given factors and the remainder will be

divided option a by 7,3,8 and 11 and check whether 6 is the remainder.

$\frac{624}{7} = remainder \: 1\\ \frac{624}{3} = 208\\ \frac{624}{8} =78 \\ \frac{624}{11} =8$

divide the second option. by 7,3,8 and 11 and check whether. 6 is the remainder?

$\frac{622}{7} = 6 \\ \frac{622}{3} = 4 \\ \frac{622}{8} = 6 \\ \frac{622}{11} = 6$

let's divide the third option by 7,3,8 and 11 and check whether 6 is the remainder?

$\frac{616}{7} = 0 \\ \frac{616}{8} = 0 \\ \frac{616}{3} = 1 \\ \frac{616}{11} = 1$

let's divide the fourth option. by 7,3,8 and 11 and check whether 6 is the remainder?

$\frac{654}{7} = 3 \\ \frac{654}{8} = 6 \\ \frac{654}{3} = 0 \\ \frac{654}{11} = 5$

Based on the above calculation it seems that no option satisfies the given condition.

Therefore, the answer to the given question is none of the above

# spj3

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