Question

find the smallest number which when divided by 66, 99 as well as 88 leaves remainder 6​

Answer 1

Answer:

798 is the answer of this question.

Step-by-step explanation:

Let us find LCM of the numbers.

66=2*3*11.

88=2*2*2*11.

99=3*3*11.

LCM=2*2*2*3*3*11=792.

As it has to leave a remainder 6 when divided by these numbers the required number is 792+6=798.

798 is the required number.

Answer 2

Answer:

798

Step-by-step explanation:

First, factor the numbers into primes:

First, factor the numbers into primes:66 = 2 x 3 x 11; 99 = 3 x 3 x 11; 88 = 2 x 2 x 2 x 11

First, factor the numbers into primes:66 = 2 x 3 x 11; 99 = 3 x 3 x 11; 88 = 2 x 2 x 2 x 11Common factors are 2 x 2; 3 x 3; and 11; so 2 x 2 x 2 x 3 x 3 x 11 or 792 is the smallest number divisible by 66; 99; and 88.

First, factor the numbers into primes:66 = 2 x 3 x 11; 99 = 3 x 3 x 11; 88 = 2 x 2 x 2 x 11Common factors are 2 x 2; 3 x 3; and 11; so 2 x 2 x 2 x 3 x 3 x 11 or 792 is the smallest number divisible by 66; 99; and 88.Add 6 to 792, gives 798, divisible by 66, 99, and 88 with remainders of 6.

Ending:-

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